2003 Spring Electromagnetism
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1. [35 points] Time-Dependent Fields

The scalar and vector potentials resulting from an arbitrary localized source are

\begin{align} V(\textbf{r},t) = \frac{1}{4\pi\epsilon_0} \int d^3\textbf{r}' \frac{\rho(\textbf{r}',t_r)}{|\textbf{r}-\textbf{r}'|},\quad \textbf{A}(\textbf{r},t) = \frac{\mu_0}{4\pi}\int \,d^3\textbf{r}' \frac{\textbf{j}(\textbf{r}',t_r)}{|\textbf{r}-\textbf{r}'|} \end{align}

A. Define $\rho,\textbf{j},\textbf{r},\textbf{r}'$. Define $t_r$ in terms of $\textbf{r},\textbf{r}',t$ and the speed of light, c.

Greg's Solution:

charge density
current density
point at which the field is being evaluated
point at which the charge and current densities are being evaluated; variable of integration

$t_r$ is the retarded time, needed to account for the fact that it takes time for information regarding changes in the configuration of charge and current to travel through space; it is given by

\begin{align} t_r := t-\frac{|\textbf{r}-\textbf{r}'|}{c} \end{align}

After time $t=0$, a time-dependent uniform current density $\textbf{K}(t)$ flows in the infinite xy plane at $z=0$ — i.e., $\textbf{K}(t)=\hat e_y K_0(t)\delta(z)$, where $\delta(z)$ is a Dirac-delta function, $\hat e_y$ is the unit-vector in the y-direction, and with $K_0(t)=0$ for $t<0$. THere are no free charges on the sheet, or anywhere in space, at any time.

B. For a point located at $x=x_0, y=y_0$ and $z=z_0$, give the contributions to $V(x_0,y_0,z_0,t)$ and $\textbf{A}(x_0,y_0,z_0,t)$ from the current in a ring of radius s and width ds in the xy-plane centered on $x=x_0,y=y_0,z=0$.

Greg's Solution:

All points along the ring are equidistant from $(x_0,y_0,z_0$ and contain the same current density, so we conclude that the contribution to the vector potential is,

\begin{align} d\textbf{A}(r,t) = \frac{\mu_0}{4\pi} \frac{2\pi s\,ds\, K_0(t_r) \hat{\textbf{e}}_y}{\sqrt{z_0^2+s^2}},\quad t_r := t-\frac{\sqrt{s^2+z_0^2}}{c} \end{align}

C. [3 points] At any time, t, what is the maximum value of s that can contribute to $V(x_0,y_0,z_0)$ and $\textbf{A}(x_0,y_0,z_0,t)$.

Greg's Solution:

Since $K(t)=0$ for $t<0$, we need $t_r\ge 0$ and thus $\frac{\sqrt{s^2+z_0^2}}{c}<t \Rightarrow s < \sqrt{c^2t^2-z_0^2}$.

D. [12 points]

Show that

\begin{align} \textbf{A}(x_0,y_0,z_0,t) = \hat{\textbf{e}}_y \theta\left(t-\frac{z_0}{c}\right)\int_0^{t-\frac{z_0}{c}}\,d\eta\,f(\eta), \end{align}

where \theta(x) is the step-function. Find $f(\eta)$.

Greg's Solution:

We integrate over the differential vector potential found in part B from $s=0$ to $s= \sqrt{c^2t^2-z_0^2}$ (the latter bound being the answer to the previous part),

\begin{aligned} \textbf{A}(x_0,y_0,z_0,t) &= \hat{\textbf{e}}_z \frac{\mu_0}{2} \theta\left(t-\frac{z_0}{c}\right) \int_0^{\sqrt{c^2t^2-z_0^2}} \frac{s\,ds\, K_0\left(t-\frac{\sqrt{s^2+z_0^2}}{c}\right) }{\sqrt{z_0^2+s^2}}, \end{aligned}

where the step-function is required since it takes a minimum of $z_0/c$ time for news that the current has been switched on to reach point $(x_0,y_0,z_0$. We make the change of variables,

\begin{aligned} \eta &= t-\frac{\sqrt{s^2+z^2}}{c} \\A d\eta &= -\frac{s\,ds}{c\sqrt{s^2+z^2}}, \\ \end{aligned}

and, noting that the interval of integration has changed from $[0,\sqrt{c^2t^2-z_0^2}]$ to $[t-\frac{z_0}{c},0]$, we see that

\begin{align} \textbf{A}(x_0,y_0,z_0,t) &= \hat{\textbf{e}}_z \theta\left(t-\frac{z_0}{c}\right) \int_0^{t-\frac{z_0}{c}} \underbrace{\frac{\mu_0 c}{2} K_0\left(\eta\right)}_{=:f(\eta)}\,d\eta \end{align}

E. [6 points] Find the electric and magnetic fields for $z>0$.

Note that $\textbf{A}$ is really only a function of z, since it is translationally invariant along the x- and y-axes. Thus we have that the magnetic field is

\begin{align} \textbf{B}(z>0) = \nabla \times \textbf{A}(z>0) = - \hat{\textbf{e}}_x \frac{\partial A_y}{z} = \hat{\textbf{e}}_x \left[\frac{1}{c}\delta\left(t-\frac{z_0}{c}\right) \int_0^{t-\frac{z_0}{c}} \frac{\mu_0 c}{2} K_0\left(\eta\right)\,d\eta + \theta\left(t-\frac{z_0}{c}\right)\frac{\mu_0}{2} K_0\left(t-\frac{z_0}{c}\right)\right], \end{align}

and since there is not potential, the electric field is

\begin{align} \textbf{E}(z>0) = \frac{\partial \textbf{A}}{\partial t}(z>0) = \hat{\textbf{e}}_y \left[\delta\left(t-\frac{z_0}{c}\right) \int_0^{t-\frac{z_0}{c}} \frac{\mu_0 c}{2} K_0\left(\eta\right)\,d\eta + \theta\left(t-\frac{z_0}{c}\right)\frac{\mu_0 c}{2} K_0\left(t-\frac{z_0}{c}\right)\right], \end{align}

2. Electrostatics

The most general solution to Laplace's equation, $\nabla^2 V=0$, assuming azimuthal symmetry, is

\begin{align} V(r,\theta) = \sum_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l(\cos\theta) \end{align}

A conducting sphere of radius $a$ at a potential $V_0$ is surrounded by a concentric non-conducting spherical shell of radius $b$ with a surface charge density $\sigma(\theta)=K\cos\theta$, where $K$ is a constant.

A. Write the allowed form of the potential in each of the two regions $a<r<b$ and $r>b$ without regard to the boundary conditions at $r=a$ and $r=n$.

Greg's Solution:

The allowed form in the region $a<r<b$ is just the form given, whereas in the region $r>b$ we require that $A_l=0$ since we want the potential to vanish at infinite. That is,

\begin{aligned} V(r<b,\theta) &= \sum_{l=0}^\infty \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l(\cos\theta) \\ V(r>b,\theta) &= \sum_{l=0}^\infty \frac{C_l}{r^{l+1}} P_l(\cos\theta) \\ \end{aligned}

B. What is the discontinuity in the electric field across the non-conducting spherical shell? Explain the physical basis of your answer.

Greg’s Solution:

We have that

\begin{align} E_r(b+)-E_r(b-)= \frac{\sigma(\theta)}{\epsilon_0} \end{align}

from Gauss's Law, drawing a small pillbox about the surface of the sphere and letting it shrink to zero size.

From the fact that, in the absense of changing fields, the electric field must be divergence free, we obtain,

\begin{align} E_\theta(b+)=E_\theta(b-). \end{align}

C. [8 points] What are the boundary conditions that must be satisfies by the potential $V(r,\theta)$?

Greg’s Solution:

There are four boundary conditions. First, it must be everywhere continous, so that in particular $V(a+,\theta)=V(a-,\theta)$. Second, it must vanish at infinity — this has already been taken care of by setting $B_l=0$ in the region $r>b$. Third we must have that $V(a,\theta)=V_0$. Finally, from the answer to part C we have that

\begin{align} \frac{\partial V}{\partial r}(b+) - \frac{\partial V}{\partial r}(b-) = \frac{K\cos\theta}{\epsilon_0} \end{align}

D. [14 points] Find the potential in both regions.

Greg’s Solution:

First, we use 14 and the fact that the Legendre functions are orthogonal to conclude that only $l=0,1$ coefficients are nonzero. Thus, $A_l=B_l=C_l=0$ for $l=\ne 0,1$. Furthermore, we allowed $l=0$ only because the derivative killed the constant $r^0$ terms, so we need to set $B_0=0$ since $r^{-1}$ is not a constant. Thus, we only have five parameters: $A_0,C_0,A_1,B_1,C_1$.

From the requirement that $V(a,\theta)=V_0$, we see that for $r=a$ the coefficient on $P_1(\cos\theta)$ must vanish so that

\begin{align} A_1 a + \frac{B_1}{a^2} = 0 \quad \Rightarrow B_1 = -A_1 a^3 \end{align}

We also see immediately that $A_0=V_0$, and by the condition of continuity at $r=b$ for $l=0$ we see that $C_0=A_0a =V_0a$.

We thus have two undetermined coefficients, but fortunately we have two more equations at our disposal: the continuity of the potential for the $l=1$ coefficient at $r=b$ and the discontinuity of the electric field:

\begin{aligned} A_1\left(1+\frac{2a^3}{b^3}\right) + \frac{2C_1}{b^2} &= K \\ A_1\left(1-\frac{a^2}{b^2}\right) - \frac{C_1}{a^2} &= 0 \\ \end{aligned}

We thus see that

\begin{aligned} A_1 &= \frac{K}{\left(1+\frac{2a^3}{b^3}\right)+\frac{2a^2}{b^2}\left(1-\frac{a^2}{b^2}\right)} \\ C_1 &= \frac{K}{\frac{2}{b^2}-\frac{1}{a^2}\frac{\left(1+\frac{2a^3}{b^3}\right)}{\left(1-\frac{a^2}{b^2}\right)} } \\ \end{aligned}

3. Energy and Momentum

Maxwell’s equations and the Lorentz force law are

\begin{aligned} \nabla\cdot F &= \rho \\ \nabla\times E &= -\frac{\partial}{\partial t}B \\ \nabla \cdot B &= 0 \\ \nabla\times H &= J + \frac{\partial}{\partial t}D \\ F &= q (E + v\times B) \\ \end{aligned}

A. Starting from the Lorentz force law, show that the work per unit time done on all charged particles in a system is given by $\frac{dW}{dt}=\int\,d^3\textbf{r}\,\textbf{E}\cdot\textbf{J}$, where $\textbf{J}$ denotes the current density of the charged particles.

Greg’s Solution:

The total power entering the system is given by,

\begin{align} \sum_i \textbf{F}_i\cdot \textbf{v}_i = \sum_i q_i(\textbf{E}_i+\textbf{v}_i\times\textbf{B}_i)\cdot \textbf{v}_i = \sum_i q_i \textbf{E}_i \cdot\textbf{v}_i \equiv \int\,d^3\textbf{r}\,\,\textbf{E}\cdot\textbf{J}, \end{align}


\begin{align} \textbf{J} = \sum_i q_i \delta(\textbf{r}_i) \end{align}

B. [8 points] A constant current I flows through a wire of finite resistance R that has a potential difference V maintained between its two ends. The wire had length L and radius a. Compute the electric and magnetic fields at the surface of the wire , and compute the Poynting vector. Comment on its direction.

Greg’s Solution:

The electric field is just the difference in potential divided by the length of the wire,

\begin{align} E = \frac{V}{L}, \end{align}

and it points parallel to the wire in the direction of the current.

The magnetic field can be found through a quick application of Ampere’s Law, with the loop just enclosing the wire:

\begin{align} H = \frac{I}{2\pi (a/2)} = \frac{I}{a\pi}; \end{align}

it points tangent to the circumference of the wire in the direction given by the right-hand rule.

Finally, the Poynting vector is just the cross of the electric field and $H$; since these two vectors are perpendicular, this product is given by

\begin{align} S = EH = \frac{IV}{La\pi} \end{align}

The Poynting vector points directly into the wire (i.e., along the radius), as given by the right-hand rule.

C. [5 points] Determine the electromagnetic energy per unit time entering the wire and compare this with naive expectations based on ohmic heating.

Integrating the magnitude of the Poynting vector everywhere on the surface of the wire (given that it is everywhere perpendicular to the surface) we obtain,

\begin{align} P = S \times L\times a\pi = IV, \end{align}

which is exactly the ohmic heating law!

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