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Table of Contents

Section 11.1 - More About the Magnetic Field
Section 11.2 - Magnetic Fields by Superposition
Section 11.3 - Magnetic Fields by Ampère's Law
Section 11.4 - Ampère's Law In Differential Form (Optional)
Section 11.5 - Induced Electric Fields
Section 11.6 - Maxwell's Equations
Section 11.7 - Electromagnetic Properties of Materials

Chapter 11. Electromagnetism

Think not that I am come to destroy the law, or the prophets: I am not come to destroy, but to fulfill. -- Matthew 5:17

11.1 More About the Magnetic Field


a / The pair of charged particles, as seen in two different frames of reference.


b / A large current is created by shorting across the leads of the battery. The moving charges in the wire attract the moving charges in the electron beam, causing the electrons to curve.


c / A charged particle and a current, seen in two different frames of reference. The second frame is moving at velocity \(v\) with respect to the first frame, so all the velocities have \(v\) subtracted from them. (As discussed in the main text, this is only approximately correct.)

11.1.1 Magnetic forces

In this chapter, I assume you know a few basic ideas about Einstein's theory of relativity, as described in sections 7.1 and 7.2. Unless your typical workday involves rocket ships or particle accelerators, all this relativity stuff might sound like a description of some bizarre futuristic world that is completely hypothetical. There is, however, a relativistic effect that occurs in everyday life, and it is obvious and dramatic: magnetism. Magnetism, as we discussed previously, is an interaction between a moving charge and another moving charge, as opposed to electric forces, which act between any pair of charges, regardless of their motion. Relativistic effects are weak for speeds that are small compared to the speed of light, and the average speed at which electrons drift through a wire is quite low (centimeters per second, typically), so how can relativity be behind an impressive effect like a car being lifted by an electromagnet hanging from a crane? The key is that matter is almost perfectly electrically neutral, and electric forces therefore cancel out almost perfectly. Magnetic forces really aren't very strong, but electric forces are even weaker.

What about the word “relativity” in the name of the theory? It would seem problematic if moving charges interact differently than stationary charges, since motion is a matter of opinion, depending on your frame of reference. Magnetism, however, comes not to destroy relativity but to fulfill it. Magnetic interactions must exist according to the theory of relativity. To understand how this can be, consider how time and space behave in relativity. Observers in different frames of reference disagree about the lengths of measuring sticks and the speeds of clocks, but the laws of physics are valid and self-consistent in either frame of reference. Similarly, observers in different frames of reference disagree about what electric and magnetic fields and forces there are, but they agree about concrete physical events. For instance, figure a/1 shows two particles, with opposite charges, which are not moving at a particular moment in time. An observer in this frame of reference says there are electric fields around the particles, and predicts that as time goes on, the particles will begin to accelerate towards one another, eventually colliding. A different observer, a/2, says the particles are moving. This observer also predicts that the particles will collide, but explains their motion in terms of both an electric field, \(\mathbf{E}\), and a magnetic field, \(\mathbf{B}\). As we'll see shortly, the magnetic field is required in order to maintain consistency between the predictions made in the two frames of reference.

To see how this really works out, we need to find a nice simple example that is easy to calculate. An example like figure a is not easy to handle, because in the second frame of reference, the moving charges create fields that change over time at any given location. Examples like figure b are easier, because there is a steady flow of charges, and all the fields stay the same over time.1 What is remarkable about this demonstration is that there can be no electric fields acting on the electron beam at all, since the total charge density throughout the wire is zero. Unlike figure a/2, figure b is purely magnetic.

To see why this must occur based on relativity, we make the mathematically idealized model shown in figure c. The charge by itself is like one of the electrons in the vacuum tube beam of figure b, and a pair of moving, infinitely long line charges has been substituted for the wire. The electrons in a real wire are in rapid thermal motion, and the current is created only by a slow drift superimposed on this chaos. A second deviation from reality is that in the real experiment, the protons are at rest with respect to the tabletop, and it is the electrons that are in motion, but in c/1 we have the positive charges moving in one direction and the negative ones moving the other way. If we wanted to, we could construct a third frame of reference in which the positive charges were at rest, which would be more like the frame of reference fixed to the tabletop in the real demonstration. However, as we'll see shortly, frames c/1 and c/2 are designed so that they are particularly easy to analyze. It's important to note that even though the two line charges are moving in opposite directions, their currents don't cancel. A negative charge moving to the left makes a current that goes to the right, so in frame c/1, the total current is twice that contributed by either line charge.

Frame 1 is easy to analyze because the charge densities of the two line charges cancel out, and the electric field experienced by the lone charge is therefore zero:

\[\begin{equation*} \mathbf{E}_1 = 0 \end{equation*}\]

In frame 1, any force experienced by the lone charge must therefore be attributed solely to magnetism.

Frame 2 shows what we'd see if we were observing all this from a frame of reference moving along with the lone charge. Why don't the charge densities also cancel in this frame? Here's where the relativity comes in. Relativity tells us that moving objects appear contracted to an observer who is not moving along with them. Both line charges are in motion in both frames of reference, but in frame 1, the line charges were moving at equal speeds, so their contractions were equal, and their charge densities canceled out. In frame 2, however, their speeds are unequal. The positive charges are moving more slowly than in frame 1, so in frame 2 they are less contracted. The negative charges are moving more quickly, so their contraction is greater now. Since the charge densities don't cancel, there is an electric field in frame 2, which points into the wire, attracting the lone charge. Furthermore, the attraction felt by the lone charge must be purely electrical, since the lone charge is at rest in this frame of reference, and magnetic effects occur only between moving charges and other moving charges.2

To summarize, frame 1 displays a purely magnetic attraction, while in frame 2 it is purely electrical.

Now we can calculate the force in frame 2, and equating it to the force in frame 1, we can find out how much magnetic force occurs. To keep the math simple, and to keep from assuming too much about your knowledge of relativity, we're going to carry out this whole calculation in the approximation where all the speeds are fairly small compared to the speed of light.3 For instance, if we find an expression such as \((v/c)^2+(v/c)^4\), we will assume that the fourth-order term is negligible by comparison. This is known as a calculation “to leading order in \(v/c\).” In fact, I've already used the leading-order approximation twice without saying so! The first time I used it implicitly was in figure c, where I assumed that the velocities of the two line charges were \(u-v\) and \(-u-v\). Relativistic velocities don't just combine by simple addition and subtraction like this, but this is an effect we can ignore in the present approximation. The second sleight of hand occurred when I stated that we could equate the forces in the two frames of reference. Force, like time and distance, is distorted relativistically when we change from one frame of reference to another. Again, however, this is an effect that we can ignore to the desired level of approximation.

Let \(\pm\lambda\) be the charge per unit length of each line charge without relativistic contraction, i.e., in the frame moving with that line charge. Using the approximation \(\gamma=(1-v^2/c^2)^{-1/2}\approx 1+v^2/2c^2\) for \(v\ll c\), the total charge per unit length in frame 2 is

\[\begin{align*} \lambda_{total,\ 2} &\approx \lambda\left[1+\frac{(u-v)^2}{2c^2}\right] -\lambda\left[1+\frac{(-u-v)^2}{2c^2}\right]\\ &= \frac{-2\lambda uv}{c^2} . \end{align*}\]

Let \(R\) be the distance from the line charge to the lone charge. Applying Gauss' law to a cylinder of radius \(R\) centered on the line charge, we find that the magnitude of the electric field experienced by the lone charge in frame 2 is

\[\begin{equation*} E = \frac{4k\lambda uv}{c^2R} , \end{equation*}\]

and the force acting on the lone charge \(q\) is

\[\begin{equation*} F = \frac{4k\lambda quv}{c^2R} . \end{equation*}\]

In frame 1, the current is \(I=2\lambda_1 u\) (see homework problem 5), which we can approximate as \(I=2\lambda u\), since the current, unlike \(\lambda_{total,\ 2}\), doesn't vanish completely without the relativistic effect. The magnetic force on the lone charge \(q\) due to the current \(I\) is

\[\begin{equation*} F = \frac{2kI qv}{c^2R} . \end{equation*}\]

Discussion Question

Resolve the following paradox concerning the argument given in this section. We would expect that at any given time, electrons in a solid would be associated with protons in a definite way. For simplicity, let's imagine that the solid is made out of hydrogen (which actually does become a metal under conditions of very high pressure). A hydrogen atom consists of a single proton and a single electron. Even if the electrons are moving and forming an electric current, we would imagine that this would be like a game of musical chairs, with the protons as chairs and the electrons as people. Each electron has a proton that is its “friend,” at least for the moment. This is the situation shown in figure c/1. How, then, can an observer in a different frame see the electrons and protons as not being paired up, as in c/2?


d / The right-hand relationship between the velocity of a positively charged particle, the magnetic field through which it is moving, and the magnetic force on it.


e / The unit of magnetic field, the tesla, is named after Serbian-American inventor Nikola Tesla.


f / A standard dipole made from a square loop of wire shorting across a battery. It acts very much like a bar magnet, but its strength is more easily quantified.


g / A dipole tends to align itself to the surrounding magnetic field.


h / The m and A vectors.


i / The torque on a current loop in a magnetic field. The current comes out of the page, goes across, goes back into the page, and then back across the other way in the hidden side of the loop.


j / A vector coming out of the page is shown with the tip of an arrowhead. A vector going into the page is represented using the tailfeathers of the arrow.


k / Dipole vectors can be added.


l / An irregular loop can be broken up into little squares.


m / The magnetic field pattern around a bar magnet is created by the superposition of the dipole fields of the individual iron atoms. Roughly speaking, it looks like the field of one big dipole, especially farther away from the magnet. Closer in, however, you can see a hint of the magnet's rectangular shape. The picture was made by placing iron filings on a piece of paper, and then bringing a magnet up underneath.

11.1.2 The magnetic field

Definition in terms of the force on a moving particle

With electricity, it turned out to be useful to define an electric field rather than always working in terms of electric forces. Likewise, we want to define a magnetic field, \(\mathbf{B}\). Let's look at the result of the preceding subsection for insight. The equation

\[\begin{equation*} F = \frac{2kI qv}{c^2R} \end{equation*}\]

shows that when we put a moving charge near other moving charges, there is an extra magnetic force on it, in addition to any electric forces that may exist. Equations for electric forces always have a factor of \(k\) in front --- the Coulomb constant \(k\) is called the coupling constant for electric forces. Since magnetic effects are relativistic in origin, they end up having a factor of \(k/c^2\) instead of just \(k\). In a world where the speed of light was infinite, relativistic effects, including magnetism, would be absent, and the coupling constant for magnetism would be zero. A cute feature of the metric system is that we have \(k/c^2=10^{-7}\ \text{N}\cdot\text{s}^2/\text{C}^2\) exactly, as a matter of definition.

Naively, we could try to work by analogy with the electric field, and define the magnetic field as the magnetic force per unit charge. However, if we think of the lone charge in our example as the test charge, we'll find that this approach fails, because the force depends not just on the test particle's charge, but on its velocity, \(v\), as well. Although we only carried out calculations for the case where the particle was moving parallel to the wire, in general this velocity is a vector, v, in three dimensions. We can also anticipate that the magnetic field will be a vector. The electric and gravitational fields are vectors, and we expect intuitively based on our experience with magnetic compasses that a magnetic field has a particular direction in space. Furthermore, reversing the current \(I\) in our example would have reversed the force, which would only make sense if the magnetic field had a direction in space that could be reversed. Summarizing, we think there must be a magnetic field vector B, and the force on a test particle moving through a magnetic field is proportional both to the B vector and to the particle's own v vector. In other words, the magnetic force vector F is found by some sort of vector multiplication of the vectors v and B. As proved on page 918, however, there is only one physically useful way of defining such a multiplication, which is the cross product.

We therefore define the magnetic field vector, B, as the vector that determines the force on a charged particle according to the following rule:

\[\begin{equation*} \mathbf{F} = q\mathbf{v}\times\mathbf{B} \text{[definition of the magnetic field]} \end{equation*}\]

From this definition, we see that the magnetic field's units are \(\text{N}\cdot\text{s}/\text{C}\cdot\text{m}\), which are usually abbreviated as teslas, \(1\ \text{T}=1\ \text{N}\cdot\text{s}/\text{C}\cdot\text{m}\). The definition implies a right-hand-rule relationship among the vectors, figure d, if the charge \(q\) is positive, and the opposite handedness if it is negative.

This is not just a definition but a bold prediction! Is it really true that for any point in space, we can always find a vector B that successfully predicts the force on any passing particle, regardless of its charge and velocity vector? Yes --- it's not obvious that it can be done, but experiments verify that it can. How? Well for example, the cross product of parallel vectors is zero, so we can try particles moving in various directions, and hunt for the direction that produces zero force; the B vector lies along that line, in either the same direction the particle was moving, or the opposite one. We can then go back to our data from one of the other cases, where the force was nonzero, and use it to choose between these two directions and find the magnitude of the B vector. We could then verify that this vector gave correct force predictions in a variety of other cases.

Even with this empirical reassurance, the meaning of this equation is not intuitively transparent, nor is it practical in most cases to measure a magnetic field this way. For these reasons, let's look at an alternative method of defining the magnetic field which, although not as fundamental or mathematically simple, may be more appealing.

Definition in terms of the torque on a dipole

A compass needle in a magnetic field experiences a torque which tends to align it with the field. This is just like the behavior of an electric dipole in an electric field, so we consider the compass needle to be a magnetic dipole. In subsection 10.1.3 on page 571, we gave an alternative definition of the electric field in terms of the torque on an electric dipole.

To define the strength of a magnetic field, however, we need some way of defining the strength of a test dipole, i.e., we need a definition of the magnetic dipole moment. We could use an iron permanent magnet constructed according to certain specifications, but such an object is really an extremely complex system consisting of many iron atoms, only some of which are aligned with each other. A more fundamental standard dipole is a square current loop. This could be little resistive circuit consisting of a square of wire shorting across a battery, f.

Applying \(\mathbf{F}=\mathbf{v}\times\mathbf{B}\), we find that such a loop, when placed in a magnetic field, g, experiences a torque that tends to align plane so that its interior “face” points in a certain direction. Since the loop is symmetric, it doesn't care if we rotate it like a wheel without changing the plane in which it lies. It is this preferred facing direction that we will end up using as our alternative definition of the magnetic field.

If the loop is out of alignment with the field, the torque on it is proportional to the amount of current, and also to the interior area of the loop. The proportionality to current makes sense, since magnetic forces are interactions between moving charges, and current is a measure of the motion of charge. The proportionality to the loop's area is also not hard to understand, because increasing the length of the sides of the square increases both the amount of charge contained in this circular “river” and the amount of leverage supplied for making torque. Two separate physical reasons for a proportionality to length result in an overall proportionality to length squared, which is the same as the area of the loop. For these reasons, we define the magnetic dipole moment of a square current loop as

\[\begin{equation*} \mathbf{m} = I\mathbf{A} , \end{equation*}\]

where the direction of the vectors is defined as shown in figure h.

We can now give an alternative definition of the magnetic field:

The magnetic field vector, B, at any location in space is defined by observing the torque exerted on a magnetic test dipole \(\mathbf{m}_{t}\) consisting of a square current loop. The field's magnitude is

\[\begin{equation*} |\mathbf{B}| = \frac{\tau}{|\mathbf{m}_{t}|\sin\theta} , \end{equation*}\]

where \(\theta\) is the angle between the dipole vector and the field. This is equivalent to the vector cross product \(\boldsymbol{\tau}=\mathbf{m}_t\times\mathbf{B}\).

Let's show that this is consistent with the previous definition, using the geometry shown in figure i. The velocity vector that point in and out of the page are shown using the convention defined in figure j. Let the mobile charge carriers in the wire have linear density \(\lambda\), and let the sides of the loop have length \(h\), so that we have \(I=\lambda v\), and \(m=h^2\lambda v\). The only nonvanishing torque comes from the forces on the left and right sides. The currents in these sides are perpendicular to the field, so the magnitude of the cross product \(\mathbf{F}=q\mathbf{v}\times\mathbf{B}\) is simply \(|\mathbf{F}|=qvB\). The torque supplied by each of these forces is \(\mathbf{r}\times\mathbf{F}\), where the lever arm \(\mathbf{r}\) has length \(h/2\), and makes an angle \(\theta\) with respect to the force vector. The magnitude of the total torque acting on the loop is therefore

\[\begin{align*} |\boldsymbol{\tau}| &=2\frac{h}{2}|\mathbf{F}|\sin\theta \\ &=h\:qvB\:\sin\theta ,\\ \text{and substituting $q=\lambda h$ and $v=m/h^2\lambda$, we have} |\boldsymbol{\tau}| &= h\:\lambda h\:\frac{m}{h^2\lambda} B\sin\theta \\ &=m B\sin\theta ,\\ \end{align*}\]

which is consistent with the second definition of the field.

It undoubtedly seems artificial to you that we have discussed dipoles only in the form of a square loop of current. A permanent magnet, for example, is made out of atomic dipoles, and atoms aren't square! However, it turns out that the shape doesn't matter. To see why this is so, consider the additive property of areas and dipole moments, shown in figure k. Each of the square dipoles has a dipole moment that points out of the page. When they are placed side by side, the currents in the adjoining sides cancel out, so they are equivalent to a single rectangular loop with twice the area. We can break down any irregular shape into little squares, as shown in figure l, so the dipole moment of any planar current loop can be calculated based on its area, regardless of its shape.

Example 1: The magnetic dipole moment of an atom

Let's make an order-of-magnitude estimate of the magnetic dipole moment of an atom. A hydrogen atom is about \(10^{-10}\) m in diameter, and the electron moves at speeds of about \(10^{-2} c\). We don't know the shape of the orbit, and indeed it turns out that according to the principles of quantum mechanics, the electron doesn't even have a well-defined orbit, but if we're brave, we can still estimate the dipole moment using the cross-sectional area of the atom, which will be on the order of \((10^{-10}\ \text{m})^2=10^{-20}\ \text{m}^2\). The electron is a single particle, not a steady current, but again we throw caution to the winds, and estimate the current it creates as \(e/\Delta t\), where \(\Delta t\), the time for one orbit, can be estimated by dividing the size of the atom by the electron's velocity. (This is only a rough estimate, and we don't know the shape of the orbit, so it would be silly, for instance, to bother with multiplying the diameter by \(\pi\) based on our intuitive visualization of the electron as moving around the circumference of a circle.) The result for the dipole moment is \(m\sim10^{-23}\ \text{A}\!\cdot\!\text{m}^2\).

Should we be impressed with how small this dipole moment is, or with how big it is, considering that it's being made by a single atom? Very large or very small numbers are never very interesting by themselves. To get a feeling for what they mean, we need to compare them to something else. An interesting comparison here is to think in terms of the total number of atoms in a typical object, which might be on the order of \(10^{26}\) (Avogadro's number). Suppose we had this many atoms, with their moments all aligned. The total dipole moment would be on the order of \(10^3\ \text{A}\!\cdot\!\text{m}^2\), which is a pretty big number. To get a dipole moment this strong using human-scale devices, we'd have to send a thousand amps of current through a one-square meter loop of wire! The insight to be gained here is that, even in a permanent magnet, we must not have all the atoms perfectly aligned, because that would cause more spectacular magnetic effects than we really observe. Apparently, nearly all the atoms in such a magnet are oriented randomly, and do not contribute to the magnet's dipole moment.

Discussion Questions

The physical situation shown in figure c on page 652 was analyzed entirely in terms of forces. Now let's go back and think about it in terms of fields. The charge by itself up above the wire is like a test charge, being used to determine the magnetic and electric fields created by the wire. In figures c/1 and c/2, are there fields that are purely electric or purely magnetic? Are there fields that are a mixture of \(\mathbf{E}\) and \(\mathbf{B}\)? How does this compare with the forces?

Continuing the analysis begun in discussion question A, can we come up with a scenario involving some charged particles such that the fields are purely magnetic in one frame of reference but a mixture of \(\mathbf{E}\) and \(\mathbf{B}\) in another frame? How about an example where the fields are purely electric in one frame, but mixed in another? Or an example where the fields are purely electric in one frame, but purely magnetic in another?


n / Example 2.


o / Magnetic forces cause a beam of electrons to move in a circle.


p / You can't isolate the poles of a magnet by breaking it in half.


q / A magnetic dipole is made out of other dipoles, not out of monopoles.

11.1.3 Some applications

Example 2: Magnetic levitation
In figure n, a small, disk-shaped permanent magnet is stuck on the side of a battery, and a wire is clasped loosely around the battery, shorting it. A large current flows through the wire. The electrons moving through the wire feel a force from the magnetic field made by the permanent magnet, and this force levitates the wire.

From the photo, it's possible to find the direction of the magnetic field made by the permanent magnet. The electrons in the copper wire are negatively charged, so they flow from the negative (flat) terminal of the battery to the positive terminal (the one with the bump, in front). As the electrons pass by the permanent magnet, we can imagine that they would experience a field either toward the magnet, or away from it, depending on which way the magnet was flipped when it was stuck onto the battery. By the right-hand rule (figure d on page 655), the field must be toward the battery.

Example 3: Nervous-system effects during an MRI scan

During an MRI scan of the head, the patient's nervous system is exposed to intense magnetic fields, and there are ions moving around in the nerves. The resulting forces on the ions can cause symptoms such as vertigo.

Example 4: A circular orbit
The magnetic force is always perpendicular to the motion of the particle, so it can never do any work, and a charged particle moving through a magnetic field does not experience any change in its kinetic energy: its velocity vector can change its direction, but not its magnitude. If the velocity vector is initially perpendicular to the field, then the curve of its motion will remain in the plane perpendicular to the field, so the magnitude of the magnetic force on it will stay the same. When an object experiences a force with constant magnitude, which is always perpendicular to the direction of its motion, the result is that it travels in a circle.

Figure o shows a beam of electrons in a spherical vacuum tube. In the top photo, the beam is emitted near the right side of the tube, and travels straight up. In the bottom photo, a magnetic field has been imposed by an electromagnet surrounding the vacuum tube; the ammeter on the right shows that the current through the electromagnet is now nonzero. We observe that the beam is bent into a circle.


Infer the direction of the magnetic field. Don't forget that the beam is made of electrons, which are negatively charged!

(answer in the back of the PDF version of the book)

Homework problem 12 is a quantitative analysis of circular orbits.

Example 5: A velocity filter

Suppose you see the electron beam in figure o, and you want to determine how fast the electrons are going. You certainly can't do it with a stopwatch! Physicists may also encounter situations where they have a beam of unknown charged particles, and they don't even know their charges. This happened, for instance, when alpha and beta radiation were discovered. One solution to this problem relies on the fact that the force experienced by a charged particle in an electric field, \(\mathbf{F}_E=q\mathbf{E}\), is independent of its velocity, but the force due to a magnetic field, \(\mathbf{F}_B=q\mathbf{v}\times\mathbf{B}\), isn't. One can send a beam of charged particles through a space containing both an electric and a magnetic field, setting up the fields so that the two forces will cancel out perfectly for a certain velocity. Note that since both forces are proportional to the charge of the particles, the cancellation is independent of charge. Such a velocity filter can be used either to determine the velocity of an unknown beam or particles, or to select from a beam of particles only those having velocities within a certain desired range. Homework problem 7 is an analysis of this application.


r / Magnetic fields have no sources or sinks.


s / Example 6.

11.1.4 No magnetic monopoles

If you could play with a handful of electric dipoles and a handful of bar magnets, they would appear very similar. For instance, a pair of bar magnets wants to align themselves head-to-tail, and a pair of electric dipoles does the same thing. (It is unfortunately not that easy to make a permanent electric dipole that can be handled like this, since the charge tends to leak.)

You would eventually notice an important difference between the two types of objects, however. The electric dipoles can be broken apart to form isolated positive charges and negative charges. The two-ended device can be broken into parts that are not two-ended. But if you break a bar magnet in half, p, you will find that you have simply made two smaller two-ended objects.

The reason for this behavior is not hard to divine from our microscopic picture of permanent iron magnets. An electric dipole has extra positive “stuff” concentrated in one end and extra negative in the other. The bar magnet, on the other hand, gets its magnetic properties not from an imbalance of magnetic “stuff” at the two ends but from the orientation of the rotation of its electrons. One end is the one from which we could look down the axis and see the electrons rotating clockwise, and the other is the one from which they would appear to go counterclockwise. There is no difference between the “stuff” in one end of the magnet and the other, q.

Nobody has ever succeeded in isolating a single magnetic pole. In technical language, we say that magnetic monopoles not seem to exist. Electric monopoles do exist --- that's what charges are.

Electric and magnetic forces seem similar in many ways. Both act at a distance, both can be either attractive or repulsive, and both are intimately related to the property of matter called charge. (Recall that magnetism is an interaction between moving charges.) Physicists's aesthetic senses have been offended for a long time because this seeming symmetry is broken by the existence of electric monopoles and the absence of magnetic ones. Perhaps some exotic form of matter exists, composed of particles that are magnetic monopoles. If such particles could be found in cosmic rays or moon rocks, it would be evidence that the apparent asymmetry was only an asymmetry in the composition of the universe, not in the laws of physics. For these admittedly subjective reasons, there have been several searches for magnetic monopoles. Experiments have been performed, with negative results, to look for magnetic monopoles embedded in ordinary matter. Soviet physicists in the 1960's made exciting claims that they had created and detected magnetic monopoles in particle accelerators, but there was no success in attempts to reproduce the results there or at other accelerators. The most recent search for magnetic monopoles, done by reanalyzing data from the search for the top quark at Fermilab, turned up no candidates, which shows that either monopoles don't exist in nature or they are extremely massive and thus hard to create in accelerators.

The nonexistence of magnetic monopoles means that unlike an electric field, a magnetic one, can never have sources or sinks. The magnetic field vectors lead in paths that loop back on themselves, without ever converging or diverging at a point, as in the fields shown in figure r. Gauss' law for magnetism is therefore much simpler than Gauss' law for electric fields:

\[\begin{equation*} \Phi_B = \sum \mathbf{B}_j\cdot\mathbf{A}_j = 0 \end{equation*}\]

The magnetic flux through any closed surface is zero.


Draw a Gaussian surface on the electric dipole field of figure r that has nonzero electric flux through it, and then draw a similar surface on the magnetic field pattern. What happens?

(answer in the back of the PDF version of the book)
Example 6: The field of a wire
\(\triangleright\) On page 654, we showed that a long, straight wire carrying current \(I\) exerts a magnetic force
\[\begin{equation*} F = \frac{2 kIqv}{ c^2 R} \end{equation*}\]
on a particle with charge \(q\) moving parallel to the wire with velocity \(v\). What, then, is the magnetic field of the wire?

\(\triangleright\) Comparing the equation above to the first definition of the magnetic field, \(\mathbf{F}=\mathbf{v}\times\mathbf{B}\), it appears that the magnetic field is one that falls off like \(1/ R\), where \(R\) is the distance from the wire. However, it's not so easy to determine the direction of the field vector. There are two other axes along which the particle could have been moving, and the brute-force method would be to carry out relativistic calculations for these cases as well. Although this would probably be enough information to determine the field, we don't want to do that much work.

Instead, let's consider what the possibilities are. The field can't be parallel to the wire, because a cross product vanishes when the two vectors are parallel, and yet we know from the case we analyzed that the force doesn't vanish when the particle is moving parallel to the wire. The other two possibilities that are consistent with the symmetry of the problem are shown in figure s. One is like a bottle brush, and the other is like a spool of thread. The bottle brush pattern, however, violates Gauss' law for magnetism. If we made a cylindrical Gaussian surface with its axis coinciding with the wire, the flux through it would not be zero. We therefore conclude that the spool-of-thread pattern is the correct one.4 Since the particle in our example was moving perpendicular to the field, we have \(| F|=|q|| v|| B|\), so

\[\begin{align*} | B| &= \frac{| F|}{|q| | v|} \\ &= \frac{2 kI}{ c^2 R}\\ \end{align*}\]


u / In this scene from Swan Lake, the choreography has a symmetry with respect to left and right.


v / C.S. Wu

11.1.5 Symmetry and handedness

Imagine that you establish radio contact with an alien on another planet. Neither of you even knows where the other one's planet is, and you aren't able to establish any landmarks that you both recognize. You manage to learn quite a bit of each other's languages, but you're stumped when you try to establish the definitions of left and right (or, equivalently, clockwise and counterclockwise). Is there any way to do it?


t / Left-handed and right-handed definitions.

If there was any way to do it without reference to external landmarks, then it would imply that the laws of physics themselves were asymmetric, which would be strange. Why should they distinguish left from right? The gravitational field pattern surrounding a star or planet looks the same in a mirror, and the same goes for electric fields. However, the magnetic field patterns shown in figure s seems to violate this principle. Could you use these patterns to explain left and right to the alien? No. If you look back at the definition of the magnetic field, it also contains a reference to handedness: the direction of the vector cross product. The aliens might have reversed their definition of the magnetic field, in which case their drawings of field patterns would look like mirror images of ours, as in the left panel of figure t.

Until the middle of the twentieth century, physicists assumed that any reasonable set of physical laws would have to have this kind of symmetry between left and right. An asymmetry would be grotesque. Whatever their aesthetic feelings, they had to change their opinions about reality when experiments by C.S. Wu et al. showed that the weak nuclear force violates right-left symmetry! It is still a mystery why right-left symmetry is observed so scrupulously in general, but is violated by one particular type of physical process.

11.2 Magnetic Fields by Superposition


a / The magnetic field of a long, straight wire.


b / A ground fault interrupter.


c / Example 8.


d / A sheet of charge.


e / A sheet of charge and a sheet of current.

11.2.1 Superposition of straight wires

In chapter 10, one of the most important goals was to learn how to calculate the electric field for a given charge distribution. The corresponding problem for magnetism would be to calculate the magnetic field arising from a given set of currents. So far, however, we only know how to calculate the magnetic field of a long, straight wire,

\[\begin{equation*} B = \frac{2kI}{c^2R} , \end{equation*}\]

with the geometry shown in figure a. Whereas a charge distribution can be broken down into individual point charges, most currents cannot be broken down into a set of straight-line currents. Nevertheless, let's see what we can do with the tools that we have.

Example 7: A ground fault interrupter

Electric current in your home is supposed to flow out of one side of the outlet, through an appliance, and back into the wall through the other side of the outlet. If that's not what happens, then we have a problem --- the current must be finding its way to ground through some other path, perhaps through someone's body. If you have outlets in your home that have “test” and “reset” buttons on them, they have a safety device built into them that is meant to protect you in this situation. The ground fault interrupter (GFI) shown in figure b, routes the outgoing and returning currents through two wires that lie very close together. The clockwise and counterclockwise fields created by the two wires combine by vector addition, and normally cancel out almost exactly. However, if current is not coming back through the circuit, a magnetic field is produced. The doughnut-shaped collar detects this field (using an effect called induction, to be discussed in section 11.5), and sends a signal to a logic chip, which breaks the circuit within about 25 milliseconds.

Example 8: An example with vector addition
\(\triangleright\) Two long, straight wires each carry current \(I\) parallel to the \(y\) axis, but in opposite directions. They are separated by a gap \(2 h\) in the \(x\) direction. Find the magnitude and direction of the magnetic field at a point located at a height \(z\) above the plane of the wires, directly above the center line.

\(\triangleright\) The magnetic fields contributed by the two wires add like vectors, which means we can add their \(x\) and \(z\) components. The \(x\) components cancel by symmetry. The magnitudes of the individual fields are equal,

\[\begin{equation*} B_1 = B_2 = \frac{2 kI}{ c^2 R} ,\\ \end{equation*}\]

so the total field in the \(z\) direction is

\[\begin{equation*} B_{z} = 2\frac{2 kI}{ c^2 R}\text{sin}\:\theta ,\\ \end{equation*}\]

where \(\theta\) is the angle the field vectors make above the \(x\) axis. The sine of this angle equals \(h/ R\), so

\[\begin{equation*} B_{z} = \frac{4 kIh}{ c^2 R^2} .\\ \end{equation*}\]

(Putting this explicitly in terms of \(z\) gives the less attractive form \(B_{z}=4 kIh/ c^2( h^2+ z^2)\).)

At large distances from the wires, the individual fields are mostly in the \(\pm x\) direction, so most of their strength cancels out. It's not surprising that the fields tend to cancel, since the currents are in opposite directions. What's more interesting is that not only is the field weaker than the field of one wire, it also falls off as \(R^{-2}\) rather than \(R^{-1}\). If the wires were right on top of each other, their currents would cancel each other out, and the field would be zero. From far away, the wires appear to be almost on top of each other, which is what leads to the more drastic \(R^{-2}\) dependence on distance.


In example 8, what is the field right between the wires, at \(z=0\), and how does this simpler result follow from vector addition?

(answer in the back of the PDF version of the book)

An alarming infinity

An interesting aspect of the \(R^{-2}\) dependence of the field in example 8 is the energy of the field. We've already established on p. 592 that the energy density of the magnetic field must be proportional to the square of the field strength, \(B^2\), the same as for the gravitational and electric fields. Suppose we try to calculate the energy per unit length stored in the field of a single wire. We haven't yet found the proportionality factor that goes in front of the \(B^2\), but that doesn't matter, because the energy per unit length turns out to be infinite! To see this, we can construct concentric cylindrical shells of length \(L\), with each shell extending from \(R\) to \(R+dR\). The volume of the shell equals its circumference times its thickness times its length, \(dv=(2\pi R)(dR)(L)=2\pi LdR\). For a single wire, we have \(B\sim R^{-1}\), so the energy density is proportional to \(R^{-2}\), and the energy contained in each shell varies as \(R^{-2}dv\sim R^{-1}dr\). Integrating this gives a logarithm, and as we let \(R\) approach infinity, we get the logarithm of infinity, which is infinite.

Taken at face value, this result would imply that electrical currents could never exist, since establishing one would require an infinite amount of energy per unit length! In reality, however, we would be dealing with an electric circuit, which would be more like the two wires of example 8: current goes out one wire, but comes back through the other. Since the field really falls off as \(R^{-2}\), we have an energy density that varies as \(R^{-4}\), which does not give infinity when integrated out to infinity. (There is still an infinity at \(R=0\), but this doesn't occur for a real wire, which has a finite diameter.)

Still, one might worry about the physical implications of the single-wire result. For instance, suppose we turn on an electron gun, like the one in a TV tube. It takes perhaps a microsecond for the beam to progress across the tube. After it hits the other side of the tube, a return current is established, but at least for the first microsecond, we have only a single current, not two. Do we have infinite energy in the resulting magnetic field? No. It takes time for electric and magnetic field disturbances to travel outward through space, so during that microsecond, the field spreads only to some finite value of \(R\), not \(R=\infty\).

This reminds us of an important fact about our study of magnetism so far: we have only been considering situations where the currents and magnetic fields are constant over time. The equation \(B = 2kI/c^2R\) was derived under this assumption. This equation is only valid if we assume the current has been established and flowing steadily for a long time, and if we are talking about the field at a point in space at which the field has been established for a long time. The generalization to time-varying fields is nontrivial, and qualitatively new effects will crop up. We have already seen one example of this on page 602, where we inferred that an inductor's time-varying magnetic field creates an electric field --- an electric field which is not created by any charges anywhere. Effects like these will be discussed in section 11.5

A sheet of current

There is a saying that in computer science, there are only three nice numbers: zero, one, and however many you please. In other words, computer software shouldn't have arbitrary limitations like a maximum of 16 open files, or 256 e-mail messages per mailbox. When superposing the fields of long, straight wires, the really interesting cases are one wire, two wires, and infinitely many wires. With an infinite number of wires, each carrying an infinitesimal current, we can create sheets of current, as in figure d. Such a sheet has a certain amount of current per unit length, which we notate \(\eta\) (Greek letter eta). The setup is similar to example 8, except that all the currents are in the same direction, and instead of adding up two fields, we add up an infinite number of them by doing an integral. For the \(y\) component, we have
\[\begin{align*} B_y &= \int \frac{2kdI}{c^2R}\cos\theta \\ &= \int_{-a}^{b} \frac{2k\etady}{c^2R}\cos\theta \\ &= \frac{2k\eta}{c^2} \int_{-a}^{b} \frac{\cos\theta}{R}dy \\ &= \frac{2k\eta}{c^2} \int_{-a}^{b}\: \frac{zdy}{y^2+z^2} \\ &= \frac{2k\eta}{c^2} \left(\tan^{-1}\frac{b}{z}-\tan^{-1}\frac{-a}{z}\right) \\ &= \frac{2k\eta\gamma}{c^2} , \end{align*}\]
where in the last step we have used the identity \(\tan^{-1}(-x)=-\tan^{-1}x\), combined with the relation \(\tan^{-1}b/z+\tan^{-1}a/z=\gamma\), which can be verified with a little geometry and trigonometry. The calculation of \(B_z\) is left as an exercise (problem 23). More interesting is what happens underneath the sheet: by the right-hand rule, all the currents make rightward contributions to the field there, so \(B_y\) abruptly reverses itself as we pass through the sheet.

Close to the sheet, the angle \(\gamma\) approaches \(\pi\), so we have

\[\begin{equation*} B_y = \frac{2\pi k\eta}{c^2} . \end{equation*}\]

Figure e shows the similarity between this result and the result for a sheet of charge. In one case the sources are charges and the field is electric; in the other case we have currents and magnetic fields. In both cases we find that the field changes suddenly when we pass through a sheet of sources, and the amount of this change doesn't depend on the size of the sheet. It was this type of reasoning that eventually led us to Gauss' law in the case of electricity, and in section 11.3 we will see that a similar approach can be used with magnetism. The difference is that, whereas Gauss' law involves the flux, a measure of how much the field spreads out, the corresponding law for magnetism will measure how much the field curls.

Is it just dumb luck that the magnetic-field case came out so similar to the electric field case? Not at all. We've already seen that what one observer perceives as an electric field, another observer may perceive as a magnetic field. An observer flying along above a charged sheet will say that the charges are in motion, and will therefore say that it is both a sheet of current and a sheet of charge. Instead of a pure electric field, this observer will experience a combination of an electric field and a magnetic one. (We could also construct an example like the one in figure c on page 652, in which the field was purely magnetic.)

11.2.2 Energy in the magnetic field

In section 10.4, I've already argued that the energy density of the magnetic field must be proportional to \(|\mathbf{B}|^2\), which we can write as \(B^2\) for convenience. To pin down the constant of proportionality, we now need to do something like the argument on page 586: find one example where we can calculate the mechanical work done by the magnetic field, and equate that to the amount of energy lost by the field itself. The easiest example is two parallel sheets of charge, with their currents in opposite directions. Homework problem 53 is such a calculation, which gives the result
\[\begin{equation*} dU_m = \frac{c^2}{8\pi k}B^2 dv . \end{equation*}\]


f / The field of any planar current loop can be found by breaking it down into square dipoles.


g / A long, straight current-carrying wire can be constructed by filling half of a plane with square dipoles.


h / Setting up the integral.


i / The field of a dipole.


j / Example 9.

11.2.3 Superposition of dipoles

To understand this subsection, you'll have to have studied section 4.2.4, on iterated integrals.

The distant field of a dipole, in its midplane

Most current distributions cannot be broken down into long, straight wires, and subsection 11.2.1 has exhausted most of the interesting cases we can handle in this way. A much more useful building block is a square current loop. We have already seen how the dipole moment of an irregular current loop can be found by breaking the loop down into square dipoles (figure l on page 658), because the currents in adjoining squares cancel out on their shared edges. Likewise, as shown in figure f, if we could find the magnetic field of a square dipole, then we could find the field of any planar loop of current by adding the contributions to the field from all the squares.

The field of a square-loop dipole is very complicated close up, but luckily for us, we only need to know the current at distances that are large compared to the size of the loop, because we're free to make the squares on our grid as small as we like. The distant field of a square dipole turns out to be simple, and is no different from the distant field of any other dipole with the same dipole moment. We can also save ourselves some work if we only worry about finding the field of the dipole in its own plane, i.e., the plane perpendicular to its dipole moment. By symmetry, the field in this plane cannot have any component in the radial direction (inward toward the dipole, or outward away from it); it is perpendicular to the plane, and in the opposite direction compared to the dipole vector. (The field inside the loop is in the same direction as the dipole vector, but we're interested in the distant field.) Letting the dipole vector be along the \(z\) axis, we find that the field in the \(x-y\) plane is of the form \(B_z=f(r)\), where \(f(r)\) is some function that depends only on \(r\), the distance from the dipole.

We can pin down the result even more without any math. We know that the magnetic field made by a current always contains a factor of \(k/c^2\), which is the coupling constant for magnetism. We also know that the field must be proportional to the dipole moment, \(m=IA\). Fields are always directly proportional to currents, and the proportionality to area follows because dipoles add according to their area. For instance, a square dipole that is 2 micrometers by 2 micrometers in size can be cut up into four dipoles that are 1 micrometer on a side. This tells us that our result must be of the form \(B_z=(k/c^2)(IA)g(r)\). Now if we multiply the quantity \((k/c^2)(IA)\) by the function \(g(r)\), we have to get units of teslas, and this only works out if \(g(r)\) has units of \(\text{m}^{-3}\) (homework problem 15), so our result must be of the form

\[\begin{equation*} B_z=\frac{\beta kIA}{c^2r^3} , \end{equation*}\]

where \(\beta\) is a unitless constant. Thus our only task is to determine \(\beta\), and we will have determined the field of the dipole (in the plane of its current, i.e., the midplane with respect to its dipole moment vector).

If we wanted to, we could simply build a dipole, measure its field, and determine \(\beta\) empirically. Better yet, we can get an exact result if we take a current loop whose field we know exactly, break it down into infinitesimally small squares, integrate to find the total field, set this result equal to the known expression for the field of the loop, and solve for \(\beta\). There's just one problem here. We don't yet know an expression for the field of any current loop of any shape --- all we know is the field of a long, straight wire. Are we out of luck? No, because, as shown in figure g, we can make a long, straight wire by putting together square dipoles! Any square dipole away from the edge has all four of its currents canceled by its neighbors. The only currents that don't cancel are the ones on the edge, so by superimposing all the square dipoles, we get a straight-line current.

This might seem strange. If the squares on the interior have all their currents canceled out by their neighbors, why do we even need them? Well, we need the squares on the edge in order to make the straight-line current. We need the second row of squares to cancel out the currents at the top of the first row of squares, and so on.

Integrating as shown in figure h, we have

\[\begin{align*} B_z &= \int_{y=0}^\infty \int_{x=-\infty}^\infty dB_z , \\ \text{where $dB_z$ is the contribution to the total magnetic field at our point of interest, which lies a distance $R$ from the wire.} B_z &= \int_{y=0}^\infty \int_{x=-\infty}^\infty \frac{\beta kIdA}{c^2r^3} \\ &= \frac{\beta kI}{c^2}\int_{y=0}^\infty \int_{x=-\infty}^\infty \frac{1}{\left[x^2+(R+y)^2\right]^{3/2}}dxdy \\ &= \frac{\beta kI}{c^2R^3}\int_{y=0}^\infty \int_{x=-\infty}^\infty \left[ \left(\frac{x}{R}\right)^2+\left(1+\frac{y}{R}\right)^2 \right]^{-3/2}dxdy \\ \text{This can be simplified with the substitutions $x=Ru$, $y=Rv$, and $dxdy=R^2dudv$ :} B_z &= \frac{\beta kI}{c^2R}\int_{v=0}^\infty \int_{u=-\infty}^\infty \frac{1}{\left[u^2+(1+v)^2\right]^{3/2}}dudv \\ \text{The $u$ integral is of the form $\int_{-\infty}^{\infty} (u^2+b)^{-3/2}du=2/b^2$, so} B_z &= \frac{\beta kI}{c^2R}\int_{v=0}^\infty \frac{1}{(1+v)^2}dv , \\ \text{and the remaining $v$ integral is equals 2, so} B_z &= \frac{2\beta kI}{c^2R} . \\ \end{align*}\]

This is the field of a wire, which we already know equals \(2kI/c^2R\), so we have \(\beta\)=1. Remember, the point of this whole calculation was not to find the field of a wire, which we already knew, but to find the unitless constant \(\beta\) in the expression for the field of a dipole. The distant field of a dipole, in its midplane, is therefore \(B_z=\beta kIA/c^2r^3= kIA/c^2r^3\), or, in terms of the dipole moment,

\[\begin{equation*} B_z=\frac{km}{c^2r^3} . \end{equation*}\]

The distant field of a dipole, out of its midplane

What about the field of a magnetic dipole outside of the dipole's midplane? Let's compare with an electric dipole. An electric dipole, unlike a magnetic one, can be built out of two opposite monopoles, i.e., charges, separated by a certain distance, and it is then straightforward to show by vector addition that the field of an electric dipole is

\[\begin{align*} E_z &= kD\left(3\cos^2\theta-1\right)r^{-3}\\ E_R &= kD\left(3\sin\theta\:\cos\theta\right)r^{-3} , \end{align*}\]

where \(r\) is the distance from the dipole to the point of interest, \(\theta\) is the angle between the dipole vector and the line connecting the dipole to this point, and \(E_z\) and \(E_R\) are, respectively, the components of the field parallel to and perpendicular to the dipole vector.

In the midplane, \(\theta\) equals \(\pi/2\), which produces \(E_z=-kDr^{-3}\) and \(E_R=0\). This is the same as the field of a magnetic dipole in its midplane, except that the electric coupling constant \(k\) replaces the magnetic version \(k/c^2\), and the electric dipole moment \(D\) is substituted for the magnetic dipole moment \(m\). It is therefore reasonable to conjecture that by using the same presto-change-o recipe we can find the field of a magnetic dipole outside its midplane:

\[\begin{align*} B_z &= \frac{km}{c^2}\left(3\cos^2\theta-1\right)r^{-3}\\ B_R &= \frac{km}{c^2}\left(3\sin\theta\:\cos\theta\right)r^{-3} . \end{align*}\]

This turns out to be correct. 5

Example 9: Concentric, counterrotating currents
\(\triangleright\) Two concentric circular current loops, with radii \(a\) and \(b\), carry the same amount of current \(I\), but in opposite directions. What is the field at the center?

\(\triangleright\) We can produce these currents by tiling the region between the circles with square current loops, whose currents all cancel each other except at the inner and outer edges. The flavor of the calculation is the same as the one in which we made a line of current by filling a half-plane with square loops. The main difference is that this geometry has a different symmetry, so it will make more sense to use polar coordinates instead of \(x\) and \(y\). The field at the center is

\[\begin{align*} B_{z} &= \int \frac{ kI}{ c^2 r^3}dA \\ &= \int_{ r= a}^{b}\frac{ kI}{ c^2 r^3}\:\cdot2\pi rdr \\ &= \frac{2\pi kI}{ c^2}\:\left(\frac{1}{a}-\frac{1}{b}\right) . \end{align*}\]

The positive sign indicates that the field is out of the page.

Example 10: Field at the center of a circular loop
\(\triangleright\) What is the magnetic field at the center of a circular current loop of radius \(a\), which carries a current \(I\)?

\(\triangleright\) This is like example 9, but with the outer loop being very large, and therefore too distant to make a significant field at the center. Taking the limit of that result as \(b\) approaches infinity, we have

\[\begin{equation*} B_{z} = \frac{2\pi kI}{ c^2 a} \end{equation*}\]

Comparing the results of examples 9 and 10, we see that the directions of the fields are both out of the page. In example 9, the outer loop has a current in the opposite direction, so it contributes a field that is into the page. This, however, is weaker than the field due to the inner loop, which dominates because it is less distant.


k / Two ways of making a current loop out of square dipoles.


l / The new method can handle non-planar currents.


m / The field of an infinite U.


n / The geometry of the Biot-Savart law. The small arrows show the result of the Biot-Savart law at various positions relative to the current segment \(d\boldsymbol{\ell}\). The Biot-Savart law involves a cross product, and the right-hand rule for this cross product is demonstrated for one case.


o / Example 12.

11.2.4 The Biot-Savart law (optional)

In section 11.2.3 we developed a method for finding the field due to a given current distribution by tiling a plane with square dipoles. This method has several disadvantages:

Figure k shows the first step in eliminating these defects: instead of spreading our dipoles out in a plane, we bring them out along an axis. As shown in figure l, this eliminates the restriction to currents that lie in a plane. Now we have to use the general equations for a dipole field from page 672, rather than the simpler expression for the field in the midplane of a dipole. This increase in complication is more than compensated for by a fortunate feature of the new geometry, which is that the infinite tube can be broken down into strips, and we can find the field of such a strip for once and for all. This means that we no longer have to do one integral inside another. The derivation of the most general case is a little messy, so I'll just present the case shown in figure m, where the point of interest is assumed to lie in the \(y-z\) plane. Intuitively, what we're really finding is the field of the short piece of length \(d\ell\) on the end of the U; the two long parallel segments are going to be canceled out by their neighbors when we assemble all the strips to make the tube. We expect that the field of this end-piece will form a pattern that circulates around the \(y\) axis, so at the point of interest, it's really the \(x\) component of the field that we want to compute:

\[\begin{align*} dB_x &= \int dB_R \cos \alpha \\ &= \int \frac{kId \elldx}{c^2s^3}(3\sin\theta\cos\theta \cos \alpha) \\ &= \frac{3kId \ell}{c^2} \int_0^\infty \frac{1}{s^3}\left(\frac{xz}{s^2}\right) dx \\ &= \frac{3kIzd \ell}{c^2} \int_0^\infty \frac{x}{(x^2+r^2)^{5/2}} dx \\ &= \frac{kId \ell\:z}{c^2r^3}\\ &= \frac{kId \ell\:\sin\phi}{c^2r^2} \end{align*}\]

In the more general case, l, the current loop is not planar, the point of interest is not in the end-planes of the U's, and the U shapes have their ends staggered, so the end-piece \(d\ell\) is not the only part of each U whose current is not canceled. Without going into the gory details, the correct general result is as follows:

\[\begin{equation*} d \mathbf{B} = \frac{kId \boldsymbol{\ell}\times\mathbf{r}}{c^2r^3} , \end{equation*}\]

which is known as the Biot-Savart law. (It rhymes with “leo bazaar.” Both t's are silent.) The distances \(d\ell\) and \( r\) are now defined as vectors, \(d\boldsymbol{\ell}\) and \(\mathbf{r}\), which point, respectively, in the direction of the current in the end-piece and the direction from the end-piece to the point of interest. The new equation looks different, but it is consistent with the old one. The vector cross product \(d\boldsymbol{\ell}\times\mathbf{r}\) has a magnitude \(rd \ell\:\sin\phi\), which cancels one of \(r\)'s in the denominator and makes the \(d \boldsymbol{\ell}\times\mathbf{r}/r^3\) into a vector with magnitude \(d \ell\:\sin\phi/r^2\).

Example 11: The field at the center of a circular loop

Previously we had to do quite a bit of work (examples 9 and 10), to calculate the field at the center of a circular loop of current of radius \(a\). It's much easier now. Dividing the loop into many short segments, each \(d\boldsymbol{\ell}\) is perpendicular to the \(\mathbf{r}\) vector that goes from it to the center of the circle, and every \(\mathbf{r}\) vector has magnitude \(a\). Therefore every cross product \(d\boldsymbol{\ell}\times\mathbf{r}\) has the same magnitude, \(ad \ell\), as well as the same direction along the axis perpendicular to the loop. The field is

\[\begin{align*} B &= \int \frac{ kIad \ell}{ c^2 a^3}\\ &= \frac{ kI}{ c^2 a^2} \int d \ell \\ &= \frac{ kI}{ c^2 a^2} (2\pi a) \\ &= \frac{2\pi kI}{ c^2 a} \\ \end{align*}\]

Example 12: Out-of-the-plane field of a circular loop
\(\triangleright\) What is the magnetic field of a circular loop of current at a point on the axis perpendicular to the loop, lying a distance \(z\) from the loop's center?

\(\triangleright\) Again, let's write \(a\) for the loop's radius. The r vector now has magnitude \(\sqrt{ a^2+ z^2}\), but it is still perpendicular to the \(d\boldsymbol{\ell}\) vector. By symmetry, the only nonvanishing component of the field is along the \(z\) axis,

\[\begin{align*} B_{z} &= \int |d\mathbf{B}|\:\text{cos}\:\alpha\\ &= \int \frac{ kI\,r\,d \ell}{ c^2 r^3}\frac{ a}{ r}\\ &= \frac{ kIa}{ c^2 r^3} \int d \ell \\ &= \frac{2\pi kIa^2}{ c^2( a^2+ z^2)^{3/2}} . \end{align*}\]

Is it the field of a particle?

We have a simple equation, based on Coulomb's law, for the electric field surrounding a charged particle. Looking at figure n, we can imagine that if the current segment \(d\ell\) was very short, then it might only contain one electron. It's tempting, then, to interpret the Biot-Savart law as a similar equation for the magnetic field surrounding a moving charged particle. Tempting but wrong! Suppose you stand at a certain point in space and watch a charged particle move by. It has an electric field, and since it's moving, you will also detect a magnetic field on top of that. Both of these fields change over time, however. Not only do they change their magnitudes and directions due to your changing geometric relationship to the particle, but they are also time-delayed, because disturbances in the electromagnetic field travel at the speed of light, which is finite. The fields you detect are the ones corresponding to where the particle used to be, not where it is now. Coulomb's law and the Biot-Savart law are both false in this situation, since neither equation includes time as a variable. It's valid to think of Coulomb's law as the equation for the field of a stationary charged particle, but not a moving one. The Biot-Savart law fails completely as a description of the field of a charged particle, since stationary particles don't make magnetic fields, and the Biot-Savart law fails in the case where the particle is moving.

If you look back at the long chain of reasoning that led to the Biot-Savart law, it all started from the relativistic arguments at the beginning of this chapter, where we assumed a steady current in an infinitely long wire. Everything that came later was built on this foundation, so all our reasoning depends on the assumption that the currents are steady. In a steady current, any charge that moves away from a certain spot is replaced by more charge coming up behind it, so even though the charges are all moving, the electric and magnetic fields they produce are constant. Problems of this type are called electrostatics and magnetostatics problems, and it is only for these problems that Coulomb's law and the Biot-Savart law are valid.

You might think that we could patch up Coulomb's law and the Biot-Savart law by inserting the appropriate time delays. However, we've already seen a clear example of a phenomenon that wouldn't be fixed by this patch: on page 602, we found that a changing magnetic field creates an electric field. Induction effects like these also lead to the existence of light, which is a wave disturbance in the electric and magnetic fields. We could try to apply another band-aid fix to Coulomb's law and the Biot-Savart law to make them deal with induction, but it won't work.

So what are the fundamental equations that describe how sources give rise to electromagnetic fields? We've already encountered two of them: Gauss' law for electricity and Gauss' law for magnetism. Experiments show that these are valid in all situations, not just static ones. But Gauss' law for magnetism merely says that the magnetic flux through a closed surface is zero. It doesn't tell us how to make magnetic fields using currents. It only tells us that we can't make them using magnetic monopoles. The following section develops a new equation, called Ampère's law, which is equivalent to the Biot-Savart law for magnetostatics, but which, unlike the Biot-Savart law, can easily be extended to nonstatic situations.

11.3 Magnetic Fields by Ampère's Law


a / The electric field of a sheet of charge, and the magnetic field of a sheet of current.


b / A Gaussian surface and an Ampèrian surface.


c / The definition of the circulation, \(\Gamma\).


d / Positive and negative signs in Ampère's law.


e / Example 13: a cutaway view of a solenoid.

11.3.1 Ampère's law

As discussed at the end of subsection 11.2.4, our goal now is to find an equation for magnetism that, unlike the Biot-Savart law, will not end up being a dead end when we try to extend it to nonstatic situations.6 Experiments show that Gauss' law is valid in both static and nonstatic situations, so it would be reasonable to look for an approach to magnetism that is similar to the way Gauss' law deals with electricity.

How can we do this? Figure a, reproduced from page 668, is our roadmap. Electric fields spread out from charges. Magnetic fields curl around currents. In figure b/1, we define a Gaussian surface, and we define the flux in terms of the electric field pointing out through this surface. In the magnetic case, b/2, we define a surface, called an Ampèrian surface, and we define a quantity called the circulation, \(\Gamma\) (uppercase Greek gamma), in terms of the magnetic field that points along the edge of the Ampèrian surface, c. We break the edge into tiny parts \(\mathbf{s}_j\), and for each of these parts, we define a contribution to the circulation using the dot product of \(d\mathbf{s}\) with the magnetic field:

\[\begin{equation*} \Gamma = \sum \mathbf{s}_j\cdot\mathbf{B}_j \end{equation*}\]

The circulation is a measure of how curly the field is. Like a Gaussian surface, an Ampèrian surface is purely a mathematical construction. It is not a physical object.

In figure b/2, the field is perpendicular to the edges on the ends, but parallel to the top and bottom edges. A dot product is zero when the vectors are perpendicular, so only the top and bottom edges contribute to \(\Gamma\). Let these edges have length \(s\). Since the field is constant along both of these edges, we don't actually have to break them into tiny parts; we can just have \(\mathbf{s}_1\) on the top edge, pointing to the left, and \(\mathbf{s}_2\) on the bottom edge, pointing to the right. The vector \(\mathbf{s}_1\) is in the same direction as the field \(\mathbf{B}_1\), and \(\mathbf{s}_2\) is in the same direction as \(\mathbf{B}_2\), so the dot products are simply equal to the products of the vectors' magnitudes. The resulting circulation is

\[\begin{align*} \Gamma &= |\mathbf{s}_1||\mathbf{B}_1|+|\mathbf{s}_2||\mathbf{B}_2| \\ &= \frac{2\pi k\eta s}{c^2}+\frac{2\pi k\eta s}{c^2} \\ &= \frac{4\pi k\eta s}{c^2} . \end{align*}\]

But \(\eta s\) is (current/length)(length), i.e., it is the amount of current that pierces the Ampèrian surface. We'll call this current \(I_{through}\). We have found one specific example of the general law of nature known as Ampère's law:

\[\begin{equation*} \Gamma = \frac{4\pi k}{c^2}\,I_{through} \end{equation*}\]

Positive and negative signs

Figures d/1 and d/2 show what happens to the circulation when we reverse the direction of the current \(I_{through}\). Reversing the current causes the magnetic field to reverse itself as well. The dot products occurring in the circulation are all negative in d/2, so the total circulation is now negative. To preserve Ampère's law, we need to define the current in d/2 as a negative number. In general, determine these plus and minus signs using the right-hand rule shown in the figure. As the fingers of your hand sweep around in the direction of the \(\mathbf{s}\) vectors, your thumb defines the direction of current which is positive. Choosing the direction of the thumb is like choosing which way to insert an ammeter in a circuit: on a digital meter, reversing the connections gives readings which are opposite in sign.

Example 13: A solenoid
\(\triangleright\) What is the field inside a long, straight solenoid of length \(\ell\) and radius \(a\), and having \(N\) loops of wire evenly wound along it, which carry a current \(I\) ?

\(\triangleright\) This is an interesting example, because it allows us to get a very good approximation to the field, but without some experimental input it wouldn't be obvious what approximation to use. Figure e/1 shows what we'd observe by measuring the field of a real solenoid. The field is nearly constant inside the tube, as long as we stay far away from the mouths. The field outside is much weaker. For the sake of an approximate calculation, we can idealize this field as shown in figure e/2. Of the edges of the Ampèrian surface shown in e/3, only AB contributes to the flux --- there is zero field along CD, and the field is perpendicular to edges BC and DA. Ampère's law gives

\[\begin{align*} \Gamma &= \frac{4\pi k}{ c^2}\, I_{ through} \\ (B)(\text{length of AB}) &= \frac{4\pi k}{ c^2}\,(\eta)(\text{length of AB})\\ B &= \frac{4\pi k\eta}{ c^2}\\ &= \frac{4\pi k NI}{ c^2 \ell}\\ \end{align*}\]


What direction is the current in figure e?

(answer in the back of the PDF version of the book)

Based on how \(\ell\) entered into the derivation in example 13, how should it be interpreted? Is it the total length of the wire?

(answer in the back of the PDF version of the book)

Surprisingly, we never needed to know the radius of the solenoid in example 13. Why is it physically plausible that the answer would be independent of the radius?

(answer in the back of the PDF version of the book)

Example 13 shows how much easier it can sometimes be to calculate a field using Ampère's law rather than the approaches developed previously in this chapter. However, if we hadn't already known something about the field, we wouldn't have been able to get started. In situations that lack symmetry, Ampère's law may make things harder, not easier. Anyhow, we will have no choice in nonstatic cases, where Ampère's law is true, and static equations like the Biot-Savart law are false.


f / A proof of Ampère's law.

11.3.2 A quick and dirty proof

Here's an informal sketch for a proof of Ampère's law, with no pretensions to rigor. Even if you don't care much for proofs, it would be a good idea to read it, because it will help to build your ability to visualize how Ampère's law works.

First we establish by a direct computation (homework problem 26) that Ampère's law holds for the geometry shown in figure f/1, a circular Ampèrian surface with a wire passing perpendicularly through its center. If we then alter the surface as in figure f/2, Ampère's law still works, because the straight segments, being perpendicular to the field, don't contribute to the circulation, and the new arc makes the same contribution to the circulation as the old one it replaced, because the weaker field is compensated for by the greater length of the arc. It is clear that by a series of such modifications, we could mold the surface into any shape, f/3.

Next we prove Ampère's law in the case shown in figure f/4: a small, square Ampèrian surface subject to the field of a distant square dipole. This part of the proof can be most easily accomplished by the methods of section 11.4. It should, for example, be plausible in the case illustrated here. The field on the left edge is stronger than the field on the right, so the overall contribution of these two edges to the circulation is slightly counterclockwise. However, the field is not quite perpendicular to the top and bottoms edges, so they both make small clockwise contributions. The clockwise and counterclockwise parts of the circulation end up canceling each other out. Once Ampère's law is established for a square surface like f/4, it follows that it is true for an irregular surface like f/5, since we can build such a shape out of squares, and the circulations are additive when we paste the surfaces together this way.

By pasting a square dipole onto the wire, f/6, like a flag attached to a flagpole, we can cancel out a segment of the wire's current and create a detour. Ampère's law is still true because, as shown in the last step, the square dipole makes zero contribution to the circulation. We can make as many detours as we like in this manner, thereby morphing the wire into an arbitrary shape like f/7.

What about a wire like f/8? It doesn't pierce the Ampèrian surface, so it doesn't add anything to \(I_{through}\), and we need to show that it likewise doesn't change the circulation. This wire, however, could be built by tiling the half-plane on its right with square dipoles, and we've already established that the field of a distant dipole doesn't contribute to the circulation. (Note that we couldn't have done this with a wire like f/7, because some of the dipoles would have been right on top of the Ampèrian surface.)

If Ampère's law holds for cases like f/7 and f/8, then it holds for any complex bundle of wires, including some that pass through the Ampèrian surface and some that don't. But we can build just about any static current distribution we like using such a bundle of wires, so it follows that Ampère's law is valid for any static current distribution.


g / Discussion question A.


h / Discussion question B.


i / Discussion question C.


j / Discussion question D.


k / Discussion question E.

11.3.3 Maxwell's equations for static fields

Static electric fields don't curl the way magnetic fields do, so we can state a version of Ampère's law for electric fields, which is that the circulation of the electric field is zero. Summarizing what we know so far about static fields, we have

\[\begin{align*} \Phi_E &= 4\pi kq_{in}\\ \Phi_B &= 0\\ \Gamma_E &= 0 \\ \Gamma_B &= \frac{4\pi k}{c^2}\,I_{through} .\\ \end{align*}\]

This set of equations is the static case of the more general relations known as Maxwell's equations. On the left side of each equation, we have information about a field. On the right is information about the field's sources.

It is vitally important to realize that these equations are only true for statics. They are incorrect if the distribution of charges or currents is changing over time. For example, we saw on page 602 that the changing magnetic field in an inductor gives rise to an electric field. Such an effect is completely inconsistent with the static version of Maxwell's equations; the equations don't even refer to time, so if the magnetic field is changing over time, they will not do anything special. The extension of Maxwell's equations to nonstatic fields is discussed in section 11.6.

Discussion Questions

Figure g/1 shows a wire with a circular Ampèrian surface drawn around its waist; in this situation, Ampère's law can be verified easily based on the equation for the field of a wire. In panel 2, a second wire has been added. Explain why it's plausible that Ampère's law still holds.

Figure h is like figure g, but now the second wire is perpendicular to the first, and lies in the plane of, and outside of, the Ampèrian surface. Carry out a similar analysis.

This discussion question is similar to questions A and B, but now the Ampèrian surface has been moved off center.

The left-hand wire has been nudged over a little. Analyze as before.

You know what to do.

11.4 Ampère's Law In Differential Form (Optional)


a / The div-meter, 1, and the curl-meter, 2 and 3.

11.4.1 The curl operator

The differential form of Gauss' law is more physically satisfying than the integral form, because it relates the charges that are present at some point to the properties of the electric field at the same point. Likewise, it would be more attractive to have a differential version of Ampère's law that would relate the currents to the magnetic field at a single point. intuitively, the divergence was based on the idea of the div-meter, a/1. The corresponding device for measuring the curliness of a field is the curl-meter, a/2. If the field is curly, then the torques on the charges will not cancel out, and the wheel will twist against the resistance of the spring. If your intuition tells you that the curlmeter will never do anything at all, then your intuition is doing a beautiful job on static fields; for nonstatic fields, however, it is perfectly possible to get a curly electric field.

Gauss' law in differential form relates \(\divg\mathbf{E}\), a scalar, to the charge density, another scalar. Ampère's law, however, deals with directions in space: if we reverse the directions of the currents, it makes a difference. We therefore expect that the differential form of Ampère's law will have vectors on both sides of the equal sign, and we should be thinking of the curl-meter's result as a vector. First we find the orientation of the curl-meter that gives the strongest torque, and then we define the direction of the curl vector using the right-hand rule shown in figure a/3.

To convert the div-meter concept to a mathematical definition, we found the infinitesimal flux, \(d\Phi\) through a tiny cubical Gaussian surface containing a volume \(dv\). By analogy, we imagine a tiny square Ampèrian surface with infinitesimal area \(d\mathbf{A}\). We assume this surface has been oriented in order to get the maximum circulation. The area vector \(d\mathbf{A}\) will then be in the same direction as the one defined in figure a/3. Ampère's law is

\[\begin{align*} d\Gamma &= \frac{4\pi k}{c^2}\,dI_{through} .\\ \text{We define a current density per unit area, $\mathbf{j}$, which is a vector pointing in the direction of the current and having magnitude $\mathbf{j}=dI/|d\mathbf{A}|$. In terms of this quantity, we have} d\Gamma &= \frac{4\pi k}{c^2}\,{j} |\mathbf{j}|\,|d\mathbf{A}| \\ \frac{d\Gamma}{|d\mathbf{A}|} &= \frac{4\pi k}{c^2}\, |\mathbf{j}| \\ \text{With this motivation, we define the magnitude of the curl as} |\curl\,\mathbf{B}| &= \frac{d\Gamma}{|d\mathbf{A}|} . \\ \text{Note that the curl, just like a derivative, has a differential divided by another differential. In terms of this definition, we find Amp To convert the div-meter concept to a mathematical definition, we found the infinitesimal flux, \(d\Phi\) through a tiny cubical Gaussian surface containing a volume \(dv\). By analogy, we imagine a tiny square Ampèrian surface with infinitesimal area \(d\mathbf{A}\). We assume this surface has been oriented in order to get the maximum circulation. The area vector \(d\mathbf{A}\) will then be in the same direction as the one defined in figure a/3. Ampère's law is {e}re's law in differential form:} \curl\,\mathbf{B} &= \frac{4\pi k}{c^2} \,\mathbf{j} \end{align*}\]

The complete set of Maxwell's equations in differential form is collected on page 920.


b / The coordinate system used in the following examples.


c / The field \(\hat{\mathbf{x}}\).


d / The field \(\hat{\mathbf{y}}\).


e / The field \(x\hat{\mathbf{y}}\).


f / The field \(- y\hat{\mathbf{x}}\).


g / Example 14.


h / Example 15.


i / A cyclic permutation of \(x\), \(y\), and \(z\).


j / Example 17.

11.4.2 Properties of the curl operator

The curl is a derivative.

As an example, let's calculate the curl of the field \(\hat{\mathbf{x}}\) shown in figure c. For our present purposes, it doesn't really matter whether this is an electric or a magnetic field; we're just getting out feet wet with the curl as a mathematical definition. Applying the definition of the curl directly, we construct an Ampèrian surface in the shape of an infinitesimally small square. Actually, since the field is uniform, it doesn't even matter very much whether we make the square finite or infinitesimal. The right and left edges don't contribute to the circulation, since the field is perpendicular to these edges. The top and bottom do contribute, but the top's contribution is clockwise, i.e., into the page according to the right-hand rule, while the bottom contributes an equal amount in the counterclockwise direction, which corresponds to an out-of-the-page contribution to the curl. They cancel, and the circulation is zero. We could also have determined this by imagining a curl-meter inserted in this field: the torques on it would have canceled out.

It makes sense that the curl of a constant field is zero, because the curl is a kind of derivative. The derivative of a constant is zero.

The curl is rotationally invariant.

Figure c looks just like figure c, but rotated by 90 degrees. Physically, we could be viewing the same field from a point of view that was rotated. Since the laws of physics are the same regardless of rotation, the curl must be zero here as well. In other words, the curl is rotationally invariant. If a certain field has a certain curl vector, then viewed from some other angle, we simply see the same field and the same curl vector, viewed from a different angle. A zero vector viewed from a different angle is still a zero vector.

As a less trivial example, let's compute the curl of the field \(\mathbf{F}=x\hat{\mathbf{y}}\) shown in figure e, at the point \((x=0,y=0)\). The circulation around a square of side \(s\) centered on the origin can be approximated by evaluating the field at the midpoints of its sides, \begin{alignat*}{4} x&=s/2 & y&=0 & F&=(s/2)\hat{y} & s_1\cdotF&=s^2/2
x&=0 & y&=s/2 & F&=0 & s_2\cdotF&=0
x&=-s/2 & y&=0 & F&=-(s/2)\hat{y} & s_3\cdotF&=s^2/2
x&=0 & y&=-s/2 & F&=0 & s_4\cdotF&=0 ,
\end{alignat*} which gives a circulation of \(s^2\), and a curl with a magnitude of \(s^2/\text{area}=s^2/s^2=1\). By the right-hand rule, the curl points out of the page, i.e., along the positive \(z\) axis, so we have

\[\begin{equation*} \curl\,x\hat{\mathbf{y}} = \hat{\mathbf{z}} . \end{equation*}\]

Now consider the field \(-y\hat{\mathbf{x}}\), shown in figure f. This is the same as the previous field, but with your book rotated by 90 degrees about the \(z\) axis. Rotating the result of the first calculation, \(\hat{\mathbf{z}}\), about the \(z\) axis doesn't change it, so the curl of this field is also \(\hat{\mathbf{z}}\).


When you're taking an ordinary derivative, you have the rule

\[\begin{equation*} \frac{d}{dx}[cf(x)] = c\frac{d}{dx}f(x) . \end{equation*}\]

In other words, multiplying a function by a constant results in a derivative that is multiplied by that constant. The curl is a kind of derivative operator, and the same is true for a curl.

Example 14: Multiplying the field by \(-1\).
\(\triangleright\) What is the curl of the field \(- x\hat{\mathbf{y}}\) at the origin?

\(\triangleright\) Using the scaling property just discussed, we can make this into a curl that we've already calculated:

\[\begin{align*} \curl\,(- x\hat{\mathbf{y}}) &= -\curl\,( x\hat{\mathbf{y}}) \\ &= -\hat{\mathbf{z}} \\ \end{align*}\]

This is in agreement with the right-hand rule.

The curl is additive.

We have only calculated each field's curl at the origin, but each of these fields actually has the same curl everywhere. In example 14, for instance, it is obvious that the curl is constant along any vertical line. But even if we move along the \(x\) axis, there is still an imbalance between the torques on the left and right sides of the curl-meter. More formally, suppose we start from the origin and move to the left by one unit. We find ourselves in a region where the field is very much as it was before, except that all the field vectors have had one unit worth of \(\hat{\mathbf{y}}\) added to them. But what do we get if we take the curl of \(- x\hat{\mathbf{y}}+\hat{\mathbf{y}}\)? The curl, like any god-fearing derivative operation, has the additive property

\[\begin{equation*} \curl(\mathbf{F}+\mathbf{G}) = \curl\,\mathbf{F}+\curl\,\mathbf{G} , \end{equation*}\]


\[\begin{equation*} \curl(- x\hat{\mathbf{y}}+\hat{\mathbf{y}}) = \curl(- x\hat{\mathbf{y}})+\curl(\hat{\mathbf{y}}) . \end{equation*}\]

But the second term is zero, so we get the same result as at the origin.

Example 15: A field that goes in a circle
\(\triangleright\) What is the curl of the field \(x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\)?

\(\triangleright\) Using the linearity of the curl, and recognizing each of the terms as one whose curl we have already computed, we find that this field's curl is a constant \(2\hat{\mathbf{z}}\). This agrees with the right-hand rule.

Example 16: The field inside a long, straight wire

\(\triangleright\) What is the magnetic field inside a long, straight wire in which the current density is \(j\)?

\(\triangleright\) Let the wire be along the \(z\) axis, so \(\mathbf{j}= j\hat{\mathbf{z}}\). Ampère's law gives

\[\begin{equation*} \curl\,\mathbf{B}= \frac{4\pi k}{ c^2} \, j\hat{\mathbf{z}} . \end{equation*}\]

In other words, we need a magnetic field whose curl is a constant. We've encountered several fields with constant curls, but the only one that has the same symmetry as the cylindrical wire is \(x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\), so the answer must be this field or some constant multiplied by it,

\[\begin{equation*} \mathbf{B}= b\left( x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\right) . \end{equation*}\]

The curl of this field is \(2 b\hat{\mathbf{z}}\), so

\[\begin{align*} 2 b &= \frac{4\pi k}{ c^2} \, j ,\\ \text{and thus} \mathbf{B}&= \frac{2\pi k}{ c^2} \, j\left( x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\right) .\\ \end{align*}\]

The curl in component form

Now consider the field

\[\begin{align*} F_x &= ax+by+c \\ F_y &= dx+ey+f ,\\ \text{i.e.,} \mathbf{F} &= ax\hat{\mathbf{x}}+by\hat{\mathbf{x}}+c\hat{\mathbf{x}}+dx\hat{\mathbf{y}}+ey\hat{\mathbf{y}}+f\hat{\mathbf{y}} . \end{align*}\]

The only terms whose curls we haven't yet explicitly computed are the \(a\), \(e\), and \(f\) terms, and their curls turn out to be zero (homework problem 50). Only the \(b\) and \(d\) terms have nonvanishing curls. The curl of this field is

\[\begin{align*} \curl\,\mathbf{F} &= \curl(by\hat{\mathbf{x}})+\curl(dx\hat{\mathbf{y}}) \ &= b\,\curl(y\hat{\mathbf{x}})+d\,\curl(x\hat{\mathbf{y}}) \text{[scaling]}\ &= b(-\hat{\mathbf{z}})+d(\hat{\mathbf{z}}) \text{[found previously]}\ &= (d-b)\hat{\mathbf{z}} .\ \end{align*}\]

But any field in the \(x-y\) plane can be approximated with this type of field, as long as we only need to get a good approximation within a small region. The infinitesimal Ampèrian surface occurring in the definition of the curl is tiny enough to fit in a pretty small region, so we can get away with this here. The \(d\) and \(b\) coefficients can then be associated with the partial derivatives \(\partial F_y/\partial x\) and \(\partial F_x/\partial y\). We therefore have

\[\begin{equation*} \curl\,\mathbf{F} = \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}} \ \end{equation*}\]

for any field in the \(x-y\) plane. In three dimensions, we just need to generate two more equations like this by doing a cyclic permutation of the variables \(x\), \(y\), and \(z\):

\[\begin{align*} (\curl\,\mathbf{F})_x &= \frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z} \ (\curl\,\mathbf{F})_y &= \frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x} \ (\curl\,\mathbf{F})_z &= \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} \ \end{align*}\]

Example 17: A sine wave
\(\triangleright\) Find the curl of the following electric field
\[\begin{equation*} \mathbf{E} = (\text{sin}\, x)\hat{\mathbf{y}} , \end{equation*}\]
and interpret the result.

\(\triangleright\) The only nonvanishing partial derivative occurring in this curl is

\[\begin{align*} (\curl\,\mathbf{E})_z &= \frac{\partial E_{y}}{\partial x} \ &= \text{cos}\, x ,\ \text{so} \curl\,\mathbf{E} &= \text{cos}\,\hat{\mathbf{z}} \end{align*}\]

This is visually reasonable: the curl-meter would spin if we put its wheel in the plane of the page, with its axle poking out along the \(z\) axis. In some areas it would spin clockwise, in others counterclockwise, and this makes sense, because the cosine is positive in some placed and negative in others.

This is a perfectly reasonable field pattern: it the electric field pattern of a light wave! But Ampère's law for electric fields says the curl of E is supposed to be zero. What's going on? What's wrong is that we can't assume the static version of Ampère's law. All we've really proved is that this pattern is impossible as a static field: we can't have a light wave that stands still.

Figure k is a summary of the vector calculus presented in the optional sections of this book. The first column shows that one function is a related to another by a kind of differentiation. The second column states the fundamental theorem of calculus, which says that if you integrate the derivative over the interior of a region, you get some information about the original function at the boundary of that region.


k / A summary of the derivative, gradient, curl, and divergence.

11.5 Induced Electric Fields


a / Faraday on a British banknote.


b / Faraday's experiment, simplified and shown with modern equipment.


c / Detail from Ascending and Descending, M.C. Escher, 1960.


d / The relationship between the change in the magnetic field, and the electric field it produces.


e / The electric circulation is the sum of the voltmeter readings.


f / A generator.


g / A transformer.

11.5.1 Faraday's experiment

Nature is simple, but the simplicity may not become evident until a hundred years after the discovery of some new piece of physics. We've already seen, on page 602, that the time-varying magnetic field in an inductor causes an electric field. This electric field is not created by charges. That argument, however, only seems clear with hindsight. The discovery of this phenomenon of induced electric fields --- fields that are not due to charges --- was a purely experimental accomplishment by Michael Faraday (1791-1867), the son of a blacksmith who had to struggle against the rigid class structure of 19th century England. Faraday, working in 1831, had only a vague and general idea that electricity and magnetism were related to each other, based on Oersted's demonstration, a decade before, that magnetic fields were caused by electric currents.

Figure b is a simplified drawing of the following experiment, as described in Faraday's original paper: “Two hundred and three feet of copper wire ... were passed round a large block of wood; [another] two hundred and three feet of similar wire were interposed as a spiral between the turns of the first, and metallic contact everywhere prevented by twine [insulation]. One of these [coils] was connected with a galvanometer [voltmeter], and the other with a battery... When the contact was made, there was a sudden and very slight effect at the galvanometer, and there was also a similar slight effect when the contact with the battery was broken. But whilst the ... current was continuing to pass through the one [coil], no ... effect ... upon the other [coil] could be perceived, although the active power of the battery was proved to be great, by its heating the whole of its own coil [through ordinary resistive heating] ...”

From Faraday's notes and publications, it appears that the situation in figure b/3 was a surprise to him, and he probably thought it would be a surprise to his readers, as well. That's why he offered evidence that the current was still flowing: to show that the battery hadn't just died. The induction effect occurred during the short time it took for the black coil's magnetic field to be established, b/2. Even more counterintuitively, we get an effect, equally strong but in the opposite direction, when the circuit is broken, b/4. The effect occurs only when the magnetic field is changing, and it appears to be proportional to the derivative \(\partial\mathbf{B}/\partial\mathbf{t}\), which is in one direction when the field is being established, and in the opposite direction when it collapses.

The effect is proportional to \(\partial\mathbf{B}/\partial\mathbf{t}\), but what is the effect? A voltmeter is nothing more than a resistor with an attachment for measuring the current through it. A current will not flow through a resistor unless there is some electric field pushing the electrons, so we conclude that the changing magnetic field has produced an electric field in the surrounding space. Since the white wire is not a perfect conductor, there must be electric fields in it as well. The remarkable thing about the circuit formed by the white wire is that as the electrons travel around and around, they are always being pushed forward by electric fields. This violates the loop rule, which says that when an electron makes a round trip, there is supposed to be just as much “uphill” (moving against the electric field) as “downhill” (moving with it). That's OK. The loop rule is only true for statics. Faraday's experiments show that an electron really can go around and around, and always be going “downhill,” as in the famous drawing by M.C. Escher shown in figure c. That's just what happens when you have a curly field.

When a field is curly, we can measure its curliness using a circulation. Unlike the magnetic circulation \(\Gamma_B\), the electric circulation \(\Gamma_E\) is something we can measure directly using ordinary tools. A circulation is defined by breaking up a loop into tiny segments, \(d\mathbf{s}\), and adding up the dot products of these distance vectors with the field. But when we multiply electric field by distance, what we get is an indication of the amount of work per unit charge done on a test charge that has been moved through that distance. The work per unit charge has units of volts, and it can be measured using a voltmeter, as shown in figure e, where \(\Gamma_E\) equals the sum of the voltmeter readings. Since the electric circulation is directly measurable, most people who work with circuits are more familiar with it than they are with the magnetic circulation. They usually refer to \(\Gamma_E\) using the synonym “emf,” which stands for “electromotive force,” and notate it as \(\mathcal{E}\). (This is an unfortunate piece of terminology, because its units are really volts, not newtons.) The term emf can also be used when the path is not a closed loop.

Faraday's experiment demonstrates a new relationship

\[\begin{equation*} \Gamma_E \propto -\frac{\partial B}{\partial t} , \end{equation*}\]

where the negative sign is a way of showing the observed left-handed relationship, d. This is similar to the structure of of Ampère's law:

\[\begin{equation*} \Gamma_B \propto I_{through} , \end{equation*}\]

which also relates the curliness of a field to something that is going on nearby (a current, in this case).

It's important to note that even though the emf, \(\Gamma_E\), has units of volts, it isn't a voltage. A voltage is a measure of the electrical energy a charge has when it is at a certain point in space. The curly nature of nonstatic fields means that this whole concept becomes nonsense. In a curly field, suppose one electron stays at home while its friend goes for a drive around the block. When they are reunited, the one that went around the block has picked up some kinetic energy, while the one who stayed at home hasn't. We simply can't define an electrical energy \(U_e=qV\) so that \(U_e+K\) stays the same for each electron. No voltage pattern, \(V\), can do this, because then it would predict the same kinetic energies for the two electrons, which is incorrect. When we're dealing with nonstatic fields, we need to think of the electrical energy in terms of the energy density of the fields themselves.

It might sound as though an electron could get a free lunch by circling around and around in a curly electric field, resulting in a violation of conservation of energy. The following examples, in addition to their practical interest, both show that energy is in fact conserved.

Example 18: The generator

A basic generator, f, consists of a permanent magnet that rotates within a coil of wire. The magnet is turned by a motor or crank, (not shown). As it spins, the nearby magnetic field changes. This changing magnetic field results in an electric field, which has a curly pattern. This electric field pattern creates a current that whips around the coils of wire, and we can tap this current to light the lightbulb.

If the magnet was on a frictionless bearing, could we light the bulb for free indefinitely, thus violating conservation of energy? No. Mechanical work has to be done to crank the magnet, and that's where the energy comes from. If we break the light-bulb circuit, it suddenly gets easier to crank the magnet! This is because the current in the coil sets up its own magnetic field, and that field exerts a torque on the magnet. If we stopped cranking, this torque would quickly make the magnet stop turning.


When you're driving your car, the engine recharges the battery continuously using a device called an alternator, which is really just a generator. Why can't you use the alternator to start the engine if your car's battery is dead?

(answer in the back of the PDF version of the book)
Example 19: The transformer

In example 18 on page 544, we discussed the advantages of transmitting power over electrical lines using high voltages and low currents. However, we don't want our wall sockets to operate at 10000 volts! For this reason, the electric company uses a device called a transformer, g, to convert everything to lower voltages and higher currents inside your house. The coil on the input side creates a magnetic field. Transformers work with alternating current, so the magnetic field surrounding the input coil is always changing. This induces an electric field, which drives a current around the output coil.

Since the electric field is curly, an electron can keep gaining more and more energy by circling through it again and again. Thus the output voltage can be controlled by changing the number of coils of wire on the output side. Changing the number of coils on the input side also has an effect (homework problem 33).

In any case, conservation of energy guarantees that the amount of power on the output side must equal the amount put in originally, \(I_{in}V_{in} = I_{out}V_{out}\), so no matter what factor the voltage is reduced by, the current is increased by the same factor.

Discussion Questions

Suppose the bar magnet in figure f on page 691 has a magnetic field pattern that emerges from its top, circling around and coming back in the bottom. This field is created by electrons orbiting atoms inside the magnet. Are these atomic currents clockwise or counterclockwise as seen from above? In what direction is the current flowing in the circuit?

We have a circling atomic current inside the circling current in the wires. When we have two circling currents like this, they will make torques on each other that will tend to align them in a certain way. Since currents in the same direction attract one another, which way is the torque made by the wires on the bar magnet? Verify that due to this torque, mechanical work has to be done in order to crank the generator.


h / It doesn't matter whether it's the coil or the permanent magnet that spins. Either way, we get a functioning generator.


i / A generator that works with linear motion.

11.5.2 Why induction?

Faraday's results leave us in the dark about several things:

We can get some guidance from the example of a car's alternator (which just means generator), referred to in the self-check on page 691. To keep things conceptually simple, I carefully avoided mentioning that in a real car's alternator, it isn't actually the permanent magnet that spins. The coil is what spins. The choice of design h/1 or h/2 is merely a matter of engineering convenience, not physics. All that matters is the relative motion of the two objects.

This is highly suggestive. As discussed at the beginning of this chapter, magnetism is a relativistic effect. From arguments about relative motion, we concluded that moving electric charges create magnetic fields. Now perhaps we can use reasoning with the same flavor to show that changing magnetic fields produce curly electric fields. Note that figure h/2 doesn't even require induction. The protons and electrons in the coil are moving through a magnetic field, so they experience forces. The protons can't flow, because the coil is a solid substance, but the electrons can, so a current is induced.7

Now if we're convinced that figure h/2 produces a current in the coil, then it seems very plausible that the same will happen in figure h/1, which implies the existence of induction effects. But this example involves circular motion, so it doesn't quite work as a way of proving that induction exists. When we say that motion is relative, we only mean straight-line motion, not circular motion.

A more ironclad relativistic argument comes from the arrangement shown in figure i. This is also a generator --- one that is impractical, but much easier to understand.

Flea 1 doesn't believe in this modern foolishness about induction. She's sitting on the bar magnet, which to her is obviously at rest. As the square wire loop is dragged away from her and the magnet, its protons experience a force out of the page, because the cross product \(\mathbf{F}=q\mathbf{v}\times\mathbf{B}\) is out of the page. The electrons, which are negatively charged, feel a force into the page. The conduction electrons are free to move, but the protons aren't. In the front and back sides of the loop, this force is perpendicular to the wire. In the right and left sides, however, the electrons are free to respond to the force. Note that the magnetic field is weaker on the right side. It's as though we had two pumps in a loop of pipe, with the weaker pump trying to push in the opposite direction; the weaker pump loses the argument.8 We get a current that circulates around the loop.9 There is no induction going on in this frame of reference; the forces that cause the current are just the ordinary magnetic forces experienced by any charged particle moving through a magnetic field.

Flea 2 is sitting on the loop, which she considers to be at rest. In her frame of reference, it's the bar magnet that is moving. Like flea 1, she observes a current circulating around the loop, but unlike flea 1, she cannot use magnetic forces to explain this current. As far as she is concerned, the electrons were initially at rest. Magnetic forces are forces between moving charges and other moving charges, so a magnetic field can never accelerate a charged particle starting from rest. A force that accelerates a charge from rest can only be an electric force, so she is forced to conclude that there is an electric field in her region of space. This field drives electrons around and around in circles, so it is apparently violating the loop rule --- it is a curly field. What reason can flea 2 offer for the existence of this electric field pattern? Well, she's been noticing that the magnetic field in her region of space has been changing, possibly because that bar magnet over there has been getting farther away. She observes that a changing magnetic field creates a curly electric field.

We therefore conclude that induction effects must exist based on the fact that motion is relative. If we didn't want to admit induction effects, we would have to outlaw flea 2's frame of reference, but the whole idea of relative motion is that all frames of reference are created equal, and there is no way to determine which one is really at rest.

This whole line of reasoning was not available to Faraday and his contemporaries, since they thought the relative nature of motion only applied to matter, not to electric and magnetic fields.10 But with the advantage of modern hindsight, we can understand in fundamental terms the facts that Faraday had to take simply as mysterious experimental observations. For example, the geometric relationship shown in figure d follows directly from the direction of the current we deduced in the story of the two fleas.


j / A new version of figure i with a tiny loop. The point of view is above the plane of the loop. In the frame of reference where the magnetic field is constant, the loop is moving to the right.


k / Example 21.


l / Example 22.

11.5.3 Faraday's law

We can also answer the other questions posed on page 692. The divide-and-conquer approach should be familiar by now. We first determine the circulation \(\Gamma_E\) in the case where the wire loop is very tiny, j. Then we can break down any big loop into a grid of small ones; we've already seen that when we make this kind of grid, the circulations add together. Although we'll continue to talk about a physical loop of wire, as in figure i, the tiny loop can really be just like the edges of an Ampèrian surface: a mathematical construct that doesn't necessarily correspond to a real object.

In the close-up view shown in figure j, the field looks simpler. Just as a tiny part of a curve looks straight, a tiny part of this magnetic field looks like the field vectors are just getting shorter by the same amount with each step to the right. Writing \(dx\) for the width of the loop, we therefore have

\[\begin{equation*} B(x+dx)-B(x)=\frac{\partial B}{\partial x}\,dx \end{equation*}\]

for the difference in the strength of the field between the left and right sides. In the frame of reference where the loop is moving, a charge \(q\) moving along with the loop at velocity \(v\) will experience a magnetic force \(\mathbf{F}_B=qvB\hat{\mathbf{y}}\). In the frame moving along with the loop, this is interpreted as an electrical force, \(\mathbf{F}_E=qE\hat{\mathbf{y}}\). Observers in the two frames agree on how much force there is, so in the loop's frame, we have an electric field \(\mathbf{E}=vB\hat{\mathbf{y}}\). This field is perpendicular to the front and back sides of the loop, BC and DA, so there is no contribution to the circulation along these sides, but there is a counterclockwise contribution to the circulation on CD, and smaller clockwise one on AB. The result is a circulation that is counterclockwise, and has an absolute value

\[\begin{align*} |\Gamma_E| &= |E(x)dy - E(x+dx)dy|\\ &= |v[B(x)-B(x+dx)]|dy\\ &= \left|v\,\frac{\partial B}{\partial x}\right|\,dxdy \\ &= \left|\frac{dx}{dt}\,\frac{\partial B}{\partial x}\right|\,dxdy \\ &= \left|\frac{\partial B}{\partial t}\right|\,dA . \end{align*}\]

Using a right-hand rule, the counterclockwise circulation is represented by pointing one's thumb up, but the vector \(\partial \mathbf{B}/\partial t\) is down. This is just a rephrasing of the geometric relationship shown in figure d on page 690. We can represent the opposing directions using a minus sign,

\[\begin{equation*} \Gamma_E = -\frac{\partial B}{\partial t}\,dA . \end{equation*}\]

Although this derivation was carried out with everything aligned in a specific way along the coordinate axes, it turns out that this relationship can be generalized as a vector dot product,

\[\begin{equation*} \Gamma_E = -\frac{\partial \mathbf{B}}{\partial t}\cdotd\mathbf{A} . \end{equation*}\]

Finally, we can take a finite-sized loop and break down the circulation around its edges into a grid of tiny loops. The circulations add, so we have

\[\begin{equation*} \Gamma_E = -\sum \frac{\partial \mathbf{B}_j}{\partial t}\cdotd\mathbf{A}_j . \end{equation*}\]

This is known as Faraday's law. (I don't recommend memorizing all these names.) Mathematically, Faraday's law is very similar to the structure of Ampère's law: the circulation of a field around the edges of a surface is equal to the sum of something that points through the

If the loop itself isn't moving, twisting, or changing shape, then the area vectors don't change over time, and we can move the derivative outside the sum, and rewrite Faraday's law in a slightly more transparent form:

\[\begin{align*} \Gamma_E &= -\frac{\partial}{\partial t}\sum \mathbf{B}_j\cdotd\mathbf{A}_j \\ &= -\frac{\partial\Phi_B}{\partial t} \end{align*}\]

A changing magnetic flux makes a curly electric field. You might think based on Gauss' law for magnetic fields that \(\Phi_B\) would be identically zero. However, Gauss' law only applies to surfaces that are closed, i.e., have no edges.


Check that the units in Faraday's law work out. An easy way to approach this is to use the fact that \( vB\) has the same units as \(E\), which can be seen by comparing the equations for magnetic and electric forces used above.

(answer in the back of the PDF version of the book)
Example 20: A pathetic generator

\(\triangleright\) The horizontal component of the earth's magnetic field varies from zero, at a magnetic pole, to about \(10^{-4}\) T near the equator. Since the distance from the equator to a pole is about \(10^7\) m, we can estimate, very roughly, that the horizontal component of the earth's magnetic field typically varies by about \(10^{-11}\) T/m as you go north or south. Suppose you connect the terminals of a one-ohm lightbulb to each other with a loop of wire having an area of 1 \(\text{m}^2\). Holding the loop so that it lies in the east-west-up-down plane, you run straight north at a speed of 10 m/s, how much current will flow? Next, repeat the same calculation for the surface of a neutron star. The magnetic field on a neutron star is typically \(10^9\) T, and the radius of an average neutron star is about \(10^4\) m.

\(\triangleright\) Let's work in the frame of reference of the running person. In this frame of reference, the earth is moving, and therefore the local magnetic field is changing in strength by \(10^{-9}\) T/s. This rate of change is almost exactly the same throughout the interior of the loop, so we can dispense with the summation, and simply write Faraday's law as

\[\begin{equation*} \Gamma_E = -\frac{\partial\mathbf{B}}{\partial t}\cdot\mathbf{A} . \end{equation*}\]

Since what we estimated was the rate of change of the horizontal component, and the vector \(\mathbf{A}\) is horizontal (perpendicular to the loop), we can find this dot product simply by multiplying the two numbers:

\[\begin{align*} \Gamma_E &= (10^{-9}\ \text{T}/\text{s})(1\ \text{m}^2) \\ &= 10^{-9}\ \text{T}\!\cdot\!\text{m}^2/\text{s}\\ &= 10^{-9}\ \text{V}\\ \end{align*}\]

This is certainly not enough to light the bulb, and would not even be easy to measure using the most sensitive laboratory instruments.

Now what about the neutron star? We'll pretend you're tough enough that its gravity doesn't instantly crush you. The spatial variation of the magnetic field is on the order of \((10^9\ \text{T}/10^4\ \text{m})=10^5\ \text{T}/\text{m}\). If you can run north at the same speed of 10 m/s, then in your frame of reference there is a temporal (time) variation of about \(10^6\) T/s, and a calculation similar to the previous one results in an emf of \(10^6\) V! This isn't just strong enough to light the bulb, it's sufficient to evaporate it, and kill you as well!

It might seem as though having access to a region of rapidly changing magnetic field would therefore give us an infinite supply of free energy. However, the energy that lights the bulb is actually coming from the mechanical work you do by running through the field. A tremendous force would be required to make the wire loop move through the neutron star's field at any significant speed.

Example 21: Speed and power in a generator
\(\triangleright\) Figure k shows three graphs of the magnetic flux through a generator's coils as a function of time. In graph 2, the generator is being cranked at twice the frequency. In 3, a permanent magnet with double the strength has been used. In 4, the generator is being cranked in the opposite direction. Compare the power generated in figures 2-4 with the the original case, 1.

\(\triangleright\) If the flux varies as \(\Phi=A\sin\omega t\), then the time derivative occurring in Faraday's law is \(\partial\Phi/\partial t=A\omega\cos\omega t\). The absolute value of this is the same as the absolute value of the emf, \(\Gamma_E\). The current through the lightbulb is proportional to this emf, and the power dissipated depends on the square of the current (\(P=I^2R\)), so \(P\propto A^2\omega^2\). Figures 2 and 3 both give four times the output power (and require four times the input power). Figure 4 gives the same result as figure 1; we can think of this as a negative amplitude, which gives the same result when squared.

Example 22: An approximate loop rule
Figure l/1 shows a simple RL circuit of the type discussed in the last chapter. A current has already been established in the coil, let's say by a battery. The battery was then unclipped from the coil, and we now see the circuit as the magnetic field in and around the inductor is beginning to collapse. I've already cautioned you that the loop rule doesn't apply in nonstatic situations, so we can't assume that the readings on the four voltmeters add up to zero. The interesting thing is that although they don't add up to exactly zero in this circuit, they very nearly do. Why is the loop rule even approximately valid in this situation?

The reason is that the voltmeters are measuring the emf \(\Gamma_E\) around the path shown in figure l/2, and the stray field of the solenoid is extremely weak out there. In the region where the meters are, the arrows representing the magnetic field would be too small to allow me to draw them to scale, so I have simply omitted them. Since the field is so weak in this region, the flux through the loop is nearly zero, and the rate of change of the flux, \(\partial\Phi_B/\partial t\), is also nearly zero. By Faraday's law, then, the emf around this loop is nearly zero.

Now consider figure l/3. The flux through the interior of this path is not zero, because the strong part of the field passes through it, and not just once but many times. To visualize this, imagine that we make a wire frame in this shape, dip it in a tank of soapy water, and pull it out, so that there is a soap-bubble film spanning its interior. Faraday's law refers to the rate of change of the flux through a surface such as this one. (The soap film tends to assume a certain special shape which results in the minimum possible surface area, but Faraday's law would be true for any surface that filled in the loop.) In the coiled part of the wire, the soap makes a three-dimensional screw shape, like the shape you would get if you took the steps of a spiral staircase and smoothed them into a ramp. The loop rule is going to be strongly violated for this path.

We can interpret this as follows. Since the wire in the solenoid has a very low resistance compared to the resistances of the light bulbs, we can expect that the electric field along the corkscrew part of loop l/3 will be very small. As an electron passes through the coil, the work done on it is therefore essentially zero, and the true emf along the coil is zero. In figure l/1, the meter on top is therefore not telling us the actual emf experienced by an electron that passes through the coil. It is telling us the emf experienced by an electron that passes through the meter itself, which is a different quantity entirely. The other three meters, however, really do tell us the emf through the bulbs, since there are no magnetic fields where they are, and therefore no funny induction effects.

11.6 Maxwell's Equations


a / James Clerk Maxwell (1831-1879)


b / Where is the moving charge responsible for this magnetic field?


c / An Ampèrian surface superimposed on the landscape.


d / An electron jumps through a hoop.


e / An alternative Ampèrian surface.

11.6.1 Induced magnetic fields

We are almost, but not quite, done figuring out the complete set of physical laws, called Maxwell's equations, governing electricity and magnetism. We are only missing one more term. For clarity, I'll state Maxwell's equations with the missing part included, and then discuss the physical motivation and experimental evidence for sticking it in:

Maxwell's equations

For any closed surface, the fluxes through the surface are

\[\begin{align*} \Phi_E &= 4\pi kq_{in} \text{and}\\ \Phi_B &= 0 .\\ \text{For any surface that is not closed, the circulations around the edges of the surface are given by} \Gamma_E &= -\frac{\partial\Phi_B}{\partial t} \text{and} \\ c^2\Gamma_B &= \frac{\partial\Phi_E}{\partial t} + 4\pi k I_{through} . \end{align*}\]

The \(\Phi_E\) equation is Gauss' law: charges make diverging electric fields. The corresponding equation for \(\Phi_B\) tells us that magnetic “charges” (monopoles) don't exist, so magnetic fields never diverge. The third equation says that changing magnetic fields induce curly electric fields, whose curliness we can measure using the emf, \(\Gamma_E\), around a closed loop. The final equation, for \(\Gamma_B\), is the only one where anything new has been added. Without the new time derivative term, this equation would simply be Ampère's law. (I've chosen to move the \(c^2\) over to the left because it simplifies the writing, and also because it more clearly demonstrates the analogous roles played by charges and currents in the \(\Phi_E\) and \(\Gamma_B\) equations.)

This new \(\partial\Phi_E/\partial t\) term says that just as a changing magnetic field can induce a curly electric field, a changing electric field can induce a curly magnetic field. Why should this be so? The following examples show that Maxwell's equations would not make sense in general without it.

Figure b shows a mysterious curly magnetic field. Magnetic fields are supposed to be made by moving charges, but there don't seem to be any moving charges in this landscape. Where are they? One reasonable guess would be that they're behind your head, where you can't see them. Suppose there's a positively charged particle about to hit you in the back of the head. This particle is like a current going into the page. We're used to dealing with currents made by many charged particles, but logically we can't have some minimum number that would qualify as a current. This is not a static current, however, because the current at a given point in space is not staying the same over time. If the particle is pointlike, then it takes zero time to pass any particular location, and the current is then infinite at that point in space. A moment later, when the particle is passing by some other location, there will be an infinite current there, and zero current in the previous location. If this single particle qualifies as a current, then it should be surrounded by a curly magnetic field, just like any other current.11

This explanation is simple and reasonable, but how do we know it's correct? Well, it makes another prediction, which is that the positively charged particle should be making an electric field as well. Not only that, but if it's headed for the back of your head, then it's getting closer and closer, so the electric field should be getting stronger over time. But this is exactly what Maxwell's equations require. There is no current \(I_{through}\) piercing the Ampèrian surface shown in figure c, so Maxwell's equation for \(\Gamma_B\) becomes \(c^2\Gamma_B = \partial\Phi_E/\partial t\). The only reason for an electric field to change is if there are charged particles making it, and those charged particles are moving. When charged particles are moving, they make magnetic fields as well.

Note that the above example is also sufficient to prove the positive sign of the \(\partial\Phi_E/\partial t\) term in Maxwell's equations, which is different from the negative sign of Faraday's \(-\partial\Phi_B/\partial t\) term.

The addition of the \(\partial\Phi_E/\partial t\) term has an even deeper and more important physical meaning. With the inclusion of this term, Maxwell's equations can describe correctly the way in which disturbances in the electric and magnetic fields ripple outwards at the speed of light. Indeed, Maxwell was the first human to understand that light was in fact an electromagnetic wave. Legend has it that it was on a starry night that he first realized this implication of his equations. He went for a walk with his wife, and told her she was the only other person in the world who really knew what starlight was.

To see how the \(\partial\Phi_E/\partial t\) term relates to electromagnetic waves, let's look at an example where we would get nonsense without it. Figure d shows an electron that sits just on one side of an imaginary Ampèrian surface, and then hops through it at some randomly chosen moment. Unadorned with the \(\partial\Phi_E/\partial t\) term, Maxwell's equation for \(\Gamma_B\) reads as \(c^2\Gamma_B = 4\pi k I_{through}\), which is Ampère's law. If the electron is a pointlike particle, then we have an infinite current \(I_{through}\) at the moment when it pierces the imaginary surface, and zero current at all other times. An infinite magnetic circulation \(\Gamma_B\) can only be produced by an infinite magnetic field, so without the \(\partial\Phi_E/\partial t\) term, Maxwell's equations predict nonsense: the edge of the surface would experience an infinite magnetic field at one instant, and zero magnetic field at all other times. Even if the infinity didn't upset us, it doesn't make sense that anything special would happen at the moment the electron passed through the surface, because the surface is an imaginary mathematical construct. We could just as well have chosen the curved surface shown in figure e, which the electron never crosses at all. We are already clearly getting nonsensical results by omitting the \(\partial\Phi_E/\partial t\) term, and this shouldn't surprise us because Ampère's law only applies to statics. More to the point, Ampère's law doesn't have time in it, so it predicts that this effect is instantaneous. According to Ampère's law, we could send Morse code signals by wiggling the electron back and forth, and these signals would be received at distant locations instantly, without any time delay at all. This contradicts the theory of relativity, one of whose predictions is that information cannot be transmitted at speeds greater than the speed of light.

Discussion Questions

Induced magnetic fields were introduced in the text via the imaginary landscape shown in figure b on page 699, and I argued that the magnetic field could have been produced by a positive charge coming from behind your head. This is a specific assumption about the number of charges (one), the direction of motion, and the sign of the charge. What are some other scenarios that could explain this field?


f / A magnetic field in the form of a sine wave.


g / The wave pattern is curly. For example, the circulation around this reactangle is nonzero and counterclockwise.


i / An impossible wave pattern.


j / Example 23. The incident and reflected waves are drawn offset from each other for clarity, but are actually on top of each other so that their fields superpose.


k / Red and blue light travel at the same speed.


l / Bright and dim light travel at the same speed.


m / A nonsinusoidal wave.


n / The magnetic field of the wave. The electric field, not shown, is perpendicular to the page.


p / A classical calculation of the momentum of a light wave. An antenna of length \(\ell\) is bathed in an electromagnetic wave. The black arrows represent the electric field, the white circles the magnetic field coming out of the page. The wave is traveling to the right.


q / A simplified drawing of the 1903 experiment by Nichols and Hull that verified the predicted momentum of light waves. Two circular mirrors were hung from a fine quartz fiber, inside an evacuated bell jar. A 150 mW beam of light was shone on one of the mirrors for 6 s, producing a tiny rotation, which was measurable by an optical lever (not shown). The force was within 0.6% of the theoretically predicted value (problem 11 on p. 441) of \(0.001\ \mu\text{N}\). For comparison, a short clipping of a single human hair weighs \(\sim 1\ \mu\text{N}\).


Discussion question A.


Discussion question B.


Discussion questions C and D.

11.6.2 Light waves

We could indeed send signals using this scheme, and the signals would be a form of light. A radio transmitting antenna, for instance, is simply a device for whipping electrons back and forth at megahertz frequencies. Radio waves are just like visible light, but with a lower frequency. With the addition of the \(\partial\Phi_E/\partial t\) term, Maxwell's equations are capable of describing electromagnetic waves. It would be possible to use Maxwell's equations to calculate the pattern of the electric and magnetic fields rippling outward from a single electron that fidgets at irregular intervals, but let's pick a simpler example to analyze.

The simplest wave pattern is a sine wave like the one shown in figure f. Let's assume a magnetic field of this form, and see what Maxwell's equations tell us about it. If the wave is traveling through empty space, then there are no charges or currents present, and Maxwell's equations become

\[\begin{align*} \Phi_E &= 0 \\ \Phi_B &= 0 \\ \Gamma_E &= -\frac{\partial\Phi_B}{\partial t} \\ c^2\Gamma_B &= \frac{\partial\Phi_E}{\partial t} . \end{align*}\]

The equation \(\Phi=0\) has already been verified for this type of wave pattern in example 36 on page 634. Even if you haven't learned the techniques from that section, it should be visually plausible that this field pattern doesn't diverge or converge on any particular point.

Geometry of the electric and magnetic fields

The equation \(c^2\Gamma_B=\partial\Phi_E/\partial t\) tells us that there can be no such thing as a purely magnetic wave. The wave pattern clearly does have a nonvanishing circulation around the edge of the surface suggested in figure g, so there must be an electric flux through the surface. This magnetic field pattern must be intertwined with an electric field pattern that fills the same space. There is also no way that the two sides of the equation could stay synchronized with each other unless the electric field pattern is also a sine wave, and one that has the same wavelength, frequency, and velocity. Since the electric field is making a flux through the indicated surface, it's plausible that the electric field vectors lie in a plane perpendicular to that of the magnetic field vectors. The resulting geometry is shown in figure h. Further justification for this geometry is given later in this subsection.


h / The geometry of an electromagnetic wave.

One feature of figure h that is easily justified is that the electric and magnetic fields are perpendicular not only to each other, but also to the direction of propagation of the wave. In other words, the vibration is sideways, like people in a stadium “doing the wave,” not lengthwise, like the accordion pattern in figure i. (In standard wave terminology, we say that the wave is transverse, not longitudinal.) The wave pattern in figure i is impossible, because it diverges from the middle. For virtually any choice of Gaussian surface, the magnetic and electric fluxes would be nonzero, contradicting the equations \(\Phi_B=0\) and \(\Phi_E=0\).12

Example 23: Reflection
The wave in figure j hits a silvered mirror. The metal is a good conductor, so it has constant voltage throughout, and the electric field equals zero inside it: the wave doesn't penetrate and is 100% reflected. If the electric field is to be zero at the surface as well, the reflected wave must have its electric field inverted (p. 362), so that the incident and reflected fields cancel there.

But the magnetic field of the reflected wave is not inverted. This is because the reflected wave has to have the correct right-handed relationship between the fields and the direction of propagation.


Two electromagnetic waves traveling in the same direction through space can differ by having their electric and magnetic fields in different directions, a property of the wave called its polarization.

The speed of light

What is the velocity of the waves described by Maxwell's equations? Maxwell convinced himself that light was an electromagnetic wave partly because his equations predicted waves moving at the velocity of light, \(c\). The only velocity that appears in the equations is \(c\), so this is fairly plausible, although a real calculation is required in order to prove that the velocity of the waves isn't something like \(2c\) or \(c/\pi\) --- or zero, which is also \(c\) multiplied by a constant! The following discussion, leading up to a proof that electromagnetic waves travel at \(c\), is meant to be understandable even if you're reading this book out of order, and haven't yet learned much about waves. As always with proofs in this book, the reason to read it isn't to convince yourself that it's true, but rather to build your intuition. The style will be visual. In all the following figures, the wave patterns are moving across the page (let's say to the right), and it usually doesn't matter whether you imagine them as representing the wave's magnetic field or its electric field, because Maxwell's equations in a vacuum have the same form for both fields. Whichever field we imagine the figures as representing, the other field is coming in and out of the page.

The velocity of the waves is not zero. If the wave pattern was standing still in space, then the right sides of the \(\Gamma\) equations would be zero, because there would be no change in the field over time at a particular point. But the left sides are not zero, so this is impossible.13

The velocity of the waves is a fixed number for a given wave pattern. Consider a typical sinusoidal wave of visible light, with a distance of half a micrometer from one peak to the next peak. Suppose this wave pattern provides a valid solution to Maxwell's equations when it is moving with a certain velocity. We then know, for instance, that there cannot be a valid solution to Maxwell's equations in which the same wave pattern moves with double that velocity. The time derivatives on the right sides of Maxwell's equations for \(\Gamma_E\) and \(\Gamma_B\) would be twice as big, since an observer at a certain point in space would see the wave pattern sweeping past at twice the rate. But the left sides would be the same, so the equations wouldn't equate.

The velocity is the same for all wave patterns. In other words, it isn't \(0.878c\) for one wave pattern, and \(1.067c\) for some other pattern. This is surprising, since, for example, water waves with different shapes do travel at different speeds. Similarly, even though we speak of “the speed of sound,” sound waves do travel at slightly different speeds depending on their pitch and loudness, although the differences are small unless you're talking about cannon blasts or extremely high frequency ultrasound. To see how Maxwell's equations give a consistent velocity, consider figure k. Along the right and left edges of the same Ampèrian surface, the more compressed wave pattern of blue light has twice as strong a field, so the circulations on the left sides of Maxwell's equations are twice as large.14 To satisfy Maxwell's equations, the time derivatives of the fields must also be twice as large for the blue light. But this is true only if the blue light's wave pattern is moving to the right at the same speed as the red light's: if the blue light pattern is sweeping over an observer with a given velocity, then the time between peaks is half as much, like the clicking of the wheels on a train whose cars are half the length.15

We can also check that bright and dim light, as shown in figure l, have the same velocity. If you haven't yet learned much about waves, then this might be surprising. A material object with more energy goes faster, but that's not the case for waves. The circulation around the edge of the Ampèrian surface shown in the figure is twice as strong for the light whose fields are doubled in strength, so the left sides of Maxwell's \(\Gamma\) equations are doubled. The right sides are also doubled, because the derivative of twice a function is twice the derivative of the original function. Thus if dim light moving with a particular velocity is a solution, then so is bright light, provided that it has the same velocity.

We can now see that all sinusoidal waves have the same velocity. What about nonsinusoidal waves like the one in figure m? There is a mathematical theorem, due to Fourier, that says any function can be made by adding together sinusoidal functions. For instance, \(3\sin x-7\cos 3x\) can be made by adding together the functions \(3\sin x\) and \(-7\cos 3x\), but Fourier proved that this can be done even for functions, like figure m, that aren't obviously built out of sines and cosines in the first place. Therefore our proof that sinusoidal waves all have the same velocity is sufficient to demonstrate that other waves also have this same velocity.

We're now ready to prove that this universal speed for all electromagnetic waves is indeed \(c\). Since we've already convinced ourselves that all such waves travel at the same speed, it's sufficient to find the velocity of one wave in particular. Let's pick the wave whose fields have magnitudes

\[\begin{align*} E &= \tilde{E}\sin(x+vt) \text{and}\\ B &= \tilde{B}\sin(x+vt) , \end{align*}\]

which is about as simple as we can get. The peak electric field of this wave has a strength \(\tilde{E}\), and the peak magnetic field is \(\tilde{B}\). The sine functions go through one complete cycle as \(x\) increases by \(2\pi=6.28...\), so the distance from one peak of this wave to the next --- its wavelength --- is 6.28... meters. This means that it is not a wave of visible light but rather a radio wave (its wavelength is on the same order of magnitude as the size of a radio antenna). That's OK. What was glorious about Maxwell's work was that it unified the whole electromagnetic spectrum. Light is simple. Radio waves aren't fundamentally any different than light waves, x-rays, or gamma rays. 16

The justification for putting \(x+vt\) inside the sine functions is as follows. As the wave travels through space, the whole pattern just shifts over. The fields are zero at \(x=0\), \(t=0\), since the sine of zero is zero. This zero-point of the wave pattern shifts over as time goes by; at any time \(t\) its location is given by \(x+vt=0\). After one second, the zero-point is located at \(x=-(1\ \text{s})v\). The distance it travels in one second is therefore numerically equal to \(v\), and this is exactly the concept of velocity: how far something goes per unit time.

The wave has to satisfy Maxwell's equations for \(\Gamma_E\) and \(\Gamma_B\) regardless of what Ampèrian surfaces we pick, and by applying them to any surface, we could determine the speed of the wave. The surface shown in figure n turns out to result in an easy calculation: a narrow strip of width \(2\ell\) and height \(h\), coinciding with the position of the zero-point of the field at \(t=0\).

Now let's apply the equation \(c^2\Gamma_B=\partial\Phi_E/\partial t\) at \(t=0\). Since the strip is narrow, we can approximate the magnetic field using \(\sin x\approx x\), which is valid for small \(x\). The magnetic field on the right edge of the strip, at \(x=\ell\), is then \(\tilde{B}\ell\), so the right edge of the strip contributes \(\tilde{B}\ell h\) to the circulation. The left edge contributes the same amount, so the left side of Maxwell's equation is

\[\begin{equation*} c^2\Gamma_B = c^2 \cdot 2 \tilde{B}\ell h . \end{equation*}\]

The other side of the equation is

\[\begin{align*} \frac{\partial\Phi_E}{\partial t} &= \frac{\partial}{\partial t}(EA) \\ &= 2\ell h \frac{\partial E}{\partial t} ,\\ \text{where we can dispense with the usual sum because the strip is narrow and there is no variation in the field as we go up and down the strip. The derivative equals $v\tilde{E}\cos(x+vt)$, and evaluating the cosine at $x=0$, $t=0$ gives} \frac{\partial\Phi_E}{\partial t} &= 2v \tilde{E}\ell h \\ \end{align*}\]

Maxwell's equation for \(\Gamma_B\) therefore results in

\[\begin{align*} 2c^2\tilde{B}\ell h &= 2\tilde{E}\ell h v\\ c^2\tilde{B} &= v\tilde{E} . \end{align*}\]

An application of \(\Gamma_E=-\partial\Phi_B/\partial t\) gives a similar result, except that there is no factor of \(c^2\)

\[\begin{equation*} \tilde{E} = v \tilde{B} . \end{equation*}\]

(The minus sign simply represents the right-handed relationship of the fields relative to their direction of propagation.)

Multiplying these last two equations by each other, we get

\[\begin{align*} c^2\tilde{B}\tilde{E} &= v^2\tilde{E}\tilde{B} \\ c^2 &= v^2 \\ v &= \pm c . \end{align*}\]

This is the desired result. (The plus or minus sign shows that the wave can travel in either direction.)

As a byproduct of this calculation, we can find the relationship between the strengths of the electric and magnetic fields in an electromagnetic wave. If, instead of multiplying the equations \(c^2\tilde{B} = v \tilde{E}\) and \(\tilde{E} = v \tilde{B}\), we divide them, we can easily show that \(\tilde{E}=c\tilde{B}\).


o / The electromagnetic spectrum.

Figure o shows the complete spectrum of light waves. The wavelength \(\lambda\) (number of meters per cycle) and frequency \(f\) (number of cycles per second) are related by the equation \(c=f\lambda\). Maxwell's equations predict that all light waves have the same structure, regardless of wavelength and frequency, so even though radio and x-rays, for example, hadn't been discovered, Maxwell predicted that such waves would have to exist. Maxwell's 1865 prediction passed an important test in 1888, when Heinrich Hertz published the results of experiments in which he showed that radio waves could be manipulated in the same ways as visible light waves. Hertz showed, for example, that radio waves could be reflected from a flat surface, and that the directions of the reflected and incoming waves were related in the same way as with light waves, forming equal angles with the surface. Likewise, light waves can be focused with a curved, dish-shaped mirror, and Hertz demonstrated the same thing with radio waves using a metal dish.

Momentum of light waves

A light wave consists of electric and magnetic fields, and fields contain energy. Thus a light wave carries energy with it when it travels from one place to another. If a material object has kinetic energy and moves from one place to another, it must also have momentum, so it is logical to ask whether light waves have momentum as well. It can be proved based on relativity that it does, and that the momentum and energy are related by the equation \(U=p/c\), where \(p\) is the magnitude of the momentum vector, and \(U=U_e+U_m\) is the sum of the energy of the electric and magnetic fields. We can now demonstrate this without explicitly referring to relativity, and connect it to the specific structure of a light wave.

The energy density of a light wave is related to the magnitudes of the fields in a specific way --- it depends on the squares of their magnitudes, \(E^2\) and \(B^2\), which are the same as the dot products \(\mathbf{E}\cdot\mathbf{E}\) and \(\mathbf{B}\cdot\mathbf{B}\). We argued on page 586 that since energy is a scalar, the only possible expressions for the energy densities of the fields are dot products like these, multiplied by some constants. This is because the dot product is the only mathematically sensible way of multiplying two vectors to get a scalar result. (Any other way violates the symmetry of space itself.)

How does this relate to momentum? Well, we know that if we double the strengths of the fields in a light beam, it will have four times the energy, because the energy depends on the square of the fields. But we then know that this quadruple-energy light beam must have quadruple the momentum as well. If there wasn't this kind of consistency between the momentum and the energy, then we could violate conservation of momentum by combining light beams or splitting them up. We therefore know that the momentum density of a light beam must depend on a field multiplied by a field. Momentum, however, is a vector, and there is only one physically meaningful way of multiplying two vectors to get a vector result, which is the cross product (see page 918). The momentum density can therefore only depend on the cross products \(\mathbf{E}\times\mathbf{E}\), \(\mathbf{B}\times\mathbf{B}\), and \(\mathbf{E}\times\mathbf{B}\). But the first two of these are zero, since the cross product vanishes when there is a zero angle between the vectors. Thus the momentum per unit volume must equal \(\mathbf{E}\times\mathbf{B}\) multiplied by some constant,

\[\begin{equation*} d\mathbf{p} = (\text{constant})\mathbf{E}\times\mathbf{B}\,dv \end{equation*}\]

This predicts something specific about the direction of propagation of a light wave: it must be along the line perpendicular to the electric and magnetic fields. We've already seen that this is correct, and also that the electric and magnetic fields are perpendicular to each other. Therefore this cross product has a magnitude

\[\begin{align*} |\mathbf{E}\times\mathbf{B}| &=|\mathbf{E}||\mathbf{B}|\sin 90° \\ &=|\mathbf{E}||\mathbf{B}| \\ &=\frac{|\mathbf{E}|^2}{c}=c|\mathbf{B}|^2 , \end{align*}\]

where in the last step the relation \(|\mathbf{E}|=c|\mathbf{B}|\) has been used.

We now only need to find one physical example in order to fix the constant of proportionality. Indeed, if we didn't know relativity, it would be possible to believe that the constant of proportionality was zero! The simplest example of which I know is as follows. Suppose a piece of wire of length \(\ell\) is bathed in electromagnetic waves coming in sideways, and let's say for convenience that this is a radio wave, with a wavelength that is large compared to \(\ell\), so that the fields don't change significantly across the length of the wire. Let's say the electric field of the wave happens to be aligned with the wire. Then there is an emf between the ends of the wire which equals \(E\ell\), and since the wire is small compared to the wavelength, we can pretend that the field is uniform, not curly, in which case voltage is a well-defined concept, and this is equivalent to a voltage difference \(\Delta V=E\ell\) between the ends of the wire. The wire obeys Ohm's law, and a current flows in response to the wave.17 Equating the expressions \(dU/dt\) and \(I\Delta V\) for the power dissipated by ohmic heating, we have

\[\begin{equation*} dU = IE\ell dt \end{equation*}\]

for the energy the wave transfers to the wire in a time interval \(dt\).

Note that although some electrons have been set in motion in the wire, we haven't yet seen any momentum transfer, since the protons are experiencing the same amount of electric force in the opposite direction. However, the electromagnetic wave also has a magnetic field, and a magnetic field transfers momentum to (exerts a force on) a current. This is only a force on the electrons, because they're what make the current. The magnitude of this force equals \(\ell IB\) (homework problem 6), and using the definition of force, \(d\mathbf{p}/dt\), we find for the magnitude of the momentum transferred:

\[\begin{equation*} dp = \ell IB dt \end{equation*}\]

We now know both the amount of energy and the amount of momentum that the wave has lost by interacting with the wire. Dividing these two equations, we find

\[\begin{align*} \frac{dp}{dU} &= \frac{B}{E} \\ &= \frac{1}{c} , \end{align*}\]

which is what we expected based on relativity. This can now be restated in the form \(d\mathbf{p} = (\text{constant})\mathbf{E}\times\mathbf{B}\,dv\) (homework problem 40).

Note that although the equations \(p=U/c\) and \(d\mathbf{p} = (\text{constant})\mathbf{E}\times\mathbf{B}\,dv\) are consistent with each other for a sine wave, they are not consistent with each other in general. The relativistic argument leading up to \(p=U/c\) assumed that we were only talking about a single thing traveling in a single direction, whereas no such assumption was made in arguing for the \(\mathbf{E}\times\mathbf{B}\) form. For instance, if two light beams of equal strength are traveling through one another, going in opposite directions, their total momentum is zero, which is consistent with the \(\mathbf{E}\times\mathbf{B}\) form, but not with \(U/c\).

Some examples were given in chapter 3 of situations where it actually matters that light has momentum. Figure q shows the first confirmation of this fact in the laboratory.

Angular momentum of light waves

For completeness, we note that since light carries momentum, it must also be possible for it to have angular momentum. If you've studied chemistry, here's an example of why this can be important. You know that electrons in atoms can exist in states labeled s, p, d, f, and so on. What you might not have realized is that these are angular momentum labels. The s state, for example, has zero angular momentum. If light didn't have angular momentum, then, for example, it wouldn't be possible for a hydrogen atom in a p state to change to the lower-energy s state by emitting light. Conservation of angular momentum requires that the light wave carry away all the angular momentum originally possessed by the electron in the p state, since in the s state it has none.

Discussion Questions

Positive charges 1 and 2 are moving as shown. What electric and magnetic forces do they exert on each other? What does this imply for conservation of momentum?

1. The figure shows a line of charges moving to the right, creating a current \(I\). An Ampèrian surface in the form of a disk has been superimposed. Use Maxwell's equations to find the field \(B\) at point P.
2. A tiny gap is chopped out of the line of charge. What happens when this gap is directly underneath the point P?

The diagram shows an electric field pattern frozen at one moment in time. Let's imagine that it's the electric part of an electromagnetic wave. Consider four possible directions in which it could be propagating: left, right, up, and down. Determine whether each of these is consistent with Maxwell's equations. If so, infer the direction of the magnetic field.

Discuss what happens to the wave pattern shown in the diagram if we switch to a frame of reference moving along with the wave.

11.7 Electromagnetic Properties of Materials

Different types of matter have a variety of useful electrical and magnetic properties. Some are conductors, and some are insulators. Some, like iron and nickel, can be magnetized, while others have useful electrical properties, e.g., dielectrics, discussed qualitatively in the discussion question on page 596, which allow us to make capacitors with much higher values of capacitance than would otherwise be possible. We need to organize our knowledge about the properties that materials can possess, and see whether this knowledge allows us to calculate anything useful with Maxwell's equations.


a / A capacitor with a dielectric between the plates.

11.7.1 Conductors

A perfect conductor, such as a superconductor, has no DC electrical resistance. It is not possible to have a static electric field inside it, because then charges would move in response to that field, and the motion of the charges would tend to reduce the field, contrary to the assumption that the field was static. Things are a little different at the surface of a perfect conductor than on the interior. We expect that any net charges that exist on the conductor will spread out under the influence of their mutual repulsion, and settle on the surface. As we saw in chapter 10, Gauss's law requires that the fields on the two sides of a sheet of charge have \(|\mathbf{E}_{\perp,1}-\mathbf{E}_{\perp,2}|\) proportional to the surface charge density, and since the field inside the conductor is zero, we infer that there can be a field on or immediately outside the conductor, with a nonvanishing component perpendicular to the surface. The component of the field parallel to the surface must vanish, however, since otherwise it would cause the charges to move along the surface.

On a hot summer day, the reason the sun feels warm on your skin is that the oscillating fields of the light waves excite currents in your skin, and these currents dissipate energy by ohmic heating. In a perfect conductor, however, this could never happen, because there is no such thing as ohmic heating. Since electric fields can't penetrate a perfect conductor, we also know that an electromagnetic wave can never pass into one. By conservation of energy, we know that the wave can't just vanish, and if the energy can't be dissipated as heat, then the only remaining possibility is that all of the wave's energy is reflected. This is why metals, which are good electrical conductors, are also highly reflective. They are not perfect electrical conductors, however, so they are not perfectly reflective. The wave enters the conductor, but immediately excites oscillating currents, and these oscillating currents dissipate the energy both by ohmic heating and by reradiating the reflected wave. Since the parts of Maxwell's equations describing radiation have time derivatives in them, the efficiency of this reradiation process depends strongly on frequency. When the frequency is high and the material is a good conductor, reflection predominates, and is so efficient that the wave only penetrates to a very small depth, called the skin depth. In the limit of poor conduction and low frequencies, absorption predominates, and the skin depth becomes much greater. In a high-frequency AC circuit, the skin depth in a copper wire is very small, and therefore the signals in such a circuit are propagated entirely at the surfaces of the wires. In the limit of low frequencies, i.e., DC, the skin depth approaches infinity, so currents are carried uniformly over the wires' cross-sections.

We can quantify how well a particular material conducts electricity. We know that the resistance of a wire is proportional to its length, and inversely proportional to its cross-sectional area. The constant of proportionality is \(1/\sigma\), where \(\sigma\) (not the same \(\sigma\) as the surface charge density) is called the electrical conductivity. Exposed to an electric field \(\mathbf{E}\), a conductor responds with a current per unit cross-sectional area \(\mathbf{J}=\sigma\mathbf{E}\). The skin depth is proportional to \(1/\sqrt{f\sigma}\), where \(f\) is the frequency of the wave.


b / A stud finder is used to locate the wooden beams, or studs, that form the frame behind the wallboard. It is a capacitor whose capacitance changes when it is brought close to a substance with a particular permittivity. Although the wall is external to the capacitor, a change in capacitance is still observed, because the capacitor has “fringing fields” that extend outside the region between its plates.

11.7.2 Dielectrics

A material with a very low conductivity is an insulator. Such materials are usually composed of atoms or molecules whose electrons are strongly bound to them; since the atoms or molecules have zero total charge, their motion cannot create an electric current. But even though they have zero charge, they may not have zero dipole moment. Imagine such a substance filling in the space between the plates of a capacitor, as in figure a. For simplicity, we assume that the molecules are oriented randomly at first, a/1, and then become completely aligned when a field is applied, a/2. The effect has been to take all of the negatively charged black ands of the molecules and shift them upward, and the opposite for the positively charged white ends. Where the black and white charges overlap, there is still zero net charge, but we have a strip of negative charge at the top, and a strip of positive charge at the bottom, a/3. The effect has been to cancel out part of the charge that was deposited on the plates of the capacitor. Now this is very subtle, because Maxwell's equations treat these charges on an equal basis, but in terms of practical measurements, they are completely different. The charge on the plates can be measured be inserting an ammeter in the circuit, and integrating the current over time. But the charges in the layers at the top and bottom of the dielectric never flowed through any wires, and cannot be detected by an ammeter. In other words, the total charge, \(q\), appearing in Maxwell's equartions is actualy \(q=q_{\text{free}}-q_{\text{bound}}\), where \(q_{\text{free}}\) is the charge that moves freely through wires, and can be detected in an ammeter, while \(q_{bound}\) is the charge bound onto the individual molecules, which can't. We will, however, detect the presence of the bound charges via their electric fields. Since their electric fields partially cancel the fields of the free charges, a voltmeter will register a smaller than expected voltage difference between the plates. If we measure \(q_{\text{free}}/V\), we have a result that is larger than the capacitance we would have expected.

Although the relationship \(\mathbf{E}\leftrightarrow q\) between electric fields and their sources is unalterably locked in by Gauss's law, that's not what we see in practical measurements. In this example, we can measure the voltage difference between the plates of the capacitor and divide by the distance between them to find \(\mathbf{E}\), and then integrate an ammeter reading to find \(q_{\text{free}}\), and we will find that Gauss's law appears not to hold. We have \(\mathbf{E}\leftrightarrow q_{\text{free}}/(\text{constant})\), where the constant fudge factor is greater than one. This constant is a property of the dielectric material, and tells us how many dipoles there are, how strong they are, and how easily they can be reoriented. The conventional notation is to incorporate this fudge factor into Gauss's law by defining an altered version of the electric field,

\[\begin{equation*} \mathbf{D}=\epsilon \mathbf{E} , \end{equation*}\]

and to rewrite Gauss's law as

\[\begin{equation*} \Phi_D = q_{\text{in, free}} . \end{equation*}\]

The constant \(\epsilon\) is a property of the material, known as its permittivity. In a vacuum, \(\epsilon\) takes on a value known as \(\epsilon_\text{o}\), defined as \(1/(4\pi k)\). In a dielectric, \(\epsilon\) is greater than \(\epsilon_\text{o}\). When a dielectric is present between the plates of a capacitor, its capacitance is proportional to \(\epsilon\) (problem 38). The following table gives some sample values of the permittivities of a few substances.

substance epsilonepsilontextupo at zero frequency

vacuum 1
air 1.00054
water 80
barium titanate 1250

A capacitor with a very high capacitance is potentially a superior replacement for a battery, but until the 1990's this was impractical because capacitors with high enough values couldn't be made, even with dielectrics having the largest known permittivities. Such supercapactors, some with values in the kilofarad range, are now available. Most of them do not use dielectric at all; the very high capacitance values are instead obtained by using electrodes that are not parallel metal plates at all, but exotic materials such as aerogels, which allows the spacing between the “electrodes” to be very small.

Although figure a/2 shows the dipoles in the dielectric being completely aligned, this is not a situation commonly encountered in practice. In such a situation, the material would be as polarized as it could possibly be, and if the field was increased further, it would not respond. In reality, a capacitor, for example, would normally be operated with fields that produced quite a small amount of alignment, and it would be under these conditions that the linear relationship \(\mathbf{D}=\epsilon\mathbf{E}\) would actually be a good approximation. Before a material's maximum polarization is reached, it may actually spark or burn up.


Suppose a parallel-plate capacitor is built so that a slab of dielectric material can be slid in or out. (This is similar to the way the stud finder in figure b works.) We insert the dielectric, hook the capacitor up to a battery to charge it, and then use an ammeter and a voltmeter to observe what happens when the dielectric is withdrawn. Predict the changes observed on the meters, and correlate them with the expected change in capacitance. Discuss the energy transformations involved, and determine whether positive or negative work is done in removing the dielectric.

(answer in the back of the PDF version of the book)


c / The magnetic version of figure a. A magnetically permeable material is placed at the center of a solenoid.


d / Example 24: a cutaway view of a solenoid.


e / Example 24: without the iron core, the field is so weak that it barely deflects the compass. With it, the deflection is nearly 90\(°\).


f / A transformer with a laminated iron core. The input and output coils are inside the paper wrapper. The iron core is the black part that passes through the coils at the center, and also wraps around them on the outside.


g / Example 25: ferrite beads. The top panel shows a clip-on type, while the bottom shows one built into a cable.


h / A frog is levitated diamagnetically by the nonuniform field inside a powerful magnet. Evidently frog has \(\mu\lt\mu_\text{o}\).


i / At a boundary between two substances with \(\mu_2>\mu_1\), the \(\mathbf{H}\) field has a continuous component parallel to the surface, which implies a discontinuity in the parallel component of the magnetic field \(\mathbf{B}\).


l / Magnetic core memory.


m / A hysteresis curve.


n / A fluxgate compass.

11.7.3 Magnetic materials

Atoms and molecules may have magnetic dipole moments as well as electric dipole moments. Just as an electric dipole contains bound charges, a magnetic dipole has bound currents, which come from the motion of the electrons as they orbit the nucleus, c/1. Such a substance, subjected to a magnetic field, tends to align itself, c/2, so that a sheet of current circulates around the externally applied field. Figure c/3 is closely analogous to figure a/3; in the central gray area, the atomic currents cancel out, but the atoms at the outer surface form a sheet of bound current. However, whereas like charges repel and opposite charges attract, it works the other way around for currents: currents in the same direction attract, and currents in opposite directions repel. Therefore the bound currents in a material inserted inside a solenoid tend to reinforce the free currents, and the result is to strengthen the field. The total current is \(I=I_{\text{free}}+I_{\text{bound}}\), and we define an altered version of the magnetic field,

\[\begin{equation*} \mathbf{H}= \frac{\mathbf{B}}{\mu} , \end{equation*}\]

and rewrite Ampère's law as

\[\begin{equation*} \Gamma_H = I_{\text{through, free}} . \end{equation*}\]

The constant \(\mu\) is the permeability, with a vacuum value of \(\mu_\text{o}=4\pi k/c^2\). Here are the magnetic permeabilities of some substances:

substance mumutextupo

vacuum 1
aluminum 1.00002
steel 700
transformer iron4,000
mu-metal 20,000
Example 24: An iron-core electromagnet
\(\triangleright\) A solenoid has 1000 turns of wire wound along a cylindrical core with a length of 10 cm. If a current of 1.0 A is used, find the magnetic field inside the solenoid if the core is air, and if the core is made of iron with \(\mu/\mu_\text{o}=4,000\).

\(\triangleright\) Air has essentially the same permability as vacuum, so using the result of example 13 on page 678, we find that the field is 0.013 T.

We now consider the case where the core is filled with iron. The original derivation in example 13 started from Ampère's law, which we now rewrite as \(\Gamma_H = I_{\text{through, free}}\). As argued previously, the only significant contributions to the circulation come from line segment AB. This segment lies inside the iron, where \(\mathbf{H}=\mathbf{B}/\mu\). The \(\mathbf{H}\) field is the same as in the air-core case, since the new form of Ampère's law only relates \(\mathbf{H}\) to the current in the wires (the free current). This means that \(\mathbf{B}=\mu\mathbf{H}\) is greater by a factor of 4,000 than in the air-core case, or 52 T. This is an extremely intense field --- so intense, in fact, that the iron's magnetic polarization would probably become saturated before we could actually get the field that high.

The electromagnet of example 24 could also be used as an inductor, and its inductance would be proportional to the permittivity of the core. This makes it possible to construct high-value inductors that are relatively compact. Permeable cores are also used in transformers.

A transformer or inductor with a permeable core does have some disadvantages, however, in certain applications. The oscillating magnetic field induces an electric field, and because the core is typically a metal, these currents dissipate energy strongly as heat. This behaves like a fairly large resistance in series with the coil. Figure f shows a method for reducing this effect. The iron core of this transformer has been constructed out of laminated layers, which has the effect of blocking the conduction of the eddy currents.

Example 25: A ferrite bead
Cables designed to carry audio signals are typically made with two adjacent conductors, such that the current flowing out through one conductor comes back through the other one. Computer cables are similar, but usually have several such pairs bundled inside the insulator. This paired arrangement is known as differential mode, and has the advantage of cutting down on the reception and transmission of interference. In terms of transmission, the magnetic field created by the outgoing current is almost exactly canceled by the field from the return current, so electromagnetic waves are only weakly induced. In reception, both conductors are bathed in the same electric and magnetic fields, so an emf that adds current on one side subtracts current from the other side, resulting in cancellation.

The opposite of differential mode is called common mode. In common mode, all conductors have currents flowing in the same direction. Even when a circuit is designed to operate in differential mode, it may not have exactly equal currents in the two conductors with \(I_1+I_2=0\), meaning that current is leaking off to ground at one end of the circuit or the other. Although paired cables are relatively immune to differential-mode interference, they do not have any automatic protection from common-mode interference.

Figure g shows a device for reducing common-mode interference called a ferrite bead, which surrounds the cable like a bead on a string. Ferrite is a magnetically permeable alloy. In this application, the ohmic properties of the ferrite actually turn put to be advantageous.

Let's consider common-mode transmission of interference. The bare cable has some DC resistance, but is also surrounded by a magnetic field, so it has inductance as well. This means that it behaves like a series L-R circuit, with an impedance that varies as \(R+i\omega L\), where both \(R\) and \(L\) are very small. When we add the ferrite bead, the inductance is increased by orders of magnitude, but so is the resistance. Neither \(R\) nor \(L\) is actually constant with respect to frequency, but both are much greater than for the bare cable.

Suppose, for example, that a signal is being transmitted from a digital camera to a computer via a USB cable. The camera has an internal impedance that is on the order of 10 \(\Omega\), the computer's input also has a \(\sim10\) \(\Omega\) impedance, and in differential mode the ferrite bead has no effect, so the cable's impedance has its low, designed value (probably also about 10 \(\Omega\), for good impedance matching). The signal is transmitted unattenuated from the camera to the computer, and there is almost no radiation from the cable.

But in reality there will be a certain amount of common-mode current as well. With respect to common mode, the ferrite bead has a large impedance, with the exact value depending on frequency, but typically on the order of 100 \(\Omega\) for frequencies in the MHz range. We now have a series circuit consisting of three impedances: 10, 100, and 10 \(\Omega\). For a given emf applied by an external radio wave, the current induced in the circuit has been attenuated by an order of magnitude, relative to its value without the ferrite bead.

Why is the ferrite is necessary at all? Why not just insert ordinary air-core inductors in the circuit? We could, for example, have two solenoidal coils, one in the outgoing line and one in the return line, interwound with one another with their windings oriented so that their differential-mode fields would cancel. There are two good reasons to prefer the ferrite bead design. One is that it allows a clip-on device like the one in the top panel of figure g, which can be added without breaking the circuit. The other is that our circuit will inevitably have some stray capacitance, and will therefore act like an LRC circuit, with a resonance at some frequency. At frequencies close to the resonant frequency, the circuit would absorb and transmit common-mode interference very strongly, which is exactly the opposite of the effect we were hoping to produce. The resonance peak could be made low and broad by adding resistance in series, but this extra resistance would attenuate the differential-mode signals as well as the common-mode ones. The ferrite's resistance, however, is actually a purely magnetic effect, so it vanishes in differential mode.

Surprisingly, some materials have magnetic permeabilities less than \(\mu_\text{o}\). This cannot be accounted for in the model above, and although there are semiclassical arguments that can explain it to some extent, it is fundamentally a quantum mechanical effect. Materials with \(\mu>\mu_\text{o}\) are called paramagnetic, while those with \(\mu\lt\mu_\text{o}\) are referred to as diamagnetic. Diamagnetism is generally a much weaker effect than paramagnetism, and is easily masked if there is any trace of contamination from a paramagnetic material. Diamagnetic materials have the interesting property that they are repelled from regions of strong magnetic field, and it is therefore possible to levitate a diamagnetic object above a magnet, as in figure h.

A complete statement of Maxwell's equations in the presence of electric and magnetic materials is as follows:

\[\begin{align*} \Phi_D &= q_\text{free} \\ \Phi_B &= 0 \\ \Gamma_E &= -\frac{d \Phi_B}{dt} \\ \Gamma_H &= \frac{d \Phi_D}{dt} + I_\text{free} \end{align*}\]

Comparison with the vacuum case shows that the speed of an electromagnetic wave moving through a substance described by permittivity and permeability \(\epsilon\) and \(\mu\) is \(1/\sqrt{\epsilon\mu}\). For most substances, \(\mu\approx\mu_\text{o}\), and \(\epsilon\) is highly frequency-dependent.

Suppose we have a boundary between two substances. By constructing a Gaussian or Ampèrian surface that extends across the boundary, we can arrive at various constraints on how the fields must behave as me move from one substance into the other, when there are no free currents or charges present, and the fields are static. An interesting example is the application of Faraday's law, \(\Gamma_H=0\), to the case where one medium --- let's say it's air --- has a low permeability, while the other one has a very high one. We will violate Faraday's law unless the component of the \(\mathbf{H}\) field parallel to the boundary is a continuous function, \(\mathbf{H}_{\parallel,1}=\mathbf{H}_{\parallel,2}\). This means that if \(\mu/\mu_\text{o}\) is very high, the component of \(\mathbf{B}=\mu\mathbf{H}\) parallel to the surface will have an abrupt discontinuity, being much stronger inside the high-permeability material. The result is that when a magnetic field enters a high-permeability material, it tends to twist abruptly to one side, and the pattern of the field tends to be channeled through the material like water through a funnel. In a transformer, a permeable core functions to channel more of the magnetic flux from the input coil to the output coil. Figure j shows another example, in which the effect is to shield the interior of the sphere from the externally imposed field. Special high-permeability alloys, with trade names like Mu-Metal, are sold for this purpose.


j / A hollow sphere with \(\mu/\mu_\text{o}=10\), is immersed in a uniform, externally imposed magnetic field. The interior of the sphere is shielded from the field. The arrows map the magnetic field \(\mathbf{B}\). (See homework problem 'hw:spherical-shielding', page 731.)

The very last magnetic phenomenon we'll discuss is probably the very first experience you ever had of magnetism. Ferromagnetism is a phenomenon in which a material tends to organize itself so that it has a nonvanishing magnetic field. It is exhibited strongly by iron and nickel, which explains the origin of the name.


k / A model of ferromagnetism.

Figure k/1 is a simple one-dimensional model of ferromagnetism. Each magnetic compass needle represents an atom. The compasses in the chain are stable when aligned with one another, because each one's north end is attracted to its neighbor's south end. The chain can be turned around, k/2, without disrupting its organization, and the compasses do not realign themselves with the Earth's field, because their torques on one another are stronger than the Earth's torques on them. The system has a memory. For example, if I want to remind myself that my friend's address is 137 Coupling Ct., I can align the chain at an angle of 137 degrees. The model fails, however, as an explanation of real ferromagnetism, because in two or more dimensions, the most stable arrangement of a set of interacting magnetic dipoles is something more like k/3, in which alternating rows point in opposite directions. In this two-dimensional pattern, every compass is aligned in the most stable way with all four of its neighbors. This shows that ferromagnetism, like diamagnetism, has no purely classical explanation; a full explanation requires quantum mechanics.

Because ferromagnetic substances “remember” the history of how they were prepared, they are commonly used to store information in computers. Figure l shows 16 bits from an ancient (ca. 1970) 4-kilobyte random-access memory, in which each doughnut-shaped iron “core” can be magnetized in one of two possible directions, so that it stores one bit of information. Today, RAM is made of transistors rather than magnetic cores, but a remnant of the old technology remains in the term “core dump,” meaning “memory dump,” as in “my girlfriend gave me a total core dump about her mom's divorce.” Most computer hard drives today do store their information on rotating magnetic platters, but the platter technology may be obsoleted by flash memory in the near future.

The memory property of ferromagnets can be depicted on the type of graph shown in figure m, known as a hysteresis curve. The y axis is the magnetization of a sample of the material --- a measure of the extent to which its atomic dipoles are aligned with one another. If the sample is initially unmagnetized, 1, and a field \(H\) is externally applied, the magnetization increases, 2, but eventually becomes saturated, 3, so that higher fields do not result in any further magnetization, 4. The external field can then be reduced, 5, and even eliminated completely, but the material will retain its magnetization. It is a permanent magnet. To eliminate its magnetization completely, a substantial field must be applied in the opposite direction. If this reversed field is made stronger, then the substance will eventually become magnetized just as strongly in the opposite direction. Since the hysteresis curve is nonlinear, and is not a function (it has more than one value of \(M\) for a particular value of \(B\)), a ferromagnetic material does not have a single, well-defined value of the permeability \(\mu\); a value like 4,000 for transformer iron represents some kind of a rough average.

Example 26: The fluxgate compass

The fluxgate compass is a type of magnetic compass without moving parts, commonly used on ships and aircraft. An AC current is applied in a coil wound around a ferromagnetic core, driving the core repeatedly around a hysteresis loop. Because the hysteresis curve is highly nonlinear, the addition of an external field such as the Earth's alters the core's behavior. Suppose, for example, that the axis of the coil is aligned with the magnetic north-south. The core will reach saturation more quickly when the coil's field is in the same direction as the Earth's, but will not saturate as early in the next half-cycle, when the two fields are in opposite directions. With the use of multiple coils, the components of the Earth's field can be measured along two or three axes, permitting the compass's orientation to be determined in two or (for aircraft) three dimensions.

11.7.4 Maxwell's equations in differential form


Homework Problems


b / Problem 18.


c / Problem 19.


e / Problem 25.


f / A nautilus shell is approximately a logarithmic spiral, of the type in problem 28.


g / Problem 32.


h / Problem 35.


i / Problem 37.


j / Problem 42.


k / Problem 54.

[Problems] \addcontentsline{toc}{section}{\protect{Problems}}

1. A particle with a charge of 1.0 C and a mass of 1.0 kg is observed moving past point P with a velocity \((1.0\ \text{m}/\text{s})\hat{\mathbf{x}}\). The electric field at point P is \((1.0\ \text{V}/\text{m})\hat{\mathbf{y}}\), and the magnetic field is \((2.0\ \text{T})\hat{\mathbf{y}}\). Find the force experienced by the particle.(answer check available at

2. For a positively charged particle moving through a magnetic field, the directions of the \(\mathbf{v}\), \(\mathbf{B}\), and \(\mathbf{F}\) vectors are related by a right-hand rule:

\[\begin{align*} \mathbf{v} & \text{along the fingers, with the hand flat}\\ \mathbf{B} & \text{along the fingers, with the knuckles bent}\\ \mathbf{F} & \text{along the thumb} \end{align*}\]

Make a three-dimensional model of the three vectors using pencils or rolled-up pieces of paper to represent the vectors assembled with their tails together. Make all three vectors perpendicular to each other. Now write down every possible way in which the rule could be rewritten by scrambling up the three symbols \(\mathbf{v}\), \(\mathbf{B}\), and \(\mathbf{F}\). Referring to your model, which are correct and which are incorrect?

3. A charged particle is released from rest. We see it start to move, and as it gets going, we notice that its path starts to curve. Can we tell whether this region of space has \(\mathbf{E}\neq 0\), or \(\mathbf{B}\neq 0\), or both? Assume that no other forces are present besides the possible electrical and magnetic ones, and that the fields, if they are present, are uniform.

4. A charged particle is in a region of space in which there is a uniform magnetic field \(\mathbf{B}=B\hat{\mathbf{z}}\). There is no electric field, and no other forces act on the particle. In each case, describe the future motion of the particle, given its initial velocity.
(a) \(\mathbf{v}_\text{o}=0\)
(b) \(\mathbf{v}_\text{o}=(1\ \text{m}/\text{s})\hat{\mathbf{z}}\)
(c) \(\mathbf{v}_\text{o}=(1\ \text{m}/\text{s})\hat{\mathbf{y}}\)

5. (a) A line charge, with charge per unit length \(\lambda\), moves at velocity \(v\) along its own length. How much charge passes a given point in time \(dt\)? What is the resulting current? \hwans{hwans:linechargecurrent}
(b) Show that the units of your answer in part a work out correctly.

This constitutes a physical model of an electric current, and it would be a physically realistic model of a beam of particles moving in a vacuum, such as the electron beam in a television tube. It is not a physically realistic model of the motion of the electrons in a current-carrying wire, or of the ions in your nervous system; the motion of the charge carriers in these systems is much more complicated and chaotic, and there are charges of both signs, so that the total charge is zero. But even when the model is physically unrealistic, it still gives the right answers when you use it to compute magnetic effects. This is a remarkable fact, which we will not prove. The interested reader is referred to E.M. Purcell, Electricity and Magnetism, McGraw Hill, 1963.

6. Two parallel wires of length \(L\) carry currents \(I_1\) and \(I_2\). They are separated by a distance \(R\), and we assume \(R\) is much less than \(L\), so that our results for long, straight wires are accurate. The goal of this problem is to compute the magnetic forces acting between the wires.
(a) Neither wire can make a force on itself. Therefore, our first step in computing wire 1's force on wire 2 is to find the magnetic field made only by wire 1, in the space occupied by wire 2. Express this field in terms of the given quantities.(answer check available at
(b) Let's model the current in wire 2 by pretending that there is a line charge inside it, possessing density per unit length \(\lambda_2\) and moving at velocity \(v_2\). Relate \(\lambda_2\) and \(v_2\) to the current \(I_2\), using the result of problem 5a. Now find the magnetic force wire 1 makes on wire 2, in terms of \(I_1\), \(I_2\), \(L\), and \(R\). \hwans{hwans:forcebetweentwowires}
(c) Show that the units of the answer to part b work out to be newtons.

7. Suppose a charged particle is moving through a region of space in which there is an electric field perpendicular to its velocity vector, and also a magnetic field perpendicular to both the particle's velocity vector and the electric field. Show that there will be one particular velocity at which the particle can be moving that results in a total force of zero on it. Relate this velocity to the magnitudes of the electric and magnetic fields. (Such an arrangement, called a velocity filter, is one way of determining the speed of an unknown particle.)

8. The following data give the results of two experiments in which charged particles were released from the same point in space, and the forces on them were measured:

q11 mutextupCvcv11 textupmtextupshatvcxvcF11 textupmNhatvcy
q22 mutextupC vcv21 textupmtextupshatvcx vcF22 textupmNhatvcy

The data are insufficient to determine the magnetic field vector; demonstrate this by giving two different magnetic field vectors, both of which are consistent with the data.

9. The following data give the results of two experiments in which charged particles were released from the same point in space, and the forces on them were measured:

q11 textupnCvcv11 textupmtextupshatvczvcF15 textuppNhatvcx2 textuppNhatvcy
q21 textupnC vcv23 textupmtextupshatvcz vcF210 textuppNhatvcx4 textuppNhatvcy

Is there a nonzero electric field at this point? A nonzero magnetic field?

10. This problem is a continuation of problem 6. Note that the answer to problem 6b is given on page 936.
(a) Interchanging the 1's and 2's in the answer to problem 6b, what is the magnitude of the magnetic force from wire 2 acting on wire 1? Is this consistent with Newton's third law?
(b) Suppose the currents are in the same direction. Make a sketch, and use the right-hand rule to determine whether wire 1 pulls wire 2 towards it, or pushes it away.
(c) Apply the right-hand rule again to find the direction of wire 2's force on wire 1. Does this agree with Newton's third law?
(d) What would happen if wire 1's current was in the opposite direction compared to wire 2's?

11. (a) In the photo of the vacuum tube apparatus in figure o on page 660, infer the direction of the magnetic field from the motion of the electron beam. (The answer is given in the answer to the self-check on that page.)
(b) Based on your answer to part a, find the direction of the currents in the coils.
(c) What direction are the electrons in the coils going?
(d) Are the currents in the coils repelling the currents consisting of the beam inside the tube, or attracting them? Check your answer by comparing with the result of problem 10.

12. A charged particle of mass \(m\) and charge \(q\) moves in a circle due to a uniform magnetic field of magnitude \(B\), which points perpendicular to the plane of the circle.
(a) Assume the particle is positively charged. Make a sketch showing the direction of motion and the direction of the field, and show that the resulting force is in the right direction to produce circular motion.
(b) Find the radius, \(r\), of the circle, in terms of \(m\), \(q\), \(v\), and \(B\).(answer check available at
(c) Show that your result from part b has the right units.
(d) Discuss all four variables occurring on the right-hand side of your answer from part b. Do they make sense? For instance, what should happen to the radius when the magnetic field is made stronger? Does your equation behave this way?
(e) Restate your result so that it gives the particle's angular frequency, \(\omega\), in terms of the other variables, and show that \(v\) drops out.(answer check available at

A charged particle can be accelerated in a circular device called a cyclotron, in which a magnetic field is what keeps them from going off straight. This frequency is therefore known as the cyclotron frequency. The particles are accelerated by other forces (electric forces), which are AC. As long as the electric field is operated at the correct cyclotron frequency for the type of particles being manipulated, it will stay in sync with the particles, giving them a shove in the right direction each time they pass by. The particles are speeding up, so this only works because the cyclotron frequency is independent of velocity.

13. Each figure represents the motion of a positively charged particle. The dots give the particles' positions at equal time intervals. In each case, determine whether the motion was caused by an electric force, a magnetic force, or a frictional force, and explain your reasoning. If possible, determine the direction of the magnetic or electric field. All fields are uniform. In (a), the particle stops for an instant at the upper right, but then comes back down and to the left, retracing the same dots. In (b), it stops on the upper right and stays there.


a / Problem 13.

14. One model of the hydrogen atom has the electron circling around the proton at a speed of \(2.2\times10^6\) m/s, in an orbit with a radius of 0.05 nm. (Although the electron and proton really orbit around their common center of mass, the center of mass is very close to the proton, since it is 2000 times more massive. For this problem, assume the proton is stationary.)
(a) Treat the circling electron as a current loop, and calculate the current.
(b) Estimate the magnetic field created at the center of the atom by the electron.(answer check available at
(c) Does the proton experience a nonzero force from the electron's magnetic field? Explain.
(d) Does the electron experience a magnetic field from the proton? Explain.
(e) Does the electron experience a magnetic field created by its own current? Explain.
(f) Is there an electric force acting between the proton and electron? If so, calculate it.(answer check available at
(g) Is there a gravitational force acting between the proton and electron? If so, calculate it.
(h) An inward force is required to keep the electron in its orbit -- otherwise it would obey Newton's first law and go straight, leaving the atom. Based on your answers to the previous parts, which force or forces (electric, magnetic and gravitational) contributes significantly to this inward force? (Based on a problem by Arnold Arons.)

15. The equation \(B_z=\beta kIA/c^2r^3\) was found on page 670 for the distant field of a dipole. Show, as asserted there, that the constant \(\beta\) must be unitless.

16. The following data give the results of three experiments in which charged particles were released from the same point in space, and the forces on them were measured:

q11 textupCvcv10 vcF11 textupNhatvcy
q21 textupC vcv21 textupmtextupshatvcx vcF21 textupNhatvcy
q31 textupC vcv31 textupmtextupshatvcz vcF30

Determine the electric and magnetic fields.(answer check available at

17. If you put four times more current through a solenoid, how many times more energy is stored in its magnetic field?(answer check available at

18. A Helmholtz coil is defined as a pair of identical circular coils lying in parallel planes and separated by a distance, \(h\), equal to their radius, \(b\). (Each coil may have more than one turn of wire.) Current circulates in the same direction in each coil, so the fields tend to reinforce each other in the interior region. This configuration has the advantage of being fairly open, so that other apparatus can be easily placed inside and subjected to the field while remaining visible from the outside. The choice of \(h=b\) results in the most uniform possible field near the center. A photograph of a Helmholtz coil was shown in figure o on page 660.
(a) Find the percentage drop in the field at the center of one coil, compared to the full strength at the center of the whole apparatus. (answer check available at
(b) What value of \(h\) (not equal to \(b)\) would make this percentage difference equal to zero?(answer check available at

19. The figure shows a nested pair of circular wire loops used to create magnetic fields. (The twisting of the leads is a practical trick for reducing the magnetic fields they contribute, so the fields are very nearly what we would expect for an ideal circular current loop.) The coordinate system below is to make it easier to discuss directions in space. One loop is in the \(y-z\) plane, the other in the \(x-y\) plane. Each of the loops has a radius of 1.0 cm, and carries 1.0 A in the direction indicated by the arrow.
(a) Calculate the magnetic field that would be produced by one such loop, at its center. (answer check available at
(b) Describe the direction of the magnetic field that would be produced, at its center, by the loop in the \(x-y\) plane alone.
(c) Do the same for the other loop.
(d) Calculate the magnitude of the magnetic field produced by the two loops in combination, at their common center. Describe its direction.(answer check available at


d / Problem 20.

20. Four long wires are arranged, as shown, so that their cross-section forms a square, with connections at the ends so that current flows through all four before exiting. Note that the current is to the right in the two back wires, but to the left in the front wires. If the dimensions of the cross-sectional square (height and front-to-back) are \(b\), find the magnetic field (magnitude and direction) along the long central axis.(answer check available at

21. In problem 16, the three experiments gave enough information to determine both fields. Is it possible to design a procedure so that, using only two such experiments, we can always find \(\mathbf{E}\) and \(\mathbf{B}\)? If so, design it. If not, why not?

22. Use the Biot-Savart law to derive the magnetic field of a long, straight wire, and show that this reproduces the result of example 6 on page 662.

23. (a) Modify the calculation on page 667 to determine the component of the magnetic field of a sheet of charge that is perpendicular to the sheet.(answer check available at
(b) Show that your answer has the right units.
(c) Show that your answer approaches zero as \(z\) approaches infinity.
(d) What happens to your answer in the case of \(a=b\)? Explain why this makes sense.

24. Consider two solenoids, one of which is smaller so that it can be put inside the other. Assume they are long enough so that each one only contributes significantly to the field inside itself, and the interior fields are nearly uniform. Consider the configuration where the small one is inside the big one with their currents circulating in the same direction, and a second configuration in which the currents circulate in opposite directions. Compare the energies of these configurations with the energy when the solenoids are far apart. Based on this reasoning, which configuration is stable, and in which configuration will the little solenoid tend to get twisted around or spit out? \hwhint{hwhint:nestedsolenoids}

25. (a) A solenoid can be imagined as a series of circular current loops that are spaced along their common axis. Integrate the result of example 12 on page 675 to show that the field on the axis of a solenoid can be written as \(B=(2\pi k\eta/c^2)(\cos\beta+\cos\gamma)\), where the angles \(\beta\) and \(\gamma\) are defined in the figure.
(b) Show that in the limit where the solenoid is very long, this exact result agrees with the approximate one derived in example 13 on page 678 using Ampère's law.
(c) Note that, unlike the calculation using Ampère's law, this one is valid at points that are near the mouths of the solenoid, or even outside it entirely. If the solenoid is long, at what point on the axis is the field equal to one half of its value at the center of the solenoid?
(d) What happens to your result when you apply it to points that are very far away from the solenoid? Does this make sense?

26. The first step in the proof of Ampère's law on page 679 is to show that Ampère's law holds in the case shown in figure f/1, where a circular Ampèrian loop is centered on a long, straight wire that is perpendicular to the plane of the loop. Carry out this calculation, using the result for the field of a wire that was established without using Ampère's law.

27. A certain region of space has a magnetic field given by \(\mathbf{B}=bx\hat{\mathbf{y}}\). Find the electric current flowing through the square defined by \(z=0\), \(0\le x \le a\), and \(0\le y \le a\).(answer check available at

28. Perform a calculation similar to the one in problem 54, but for a logarithmic spiral, defined by \(r=we^{u\theta}\), and show that the field is \(B=(kI/c^2u)(1/a-1/b)\). Note that the solution to problem 54 is given in the back of the book.

29. (a) For the geometry described in example 8 on page 665, find the field at a point the lies in the plane of the wires, but not between the wires, at a distance \(b\) from the center line. Use the same technique as in that example.
(b) Now redo the calculation using the technique demonstrated on page 670. The integrals are nearly the same, but now the reasoning is reversed: you already know \(\beta=1\), and you want to find an unknown field. The only difference in the integrals is that you are tiling a different region of the plane in order to mock up the currents in the two wires. Note that you can't tile a region that contains a point of interest, since the technique uses the field of a distant dipole.(answer check available at

30. (a) A long, skinny solenoid consists of \(N\) turns of wire wrapped uniformly around a hollow cylinder of length \(\ell\) and cross-sectional area \(A\). Find its inductance.(answer check available at
(b) Show that your answer has the right units to be an inductance.

31. Consider two solenoids, one of which is smaller so that it can be put inside the other. Assume they are long enough to act like ideal solenoids, so that each one only contributes significantly to the field inside itself, and the interior fields are nearly uniform. Consider the configuration where the small one is partly inside and partly hanging out of the big one, with their currents circulating in the same direction. Their axes are constrained to coincide.
(a) Find the difference in the magnetic energy between the configuration where the solenoids are separate and the configuration where the small one is inserted into the big one. Your equation will include the length \(x\) of the part of the small solenoid that is inside the big one, as well as other relevant variables describing the two solenoids.(answer check available at
(b) Based on your answer to part a, find the force acting between the solenoids.(answer check available at

32. Verify Ampère's law in the case shown in the figure, assuming the known equation for the field of a wire. A wire carrying current \(I\) passes perpendicularly through the center of the rectangular Ampèrian surface. The length of the rectangle is infinite, so it's not necessary to compute the contributions of the ends.

33. The purpose of this problem is to find how the gain of a transformer depends on its construction.
(a) The number of loops of wire, \(N\), in a solenoid is changed, while keeping the length constant. How does the impedance depend on \(N\)? State your answer as a proportionality, e.g., \(Z\propto N^3\) or \(Z\propto N^{-5}\).
(b) For a given AC voltage applied across the inductor, how does the magnetic field depend on \(N\)? You need to take into account both the dependence of a solenoid's field on \(N\) for a given current and your answer to part a, which affects the current.
(c) Now consider a transformer consisting of two solenoids. The input side has \(N_1\) loops, and the output \(N_2\). We wish to find how the output voltage \(V_2\) depends on \(N_1\), \(N_2\), and the input voltage \(V_1\). The text has already established \(V_2\propto V_1N_2\), so it only remains to find the dependence on \(N_1\). Use your result from part b to accomplish this. The ratio \(V_2/V_1\) is called the voltage gain.

34. Problem 33 dealt with the dependence of a transformer's gain on the number of loops of wire in the input solenoid. Carry out a similar analysis of how the gain depends on the frequency at which the circuit is operated.

35. A U-shaped wire makes electrical contact with a second, straight wire, which rolls along it to the right, as shown in the figure. The whole thing is immersed in a uniform magnetic field, which is perpendicular to the plane of the circuit. The resistance of the rolling wire is much greater than that of the U.
(a) Find the direction of the force on the wire based on conservation of energy.
(b) Verify the direction of the force using right-hand rules.
(c) Find magnitude of the force acting on the wire. There is more than one way to do this, but please do it using Faraday's law (which works even though it's the Ampèrian surface itself that is changing, rather than the field).(answer check available at
(d) Consider how the answer to part a would have changed if the direction of the field had been reversed, and also do the case where the direction of the rolling wire's motion is reversed. Verify that this is in agreement with your answer to part c.

36. A charged particle is in motion at speed \(v\), in a region of vacuum through which an electromagnetic wave is passing. In what direction should the particle be moving in order to minimize the total force acting on it? Consider both possibilities for the sign of the charge. (Based on a problem by David J. Raymond.)

37. A wire loop of resistance \(R\) and area \(A\), lying in the \(y-z\) plane, falls through a nonuniform magnetic field \(\mathbf{B}=kz\hat{\mathbf{x}}\), where \(k\) is a constant. The \(z\) axis is vertical.
(a) Find the direction of the force on the wire based on conservation of energy.
(b) Verify the direction of the force using right-hand rules.
(c) Find the magnetic force on the wire.(answer check available at

38. A capacitor has parallel plates of area \(A\), separated by a distance \(h\). If there is a vacuum between the plates, then Gauss's law gives \(E=4\pi k\sigma=4\pi kq/A\) for the field between the plates, and combining this with \(E=V/h\), we find \(C=q/V=(1/4\pi k)A/h\). (a) Generalize this derivation to the case where there is a dielectric between the plates. (b) Suppose we have a list of possible materials we could choose as dielectrics, and we wish to construct a capacitor that will have the highest possible energy density, \(U_e/v\), where \(v\) is the volume. For each dielectric, we know its permittivity \(\epsilon\), and also the maximum electric field \(E\) it can sustain without breaking down and allowing sparks to cross between the plates. Write the maximum energy density in terms of these two variables, and determine a figure of merit that could be used to decide which material would be the best choice.

39. (a) For each term appearing on the right side of Maxwell's equations, give an example of an everyday situation it describes.
(b) Most people doing calculations in the SI system of units don't use \(k\) and \(k/c^2\). Instead, they express everything in terms of the constants

\[\begin{align*} \epsilon_\text{o}&=\frac{1}{4\pi k} \text{and} \\ \mu_\text{o}&=\frac{4\pi k}{c^2} . \end{align*}\]

Rewrite Maxwell's equations in terms of these constants, eliminating \(k\) and \(c\) everywhere.

40. (a) Prove that in an electromagnetic plane wave, half the energy is in the electric field and half in the magnetic field.
(b) Based on your result from part a, find the proportionality constant in the relation \(d\mathbf{p}\propto\mathbf{E}\times\mathbf{B}dv\), where \(d \mathbf{p}\) is the momentum of the part of a plane light wave contained in the volume \(dv\). The vector \(\mathbf{E}\times\mathbf{B}\) is known as the Poynting vector. (To do this problem, you need to know the relativistic relationship between the energy and momentum of a beam of light.)(answer check available at

41. (a) A beam of light has cross-sectional area \(A\) and power \(P\), i.e., \(P\) is the number of joules per second that enter a window through which the beam passes. Find the energy density \(U/v\) in terms of \(P\), \(A\), and universal constants.
(b) Find \(\tilde{\mathbf{E}}\) and \(\tilde{\mathbf{B}}\), the amplitudes of the electric and magnetic fields, in terms of \(P\), \(A\), and universal constants (i.e., your answer should not include \(U\) or \(v\)). You will need the result of problem 40a. A real beam of light usually consists of many short wavetrains, not one big sine wave, but don't worry about that.(answer check available at\hwhint{hwhint:solarconstant}
(c) A beam of sunlight has an intensity of \(P/A=1.35\times10^3\ \text{W}/\text{m}^2\), assuming no clouds or atmospheric absorption. This is known as the solar constant. Compute \(\tilde{\mathbf{E}}\) and \(\tilde{\mathbf{B}}\), and compare with the strengths of static fields you experience in everyday life: \(E \sim 10^6\ \text{V}/\text{m}\) in a thunderstorm, and \(B \sim 10^{-3}\) T for the Earth's magnetic field.(answer check available at

42. The circular parallel-plate capacitor shown in the figure is being charged up over time, with the voltage difference across the plates varying as \(V=st\), where \(s\) is a constant. The plates have radius \(b\), and the distance between them is \(d\). We assume \(d \ll b\), so that the electric field between the plates is uniform, and parallel to the axis. Find the induced magnetic field at a point between the plates, at a distance \(R\) from the axis. \hwhint{hwhint:circularcap}(answer check available at

43. A positively charged particle is released from rest at the origin at \(t=0\), in a region of vacuum through which an electromagnetic wave is passing. The particle accelerates in response to the wave. In this region of space, the wave varies as \(\mathbf{E}=\hat{\mathbf{x}}\tilde{E}\sin\omega t\), \(\mathbf{B}=\hat{\mathbf{y}}\tilde{B}\sin\omega t\), and we assume that the particle has a relatively large value of \(m/q\), so that its response to the wave is sluggish, and it never ends up moving at any speed comparable to the speed of light. Therefore we don't have to worry about the spatial variation of the wave; we can just imagine that these are uniform fields imposed by some external mechanism on this region of space.
(a) Find the particle's coordinates as functions of time.(answer check available at
(b) Show that the motion is confined to \(-z_{max}\leq z \leq z_{max}\), where \(z_{max} = 1.101\left(q^2\tilde{E}\tilde{B}/m^2\omega^3\right)\).

44. Electromagnetic waves are supposed to have their electric and magnetic fields perpendicular to each other. (Throughout this problem, assume we're talking about waves traveling through a vacuum, and that there is only a single sine wave traveling in a single direction, not a superposition of sine waves passing through each other.) Suppose someone claims they can make an electromagnetic wave in which the electric and magnetic fields lie in the same plane. Prove that this is impossible based on Maxwell's equations.

45. Repeat the self-check on page 714, but with one change in the procedure: after we charge the capacitor, we open the circuit, and then continue with the observations.

46. On page 717, I proved that \(\mathbf{H}_{\parallel,1}=\mathbf{H}_{\parallel,2}\) at the boundary between two substances if there is no free current and the fields are static. In fact, each of Maxwell's four equations implies a constraint with a similar structure. Some are constraints on the field components parallel to the boundary, while others are constraints on the perpendicular parts. Since some of the fields referred to in Maxwell's equations are the electric and magnetic fields \(\mathbf{E}\) and \(\mathbf{B}\), while others are the auxiliary fields \(\mathbf{D}\) and \(\mathbf{H}\), some of the constraints deal with \(\mathbf{E}\) and \(\mathbf{B}\), others with \(\mathbf{D}\) and \(\mathbf{H}\). Find the other three constraints.

47. (a) Figure j on page 718 shows a hollow sphere with \(\mu/\mu_\text{o}=x\), inner radius \(a\), and outer radius \(b\), which has been subjected to an external field \(\mathbf{B}_\text{o}\). Finding the fields on the exterior, in the shell, and on the interior requires finding a set of fields that satisfies five boundary conditions: (1) far from the sphere, the field must approach the constant \(\mathbf{B}_\text{o}\); (2) at the outer surface of the sphere, the field must have \(\mathbf{H}_{\parallel,1}=\mathbf{H}_{\parallel,2}\), as discussed on page 717; (3) the same constraint applies at the inner surface of the sphere; (4) and (5) there is an additional constraint on the fields at the inner and outer surfaces, as found in problem 46. The goal of this problem is to find the solution for the fields, and from it, to prove that the interior field is uniform, and given by

\[\begin{equation*} \mathbf{B} = \left[\frac{9x}{(2x+1)(x+2)-2\frac{a^3}{b^3}(x-1)^2}\right]\mathbf{B}_\text{o} . \end{equation*}\]

This is a very difficult problem to solve from first principles, because it's not obvious what form the fields should have, and if you hadn't been told, you probably wouldn't have guessed that the interior field would be uniform. We could, however, guess that once the sphere becomes polarized by the external field, it would become a dipole, and at \(r\gg b\), the field would be a uniform field superimposed on the field of a dipole. It turns out that even close to the sphere, the solution has exactly this form. In order to complete the solution, we need to find the field in the shell (\(a\ltr\ltb\)), but the only way this field could match up with the detailed angular variation of the interior and exterior fields would be if it was also a superposition of a uniform field with a dipole field. The final result is that we have four unknowns: the strength of the dipole component of the external field, the strength of the uniform and dipole components of the field within the shell, and the strength of the uniform interior field. These four unknowns are to be determined by imposing constraints (2) through (5) above.
(b) Show that the expression from part a has physically reasonable behavior in its dependence on \(x\) and \(a/b\).

48. Two long, parallel strips of thin metal foil form a configuration like a long, narrow sandwich. The air gap between them has height \(h\), the width of each strip is \(w\), and their length is \(\ell\). Each strip carries current \(I\), and we assume for concreteness that the currents are in opposite directions, so that the magnetic force, \(F\), between the strips is repulsive.
(a) Find the force in the limit of \(w\gg h\).(answer check available at
(b) Find the force in the limit of \(w\ll h\), which is like two ordinary wires.
(c) Discuss the relationship between the two results.

49. Suppose we are given a permanent magnet with a complicated, asymmetric shape. Describe how a series of measurements with a magnetic compass could be used to determine the strength and direction of its magnetic field at some point of interest. Assume that you are only able to see the direction to which the compass needle settles; you cannot measure the torque acting on it.

50. On page 684, the curl of \(x\hat{\mathbf{y}}\) was computed. Now consider the fields \(x\hat{\mathbf{x}}\) and \(y\hat{\mathbf{y}}\).
(a) Sketch these fields.
(b) Using the same technique of explicitly constructing a small square, prove that their curls are both zero. Do not use the component form of the curl; this was one step in deriving the component form of the curl.

51. If you watch a movie played backwards, some vectors reverse their direction. For instance, people walk backwards, with their velocity vectors flipped around. Other vectors, such as forces, keep the same direction, e.g., gravity still pulls down. An electric field is another example of a vector that doesn't turn around: positive charges are still positive in the time-reversed universe, so they still make diverging electric fields, and likewise for the converging fields around negative charges.
(a) How does the momentum of a material object behave under time-reversal?(solution in the pdf version of the book)
(b) The laws of physics are still valid in the time-reversed universe. For example, show that if two material objects are interacting, and momentum is conserved, then momentum is still conserved in the time-reversed universe.(solution in the pdf version of the book)
(c) Discuss how currents and magnetic fields would behave under time reversal. \hwhint{hwhint:timereversalem}
(d) Similarly, show that the equation \(d\mathbf{p}\propto\mathbf{E}\times\mathbf{B}\) is still valid under time reversal.

52. This problem is a more advanced exploration of the time-reversal ideas introduced in problem 51.
(a) In that problem, we assumed that charge did not flip its sign under time reversal. Suppose we make the opposite assumption, that charge does change its sign. This is an idea introduced by Richard Feynman: that antimatter is really matter traveling backward in time! Determine the time-reversal properties of \(\mathbf{E}\) and \(\mathbf{B}\) under this new assumption, and show that \(d\mathbf{p}\propto\mathbf{E}\times\mathbf{B}\) is still valid under time-reversal.
(b) Show that Maxwell's equations are time-reversal symmetric, i.e., that if the fields \(\mathbf{E}(x,y,z,t)\) and \(\mathbf{B}(x,y,z,t)\) satisfy Maxwell's equations, then so do \(\mathbf{E}(x,y,z,-t)\) and \(\mathbf{B}(x,y,z,-t)\). Demonstrate this under both possible assumptions about charge, \(q\rightarrow q\) and \(q\rightarrow -q\).

53. The purpose of this problem is to prove that the constant of proportionality \(a\) in the equation \(dU_m=aB^2 dv\), for the energy density of the magnetic field, is given by \(a=c^2/8\pi k\) as asserted on page 669. The geometry we'll use consists of two sheets of current, like a sandwich with nothing in between but some vacuum in which there is a magnetic field. The currents are in opposite directions, and we can imagine them as being joined together at the ends to form a complete circuit, like a tube made of paper that has been squashed almost flat. The sheets have lengths \(L\) in the direction parallel to the current, and widths \(w\). They are separated by a distance \(d\), which, for convenience, we assume is small compared to \(L\) and \(w\). Thus each sheet's contribution to the field is uniform, and can be approximated by the expression \(2\pi k\eta/c^2\).
(a) Make a drawing similar to the one in figure 11.2.1 on page 668, and show that in this opposite-current configuration, the magnetic fields of the two sheets reinforce in the region between them, producing double the field, but cancel on the outside.
(b) By analogy with the case of a single strand of wire, one sheet's force on the other is \(ILB_1\), were \(I=\eta w\) is the total current in one sheet, and \(B_1=B/2\) is the field contributed by only one of the sheets, since the sheet can't make any net force on itself. Based on your drawing and the right-hand rule, show that this force is repulsive.
For the rest of the problem, consider a process in which the sheets start out touching, and are then separated to a distance \(d\). Since the force between the sheets is repulsive, they do mechanical work on the outside world as they are separated, in much the same way that the piston in an engine does work as the gases inside the cylinder expand. At the same time, however, there is an induced emf which would tend to extinguish the current, so in order to maintain a constant current, energy will have to be drained from a battery. There are three types of energy involved: the increase in the magnetic field energy, the increase in the energy of the outside world, and the decrease in energy as the battery is drained. (We assume the sheets have very little resistance, so there is no ohmic heating involved.)(answer check available at
(c) Find the mechanical work done by the sheets, which equals the increase in the energy of the outside world. Show that your result can be stated in terms of \(\eta\), the final volume \(v=wLd\), and nothing else but numerical and physical constants.(answer check available at
(d) The power supplied by the battery is \(P=I\Gamma_E\) (like \(P=I\Delta V\), but with an emf instead of a voltage difference), and the circulation is given by \(\Gamma=-d\Phi_B/dt\). The negative sign indicates that the battery is being drained. Calculate the energy supplied by the battery, and, as in part c, show that the result can be stated in terms of \(\eta\), \(v\), and universal constants.(answer check available at
(e) Find the increase in the magnetic-field energy, in terms of \(\eta\), \(v\), and the unknown constant \(a\).(answer check available at
(f) Use conservation of energy to relate your answers from parts c, d, and e, and solve for \(a\).(answer check available at

54. Magnet coils are often wrapped in multiple layers. The figure shows the special case where the layers are all confined to a single plane, forming a spiral. Since the thickness of the wires (plus their insulation) is fixed, the spiral that results is a mathematical type known as an Archimedean spiral, in which the turns are evenly spaced. The equation of the spiral is \(r=w\theta\), where \(w\) is a constant. For a spiral that starts from \(r=a\) and ends at \(r=b\), show that the field at the center is given by \((kI/c^2w)\ln b/a\).(solution in the pdf version of the book)


Exercise B: Polarization


calcite (Iceland spar) crystal

polaroid film

1. Lay the crystal on a piece of paper that has print on it. You will observe a double image. See what happens if you rotate the crystal.

Evidently the crystal does something to the light that passes through it on the way from the page to your eye. One beam of light enters the crystal from underneath, but two emerge from the top; by conservation of energy the energy of the original beam must be shared between them. Consider the following three possible interpretations of what you have observed:

(a) The two new beams differ from each other, and from the original beam, only in energy. Their other properties are the same.

(b) The crystal adds to the light some mysterious new property (not energy), which comes in two flavors, X and Y. Ordinary light doesn't have any of either. One beam that emerges from the crystal has some X added to it, and the other beam has Y.

(c) There is some mysterious new property that is possessed by all light. It comes in two flavors, X and Y, and most ordinary light sources make an equal mixture of type X and type Y light. The original beam is an even mixture of both types, and this mixture is then split up by the crystal into the two purified forms.

In parts 2 and 3 you'll make observations that will allow you to figure out which of these is correct.

2. Now place a polaroid film over the crystal and see what you observe. What happens when you rotate the film in the horizontal plane? Does this observation allow you to rule out any of the three interpretations?

3. Now put the polaroid film under the crystal and try the same thing. Putting together all your observations, which interpretation do you think is correct?

4. Look at an overhead light fixture through the polaroid, and try rotating it. What do you observe? What does this tell you about the light emitted by the lightbulb?

5. Now position yourself with your head under a light fixture and directly over a shiny surface, such as a glossy tabletop. You'll see the lamp's reflection, and the light coming from the lamp to your eye will have undergone a reflection through roughly a 180-degree angle (i.e., it very nearly reversed its direction). Observe this reflection through the polaroid, and try rotating it. Finally, position yourself so that you are seeing glancing reflections, and try the same thing. Summarize what happens to light with properties X and Y when it is reflected. (This is the principle behind polarizing sunglasses.)

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

[1] For a more practical demonstration of this effect, you can put an ordinary magnet near a computer monitor. The picture will be distorted. Make sure that the monitor has a demagnetizing (“degaussing”) button, however! Otherwise you may permanently damage it. Don't use a television tube, because TV tubes don't have demagnetizing buttons.
[2] One could object that this is circular reasoning, since the whole purpose of this argument is to prove from first principles that magnetic effects follow from the theory of relativity. Could there be some extra interaction which occurs between a moving charge and any other charge, regardless of whether the other charge is moving or not? We can argue, however, that such a theory would lack self-consistency, since we have to define the electric field somehow, and the only way to define it is in terms of \(F/q\), where \(F\) is the force on a test charge \(q\) which is at rest. In other words, we'd have to say that there was some extra contribution to the electric field if the charge making it was in motion. This would, however, violate Gauss' law, and Gauss' law is amply supported by experiment, even when the sources of the electric field are moving. It would also violate the time-reversal symmetry of the laws of physics.
[3] The reader who wants to see the full relativistic treatment is referred to E.M. Purcell, Electricity and Magnetism, McGraw Hill, 1985, p. 174.
[4] Strictly speaking, there is a hole in this logic, since I've only ruled out a field that is purely along one of these three perpendicular directions. What if it has components along more than one of them? A little more work is required to eliminate these mixed possibilities. For example, we can rule out a field with a nonzero component parallel to the wire based on the following symmetry argument. Suppose a charged particle is moving in the plane of the page directly toward the wire. If the field had a component parallel to the wire, then the particle would feel a force into or out of the page, but such a force is impossible based on symmetry, since the whole arrangement is symmetric with respect to mirror-reflection across the plane of the page.
[5] If you've taken a course in differential equations, this won't seem like a very surprising assertion. The differential form of Gauss' law is a differential equation, and by giving the value of the field in the midplane, we've specified a boundary condition for the differential equation. Normally if you specify the boundary conditions, then there is a unique solution to the differential equation. In this particular case, it turns out that to ensure uniqueness, we also need to demand that the solution satisfy the differential form of Ampère's law, which is discussed in section 11.4.
[6] If you didn't read this optional subsection, don't worry, because the point is that we need to try a whole new approach anyway.
[7] Note that the magnetic field never does work on a charged particle, because its force is perpendicular to the motion; the electric power is actually coming from the mechanical work that had to be done to spin the coil. Spinning the coil is more difficult due to the presence of the magnet.
[8] If the pump analogy makes you uneasy, consider what would happen if all the electrons moved into the page on both sides of the loop. We'd end up with a net negative charge at the back side, and a net positive charge on the front. This actually would happen in the first nanosecond after the loop was set in motion. This buildup of charge would start to quench both currents due to electrical forces, but the current in the right side of the wire, which is driven by the weaker magnetic field, would be the first to stop. Eventually, an equilibrium will be reached in which the same amount of current is flowing at every point around the loop, and no more charge is being piled up.
[9] The wire is not a perfect conductor, so this current produces heat. The energy required to produce this heat comes from the hands, which are doing mechanical work as they separate the magnet from the loop.
[10] They can't be blamed too much for this. As a consequence of Faraday's work, it soon became apparent that light was an electromagnetic wave, and to reconcile this with the relative nature of motion requires Einstein's version of relativity, with all its subversive ideas how space and time are not absolute.
[11] One way to prove this rigorously is that in a frame of reference where the particle is at rest, it has an electric field that surrounds it on all sides. If the particle has been moving with constant velocity for a long time, then this is just an ordinary Coulomb's-law field, extending off to very large distances, since disturbances in the field ripple outward at the speed of light. In a frame where the particle is moving, this pure electric field is experienced instead as a combination of an electric field and a magnetic field, so the magnetic field must exist throughout the same vast region of space.
[12] Even if the fields can't be parallel to the direction of propagation, one might wonder whether they could form some angle other than 90 degrees with it. No. One proof is given on page 707. A alternative argument, which is simpler but more esoteric, is that if there was such a pattern, then there would be some other frame of reference in which it would look like figure i.
[13] A young Einstein worried about what would happen if you rode a motorcycle alongside a light wave, traveling at the speed of light. Would the light wave have a zero velocity in this frame of reference? The only solution lies in the theory of relativity, one of whose consequences is that a material object like a student or a motorcycle cannot move at the speed of light.
[14] Actually, this is only exactly true of the rectangular strip is made infinitesimally thin.
[15] You may know already that different colors of light have different speeds when they pass through a material substance, such as the glass or water. This is not in contradiction with what I'm saying here, since this whole analysis is for light in a vacuum.
[16] What makes them appear to be unrelated phenomena is that we experience them through their interaction with atoms, and atoms are complicated, so they respond to various kinds of electromagnetic waves in complicated ways.
[17] This current will soon come to a grinding halt, because we don't have a complete circuit, but let's say we're talking about the first picosecond during which the radio wave encounters the wire. This is why real radio antennas are not very short compared to a wavelength!