Problem Set 4

(6.1)

Prove the BAC–CAB rule $A \times (B \times C) = B(A \cdot C) - C (A \cdot B)$ by writing it out in the summations convention, and use it to show that $\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \nabla^2 E$ .

I’m going to use the Levi-Civita symbol. I’m also going to use $i_+$ to represent $(i + 1) \text{ mod } 3$ for any coordinate index $i$ . So if $i$ is 2, then $i_+ = 3$ and $i_{+ +} = 1$ .

The $i$ th coordinate of $A \times (B \times C)$ is

$\begin{align*} \epsilon_{i j k} A_j \epsilon_{k l m} B_l C_m &= A_{i_+} \epsilon_{i_{+ +} l m} B_l C_m - A_{i_{+ +}} \epsilon_{i_+ l m} B_l C_m \\ &= A_{i_+} (B_i C_{i_+} - B_{i_+} C_i) - A_{i_{+ +}} (B_{i_{+ +}} C_i - B_i C_{i_{+ +}}) \\ &= A_i B_i C_i + A_{i_+} B_i C_{i_+} + A_{i_{+ +}} B_i C_{i_{+ +}} - \\ &\phantom{x=} A_i B_i C_i - A_{i_+} B_{i_+} C_i - A_{i_{+ +}} B_{i_{+ +}} C_i \\ &= B_i A_j C_j - C_i A_j B_j \end{align*}$

which is the $i$ th coordinate of $B(A \cdot C) - C (A \cdot B)$ .

If we take $\nabla$ to be a three-vector of partial differentiation operators, then this shows that

$\begin{align*} \nabla \times (\nabla \times E) &= \nabla (\nabla \cdot E) - E (\nabla \cdot \nabla) \\ &= \nabla (\nabla \cdot E) - \nabla^2 E \end{align*}$

(6.2)

(a)

Use Gauss’ Law to find the capacitance between two parallel plates of area A at a potential difference V and with a spacing d. Neglect the fringing fields by assuming that this is a section of an infinite capacitor.

The capacitance between two objects is the ratio of their charge to their potential difference. (This ratio does not depend on the charge because of the linearity of Maxwell’s equations.) So to find the capacitance between two plates with a potential difference V, we need to know what charge is required to generate that potential difference.

Consider a single infinite plane with a charge density $\sigma$ . By symmetry the field must be perpendicular to the plane everywhere. For energy to be conserved this means that the field doesn’t vary with distance. Consider a section of area $A$ on the plane. If we expand this symmetrically to a box that is bisected by the plane, Gauss’ Law tells us that $2 A E = \sigma A / \epsilon_0$ , so the field strength $E$ is $\sigma / (2 \epsilon_0)$ .

Thus between two planes with opposite charge densities $\sigma$ and $-\sigma$ , the field is $\sigma / \epsilon_0$ everywhere. (Outside the field is zero, since the fields from the two planes cancel.) This means the potential difference between the planes is $d \sigma / \epsilon_0$ . As such the capacitance is $Q / V = A \sigma \epsilon_0 / (d \sigma) = A \epsilon_0 / d$ .

(b)

Show that when a current flows through the capacitor, the integral over the internal displacement current is equal to the external electrical current.

The displacement current is the derivative of $D$ with respect to time. By definition $D = \epsilon_0 E$ , and in the last problem we found that $E = \sigma / \epsilon_0$ . So here $D = \sigma$ . Since the charge per area $A$ on one of the capacitor’s plates is $Q = A \sigma$ , we can deduce that $I = d Q / d t = A d \sigma / d t$ . Thus $\partial D / \partial t = I / A$ , so if we integrate over area $A$ we find that the displacement current is $I$ as expected.

(c)

Integrate the energy density to find the stored energy at a fixed potential. The answer should be expressed in terms of the capacitance.

$U = \frac{1}{2} (E \cdot D + B \cdot H)$

At a fixed potential $V$ all charge is static, so there is no magnetic field. So since $D = \epsilon_0 E$ , $U = \epsilon_0 E^2 / 2$ . Plugging in $E = \sigma / \epsilon_0$ and integrating over a volume $A d$ we find that the energy (in Joules) stored in the electric field is

$\begin{align*} \frac{A d \sigma^2}{2 \epsilon_0} &= \frac{1}{2} \frac{A \epsilon_0}{d} \frac{d^2 \sigma^2}{\epsilon_0^2} \\ &= \frac{1}{2} C V^2 \end{align*}$

(d)

Batteries are rated by amp-hours, the current they can supply at the design voltage for an hour. Consider a 10 V laptop battery that provides 10 A · h. Assuming a plate spacing of $10^{−6} \si{m} \equiv 1 \si{\mu m}$ and a vacuum dielectric, what area would a capacitor need to be able to store this amount of energy? If such plates were 10 cm on a side and stacked vertically, how tall would the stack have to be to provide this total area?

The battery can provide $10 \si{V} \cdot 10 \si{A} \cdot 1 \si{hour} = 360 \si{kJ}$ of energy. Assuming we’re charging the capacitor to $10 \si{V}$ ,

$\begin{align*} C &= \frac{2 E}{V^2} \\ &= \frac{2 \cdot \num{3.6e5} \si{J}}{100 \si{V^2}} \\ &= 7.2 \si{kF} \end{align*}$

I don’t think I’ve ever seen kilofarads used before. To get this much capacitance we’d need an area of

$\begin{align*} A &= \frac{C d}{\epsilon_0} \\ &= \frac{\num{7.2e3} \si{F} \cdot 10^{-6} \si{m}}{\num{8.85e-12} \si{F/m}} \\ &= \num{8e8} \si{m^2} \end{align*}$

This is one quarter of the area of Rhode Island. We’d need $\num{8e10}$ squares with edge length 10cm to get this area. If the plates have negligible thickness (relative to their spacing), they’d be 81 kilometers high when stacked on top of each other.

(6.3)

(a)

Use Stokes’ Law to find the magnetic field of an infinite solenoid carrying a current I with n turns/meter.

Let the solenoid lie along the x axis, with right handed current. Consider an axis aligned square in the xy (or xz) plane, where the edges aligned with the x axis are one meter long, with one fully inside the solenoid one fully outside. By Ampère’s Law, the integral of the magnetic field along the boundary of this square is nI. This is independent of the length of the y (or z) aligned edges. So the field must be constant inside and out. Very far away from the solenoid the field should be zero, since the fields from the opposing sides cancel. Thus the field outside the solenoid is zero everywhere, and the field inside the solenoid is nI everywhere (pointing along the x axis). So assuming a vacuum inside the solenoid, we have $H = nI$ and $B = \mu_0 n I$ .

(b)

Integrate the energy density to find the energy stored in a solenoid of radius r and length l, once again neglecting fringing fields.

In this case we have no electric field, so $U = B \cdot H / 2 = \mu_0 n^2 I^2 / 2$ . Integrated over a cylinder of radius $r$ and length $l$ , we have a total energy of

$\frac{1}{2} \mu_0 n^2 I^2 \pi r^2 l$

(c)

Consider a 10 T MRI magnet (Section 10.4) with a bore diameter of 1 m and a length of 2 m. What is the outward force on the magnet? Remember – force is the gradient of potential for a conservative force.

$\begin{align*} \frac{\partial}{\partial r} U &= \frac{\partial}{\partial r} \frac{1}{2} \frac{B^2}{\mu_0} \pi r^2 l \\ &= \frac{B^2}{\mu_0} \pi r l \\ &= \frac{100 \si{T^2}}{4 \pi \times 10^{-7} \si{H/m}} \cdot \pi \cdot 0.5 \si{m} \cdot 2 \si{m} \\ &= \num{2.5e8} \si{N} \end{align*}$

Yes, there is a sign error that I’m ignoring for now. The force should be the negative of the potential gradient. I think if we explicitly calculated the potential of the solenoid loop by loop (i.e. constructing it starting with all the loops infinitely far apart), we’d find that the potential is -1 times what we found above, thus fixing the problem.

(6.4)

The ampere was formerly defined [BIPM, 2014] as “The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to $\num{2e-7}$ newton per metre of length.”

(a)

Show that that current at that distance produces that force.

First let’s find the magnetic field of an infinitely long straight conductor. Let’s use a cylindrical coordinate system along this axis. Considering the Biot-Savart Law and the symmetry of this problem, the magnetic field must be oriented along $\hat{\mathrm{d} \theta}$ , with a magnitude that depends only on $r$ . Consider then a circle of radius $r$ centered on the wire. Ampère’s Law tells us that the magnitude of the field at any point on this circle is $I / (2 \pi r)$ .

The differential force exerted by this field on a differential piece of current is $dF = I (dl \times B)$ . In this case the direction of the current and the magnetic field are perpendicular, so the direction of the cross product is always toward the other wire. The magnitude of the force per meter is

$\begin{align*} F &= \frac{\mu_0 I^2}{2 \pi r} \\ &= \frac{4 \pi \times 10^{-7} \si{H/m} \cdot 1 \si{A^2}}{2 \pi \cdot 1 \si{m}} \\ &= \num{2e-7} \si{N} \end{align*}$

Note that we don’t need to multiply this by two to account for the force wire two exerts on wire one. Newton’s third law guarantees that this is equal and opposite, so it only makes sense to talk about the force as a pair. Naturally Maxwell’s equations respect this law (otherwise one could build an electromagnetic perpetual motion machine).

(b)

What is the problem with defining the ampere this way?

It’s not a very practical experiment. No wires are infinitely long, so you’ll necessarily have fringe fields to factor into your results. No conductors are infinitely thin, so their response to an applied field depends on their specific geometry. And you have to hold the wires and supply current to them somehow, so they can’t be totally surrounded by a vacuum. Not to mention we can’t create perfect vacuums anyway.

(6.5)

The definition of the kilogram [BIPM, 2019] based on fundamental constants [Stock et al., 2019] is realized with a Kibble balance [Kibble, 1976]. This has a coil in a spatially inhomogeneous magnetic field on one side of the balance, with a weighing pan on the other side. The measurement is performed in two stages.

(a)

In the static phase, a current is passed through the coil to generate a force equal to the gravitational force on the weight. Calculate how the vertical component of the force on the coil relates to the current through it and the vertical gradient of the magnetic field (hint: use the definition of the force on an infinitesimal current, Green’s theorem, and the formula for the divergence of a magnetic field).

Instead of a coil, I’ll just consider one current loop. A coil can then be modeled as a stack of such loops. Assume that our loop is centered at the origin and lies in the xy plane. The differential force generated on a differential current by a magnetic field is $dF = I (dl \times B)$ . In a uniform magnetic field the forces cancel out, so let’s consider a linearly nonuniform magnetic field

$B(x) = B_0 + D[B] x$

where $B(x)$ , $x$ , and $B_0$ are 3-vectors and $D[B]$ is the (3x3) Jacobian matrix of $B(x)$ .

Then the $z$ component of $I (dl \times B)$ at any location $(x, y, 0)$ on our loop is

$I dl_x \left( B_{0x} + \frac{\partial B_y}{\partial x} x + \frac{\partial B_y}{\partial y} y \right) - I dl_y \left( B_{0y} + \frac{\partial B_x}{\partial x} x + \frac{\partial B_x}{\partial y} y \right)$

Let’s move to radial coordinates to make the inevitable integration easier. Let’s say our loop has radius $r$ . For a point on our loop at angular position $\theta$ ,

$dl = (-r \mathrm{d} \theta \sin{\theta}, r \mathrm{d} \theta \cos{\theta}, 0)$

Thus the $z$ component of $I (dl \times B)$ is

$-I r \mathrm{d} \theta \sin{\theta} \left( B_{0x} + \frac{\partial B_y}{\partial x} \cos{\theta} + \frac{\partial B_y}{\partial y} \sin{\theta} \right) \\ - I r \mathrm{d} \theta \cos{\theta} \left( B_{0y} + \frac{\partial B_x}{\partial x} \cos{\theta} + \frac{\partial B_x}{\partial y} \sin{\theta} \right)$

We need to integrate this with respect to $\theta$ from 0 to $2 \pi$ . Though the expression above looks complicated, the integral ends up being easy. To spare all the algebra, any term that only involves $\cos{\theta}$ or $\sin{\theta}$ integrates to zero since we’re integrating over one complete period. The same happens for both terms that have $\cos{\theta} \sin{\theta}$ . So there are only two terms left, and they reduce to $\cos^2{\theta}$ and $\sin^2{\theta}$ , both of which integrate to $\pi$ . So in the end the $z$ component of the force on the loop is

$-I \pi r^2 \left( \frac{\partial B_y}{\partial y} + \frac{\partial B_x}{\partial x} \right) = I \pi r^2 \frac{\partial B_z}{\partial z}$

since $\nabla \cdot B = 0$ .

The current is adjusted until this force exactly balances the weight of the mass, so we have

$mg = I \pi r^2 \frac{\partial B_z}{\partial z}$

where $g = 9.8 \si{m/s^2}$ is the acceleration due to gravity. (To get a really precise measurement of $m$ , you’d want to measure $g$ at your lab specifically since it varies a bit with local geography and geology.)

(b)

In the dynamic phase, the coil is moved at a constant vertical velocity and the voltage across it is measured. Calculate how the voltage depends on the velocity and vertical gradient of the magnetic field.

The induced voltage is proportional to the time rate of change of the magnetic flux through the current loop. The flux is $\pi r^2 B_z$ . If we move the loop along the z axis with a velocity $v$ , the rate of change of the flux is

$\pi r^2 v \frac{\partial B_z}{\partial z}$

So the voltage is

$V = - \pi r^2 v \frac{\partial B_z}{\partial z}$

(c)

Combine these results to express the weight in terms of the voltage and current.

From part (a) we have

$\pi r^2 \frac{\partial B_z}{\partial z} = \frac{mg}{I}$

and from part (b)

$\pi r^2 \frac{\partial B_z}{\partial z} = -\frac{V}{v}$

Thus

$m = -\frac{I V}{g v}$

If we use a coil with $n$ windings instead of a single loop then we just need to replace $I$ with $n I$ .

(a), (b), and (c) without the assumption of linearity

Now let’s drop the assumption that the magnetic field varies linearly with distance. After all, though this assumption is reasonable over small distances, it’s hard to justify over the whole macroscopic length of our current loop. Instead, let’s assume that the field is radially symmetric. This is a reasonable assumption since if the field isn’t at least periodically radially symmetric then there will be a torque on the loop.

Call the radial component of the field experienced by our loop $B_r(z)$ . (There must also be some z component, but we won’t end up needing it so we won’t give it a name.)

Now calculating the force on the loop is easy. The z component of $B$ doesn’t matter since it just pulls the loop outward. And the radial component produces a force along z. So the force is $-2 \pi r I B_r(z)$ . So when the scale is balanced

$mg = -2 \pi r I B_r(z)$

But how can we calculate the flux? Consider a z-aligned cylinder of radius $r$ and height $z$ , with the center of its base at the origin. The magnetic flux through the bottom is some constant amount $\Phi_\text{bottom}$ . (We could arrange for this to be zero, either by placing our origin very far away, or by using two magnets in a mirrored configuration, as is the case with NIST’s balance. But it doesn’t matter for this analysis.) The flux through the walls is

$\Phi_\text{walls}(z) = 2 \pi r \int_0^z B_r(z') \mathrm{d}z'$

Then since the divergence of $B$ is zero everywhere, Green’s theorem tells us that

$\Phi_\text{top}(z) = -\Phi_\text{walls}(z) - \Phi_\text{bottom}$

So by the fundamental theorem of calculus

$\frac{\partial}{\partial z} \Phi_\text{top}(z) = -2 \pi r B_r(z)$

Thus when our loop travels at a fixed velocity $v$ the induced voltage is

$2 \pi r v B_r(z)$

Solving these equations for $m$ yields

$m = -\frac{I V}{g v}$

as before.

(d)

Why are the voltage and current measured separately, rather than by doing a measurement of both at the same time?

Well the voltage is only generated when the coil is moving, but the current is defined as that which keeps the coil still. So you can’t do both at the same time. You could start the coil moving and then measure the current that keeps it moving at a constant velocity, simultaneously measuring the resulting voltage. But in this scenario some of the current is probably being used to overcome friction in the mass balance, rather than just counteract gravity. So you’d have to have a really accurate model of that, which isn’t practically possible.

At the end of the day the point is to have the effects of the magnetic field cancel out, so it doesn’t need to be perfect as it did in the previous definition of the ampere. In other words, we just want to measure a voltage and a current and have that be enough. We don’t want to worry about the specifics of the magnetic field, or the dynamic response of the mass balance, etc.

(6.6)

(a)

Assume that sunlight has a power energy density of $1 \si{kW/m^2}$ (this is a peak number; the typical average value in the continental USA is $\approx 200 \si{W/m^2}$ ). Estimate the electric field strength associated with this radiation.

We need a relation between power, area, and electric field strength. We should be able to derive this from a volumetric integral over U, or a surface integral over P. Let’s try both.

Using U

The energy density of light in a vacuum is

$U = \frac{1}{2} (E \cdot D + B \cdot H)$

For light propagating through free space, we know $\vert E \vert / \vert H \vert = \sqrt{\mu_0/\epsilon_0}$ . So we can simplify to

$\begin{align*} U &= \frac{1}{2} \left( \epsilon_0 |E|^2 + \mu_0 |H|^2 \right) \\ &= \frac{1}{2} \left( \epsilon_0 |E|^2 + \epsilon_0 |E|^2 \right) \\ &= \epsilon_0 |E|^2 \end{align*}$

Since the strength of the electric field is oscillating sinusoidally, on average the squared field strength is one half the squared field strength squared:

$\langle U \rangle = \epsilon_0 \langle |E|^2 \rangle = \frac{\epsilon_0}{2} E_\text{max}^2$

The energy delivered per square meter per second by light is the energy stored in a 1 meter by 1 meter by 1 light second volume. So the power W (in watts) delivered to an area A (in square meters) is

$W = \frac{A c \epsilon_0}{2} E_\text{max}^2$

The speed of light is $c = (\epsilon_0 \mu_0)^{-1/2}$ , so we can rewrite this as

$W = \frac{A}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2$

Using P

The Poynting Vector gives the direction and magnitude of energy flux due to the electric and magnetic fields. In MKS units it is measured in units of watts per square meter and is defined as $P = E \times H$ .

For light in free space, E and H are perpendicular to each other and to the direction of travel. Thus the Poynting vector points in the direction of travel. We also know that $\vert E \vert / \vert H \vert = \sqrt{\mu_0/\epsilon_0}$ , so its magnitude is

$|P| = |E| |H| = \sqrt{\frac{\epsilon_0}{\mu_0}} |E|^2$

The E field oscillates sinusoidally, so the time average of its squared magnitude is one half its peak magnitude squared.

$\langle |P| \rangle = \sqrt{\frac{\epsilon_0}{\mu_0}} \langle |E|^2 \rangle = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2$

So the average power W (in watts) delivered to an area A (in square meters) will be

$W = \frac{A}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2$

as expected.

The field strength of sunlight

Solving for the max field strength, we find that

$E_\text{max}^2 = \frac{2 W}{A} \sqrt{\frac{\mu_0}{\epsilon_0}}$

So given that strong sunlight is one kilowatt per square meter, we can compute

$\begin{align*} E_\text{max}^2 &= \frac{2 \cdot 10^3 \si{(W)}}{1 \si{(m^2)}} \sqrt{\frac{\mu_0}{\epsilon_0}} \si{(\ohm)} \\ &= 753 \si{V^2/m^2} \\ \end{align*}$

So the peak field strength is $868 \si{V/m}$ .

(b)

If 1 W of power is focused in a laser beam to a square millimeter, what is the field strength? What about if it is focused to the diffraction limit of $\approx 1 \si{\mu m^2}$ ?

For one watt in one square millimeter,

$\begin{align*} E_\text{max}^2 &= \frac{2 \cdot 1 \si{W}}{10^{-6} \si{m^2}} \sqrt{\frac{\mu_0}{\epsilon_0}} \\ &= \num{7.5e8} \si{V^2/m^2} \\ E_\text{max} &= 27 \si{kV/m} \end{align*}$

For one watt in one square micrometer,

$\begin{align*} E_\text{max}^2 &= \frac{2 \cdot 1 \si{W}}{10^{-12} \si{m^2}} \sqrt{\frac{\mu_0}{\epsilon_0}} \\ &= \num{7.5e14} \si{V^2/m^2} \\ E_\text{max} &= \num{2.7e7} \si{V/m} \end{align*}$