Publish fourth pset

_psets/4.md

+---
+title: Problem Set 4
+---
+
+## (6.1)
+
+{:.question}
+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](https://en.wikipedia.org/wiki/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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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](https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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](https://en.wikipedia.org/wiki/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)
+
+{:.question}
+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)
+
+{:.question}
+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](https://en.wikipedia.org/wiki/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)
+
+{:.question}
+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,
+
+$$
+\begin{align*}
+E_\text{max}^2
+&= \frac{2 \cdot 1 \si{W}}{10^{-12} \si{m^2}} \sqrt{\frac{\mu_0}{\epsilon_0}} \\
+&= \num{7.5e14} \si{V^2/m^2} \\
+E_\text{max} &= \num{2.7e7} \si{V/m}
+\end{align*}
+$$