2003 Spring Electromagnetism

A thin, straight, conducting wire is centered at the origin, oriented along the z-axis, and carries a current $\textbf{I}=I_0\cos(\omega_0 t)\hat e_z$ everywhere along its length L. Define $\lambda_0:=2\pi c/\omega_0$.

A. [8 points] What is the electric dipole moment of the wire as a function of time?

Greg's Solution:

Observe that,

(1)
\begin{align} \int \textbf{J} = -\int \textbf{r} \vec{\nabla}\cdot\textbf{J}, \end{align}

since the current density must vanish at infinity. But since the continuity condition requires that $\frac{\partial \rho}{\partial t} + \nabla\cdot\textbf{J} = 0$, we have

(2)
\begin{align} \int \textbf{J} = \int \textbf{r} \frac{\partial \rho}{\partial t} = \frac{\partial\textbf{p}}{\partial t}, \end{align}

where $\textbf{p}$ is the dipole moment. Thus, we have that

(3)
\begin{aligned} \textbf{p} &= \int_{\text{time}}\int_{\text{space}} \textbf{J} \\ &= \int_{\text{time}}\int_{\text{space}} I_0 \cos(\omega_0 t)\hat e_z \delta(x)\delta(y) \\ &= \frac{L I_0}{\omega_0} \sin(\omega_0 t)\hat e_z \\ \end{aligned}

B. [4 points] What are the scalar and vector potentials, $\phi(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$, everywhere outside an extended charge distribution, $\rho(\textbf{r},t)$, and current distribution, $\textbf{j}(\textbf{r},t)$? Define the gague you have used.

Greg's Solution:

Working with in the Lorentz gauge, we have that

(4)
\begin{align} \left(\phi \atop \textbf{A}\right)(\textbf{r},t) = \frac{1}{4\pi}\int \left(\rho/\epsilon_0 \atop \mu_0\textbf{j}\right)(\textbf{r'},t') \frac{d\textbf{x}'\,dt'\,\delta\left(t-t'-\frac{\textbf{r}-\textbf{r}'}{c}\right)}{|\textbf{r}-\textbf{r}'|}. \end{align}

By Fourier tranforming both sides, we see that

(5)
\begin{align} \begin{aligned} \left(\phi \atop \textbf{A}\right)(\textbf{r},\omega) &= \frac{1}{4\pi}\int \left(\rho/\epsilon_0 \atop \mu_0\textbf{j}\right)(\textbf{r'},t') \frac{d\textbf{x}'\,dt'\,\delta\left(t-t'-\frac{\textbf{r}-\textbf{r}'}{c}\right)}{|\textbf{r}-\textbf{r}'|} \,dt\,e^{i\omega t} \\ &= \frac{1}{4\pi}\int \left(\rho/\epsilon_0 \atop \mu_0\textbf{j}\right)(\textbf{r'},t') \frac{d\textbf{x}'\,dt'\ e^{i\om\left(t'+\frac{\textbf{r}-\textbf{r}'}{c}\right)}}{|\textbf{r}-\textbf{r}'|} \\ &= \frac{1}{4\pi}\int \left(\rho/\epsilon_0 \atop \mu_0\textbf{j}\right)(\textbf{r'},\omega) \frac{d\textbf{x}'\, e^{i\om\frac{\textbf{r}-\textbf{r}'}{c}}}{|\textbf{r}-\textbf{r}'|}, \\ \end{align}

so that we obtain an expression for the potentials in the frequency domain.

C. [10 points]

Problem:

Deteremine $\phi(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$ for the current carrying wire described above when $\textbf{r}\gg L$ and NO assumption is made about the side of $\lambda_0$. Show that,

(6)
\begin{aligned} \phi(\textbf{r},t) &= \frac{I_0 e^{-i\om t}e^{i\omega|\textbf{r}|/c}}{4\pi\epsilon_0 \om|\textbf{r}|}\sin\left(\frac{kL\cos\theta}{2}\right) \\ \textbf{A}(\textbf{r},t) &= \frac{\mu_0 I_0Le^{-i\omega t}e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|}\frac{2}{kL\cos\theta}\sin\left(\frac{kL\cos\theta}{2}\right)\hat e_z. \end{aligned}

Greg's Solution:

Since $|\textbf{r}|\gg L$, we can make the approximation that $|\textbf{r}-\textbf{r'}|\approx r - \hat r \cdot \textbf{r'}$ so that we have

(7)
\begin{align} \left(\phi \atop \textbf{A}\right)(\textbf{r},\omega) = \frac{e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|}\int \left(\rho/\epsilon_0 \atop \mu_0\textbf{j}\right)(\textbf{r'},\omega) e^{- i\om \hat r \cdot \textbf{r'}/c}d\textbf{r}'. \end{align}

It will be easiest to first find $\textbf{A}(\textbf{r},\omega)$ since we can then find $\phi(\textbf{r},\omega)$ from the Lorentz gauge condition,

(8)
\begin{align} \frac{1}{c^2}\frac{\phi}{\partial t} + \nabla\cdot\textbf{A} = 0 \Rightarrow \phi = \frac{i\nabla\cdot A}{c^2 \omega} \end{align}

Plugging in our expression for the current density (expressed in the frequency domain), $\textbf{j}(\textbf{r},\omega)=I_0 \delta(\omega-\omega_0) \delta(x)\delta(y)\hat e_z$, we see that

(9)
\begin{aligned} \textbf{A}(\textbf{r},\omega) &= \frac{e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|}\int \mu_0 I_0 \delta(\omega-\omega_0) \delta(x)\delta(y)\hat e_z e^{-i\om \hat r \cdot \textbf{r'}/c}d\textbf{r}' \\ &= \frac{e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|}\int \mu_0 I_0 \delta(\omega-\omega_0) \hat e_z e^{-i k z \cos\theta}d\textbf{r}' \,dz \\ &= \frac{\mu_0 I_0 \delta(\omega-\omega_0) e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|} \frac{ 2 e^{-i (k L \cos\theta)/2}}{-i (k \cos\theta)} \hat e_z \\ &= \frac{\mu_0 L I_0 \delta(\omega-\omega_0) e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|} \frac{ e^{-i (k L \cos\theta)/2}}{-i (k L\cos\theta)/2} \hat e_z \\ \end{aligned}

We use this expression to obtain the potential,

(10)
\begin{aligned} \phi(\textbf{r},\omega) &= \frac{i\nabla\cdot A}{c^2 \om} \\ &= \frac{i}{c^2\om}\frac{\mu_0 L I_0 \delta(\omega-\omega_0) }{4\pi} \frac{e^{i\omega|\textbf{r}|/c}r\cos\theta\left(i\omega/c-1\right)}{|\textbf{r}|^2}\frac{ 2 e^{-i (k L \cos\theta)/2}}{-i (k L\cos\theta)} \\ &= \frac{1-i\omega}{\omega^2}\frac{I_0 \delta(\omega-\omega_0) e^{i\omega|\textbf{r}|/c}}{4\pi\epsilon_0 |\textbf{r}|/2} \\ \end{aligned}

D. [10 points] Determine the magnetic field resulting from $\phi(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$ and then find the pattern fo radiation from this system for $|\textbf{r}|\gg L \gg \lambda_0$.

The magnetic field is given by,

(11)
\begin{aligned} \textbf{B} = \nabla \times \textbf{B} &= \frac{\mu_0 I_0Le^{-i\omega t}e^{i\om|\textbf{r}|/c}\left(ik-\frac{1}{|\textbf{r}|}\right)}{4\pi|\textbf{r}|}\frac{2}{kL\cos\theta}\sin\left(\frac{kL\cos\theta}{2}\right) \hat r \times \hat e_z \\ &= \frac{\mu_0 I_0Le^{-i\omega t}e^{i\om|\textbf{r}|/c}\left(ik-\frac{1}{|\textbf{r}|}\right)}{4\pi|\textbf{r}|}\frac{2}{kL}\sin\left(\frac{kL\cos\theta}{2}\right) \hat e_y \\ \end{aligned}

If $|\textbf{r}|\gg \lambda_0$, then equivalently we have that $1/|\textbf{r}|\ll k$ so that

(12)
\begin{align} \textbf{B} \approx \frac{\mu_0 I_0Le^{-i\omega t}e^{i\om|\textbf{r}|/c}}{4\pi|\textbf{r}|}\frac{2 i}{L}\sin\left(\frac{kL\cos\theta}{2}\right) \hat e_y \\ \end{align}

Observe that whenever

(13)
\begin{align} \frac{kL\cos\theta}{2} = n\pi \end{align}

we have a node. The number of nodes is given by

(14)
\begin{align} n = \frac{kL}{2\pi} = \frac{L}{\lambda}, \end{align}

so that for example if the length of the wire is twice the wavelength of the system, then we'll get nodes at $\theta=0,\pi/2,\pi$. Indeed, the radiation pattern is similar to what one would get if one had two slits emitting radiation with wave number $k$ and separated by a distance $L/2$.

E. [8 points] Compare this with the standard dipole case where $|\textbf{r}|\gg \lambda_0 \gg L.$ Explain the physical origin of any differences.

In the dipole case there are only nodes at $\theta=0,\pi$, where in the case of our system there can be as many dipoles as $L$ is a multiply of the wavelength. Physically, this comes from the fact that the wave is being generated over a region of space, rather than at a single point, so that essentially waves which come from different parts of the wire interfere with each other. In fact, an equivalent system which has the same radiation pattern is just light passing through a slit of length $L$!

2. [30 points total] Electrostatics

An infinitely long hollow cylindrical conductor of radius a is divided into two parts by a plane through the axis, and the parts are separated by a negligibly small interval. The parts are kept at potentials $V_1$ and $V_2$.

Formula that may be of use:

(15)
\begin{align} \nabla^2 f = \frac{\partial^2 f}{\partial r^2} + \frac{1}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2} + \frac{\partial^2 f}{\partial z^2} \end{align}

A. [15 points] Show that the most general form for the potential inside the conducting cylinder is,

(16)
\begin{align} V(r,\theta) = \sum_{m=0}^\infty \left(\frac{r}{a}\right)^m \left[A_m\cos(m\theta) + B_m\sin(m\theta)\right] \end{align}

Greg's Solution:

We make the Ansatz that $V$ is separable and that (due to the translational symmetry in the problem) it does not depend on $z$ — that is,

(17)
\begin{align} V(r,\theta) = R(r)\Theta(\theta) \end{align}

Since there are not charges in the system, the potential must satisfy $\nabla^2 V =0$. Thus we have that,

(18)
\begin{aligned} 0 &= \nabla^2 V \\ &= \frac{\partial^2 V}{\partial r^2} + \frac{1}{r}\frac{\partial V}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2} \\ &= R''(r)\Theta(\theta) + \frac{1}{r}R'(r)\Theta(\theta) + \frac{1}{r^2}R(r)\Theta''(\theta) \\ &= \underbrace{r^2\frac{R''(r)}{R(r)} + r\frac{R'(r)}{R(r)}}_{=m^2} + \underbrace{\frac{\Theta''(\theta)}{\Theta(\theta)}}_{=-m^2} \\ \end{aligned}

Each term must be equal to a constant (as can be seen trivially by taking partial derivatives).

The $\theta$ dependent term gives us,

(19)
\begin{align} \Theta''(\theta) = -m^2\Theta(\theta) \Rightarrow \Theta(\theta) = A\cos(m\theta) + B\sin(m\theta). \end{align}

The $r$ dependent term gives us,

(20)
\begin{equation} r^2 R''(r) + r R'(r) - m^2 R(r) = 0 \end{equation}

We make the Ansatz that $R(r)=r^\alpha$. Then we see that,

(21)
\begin{align} \alpha^2-\alpha + \alpha - m^2 = 0 \Rightarrow m=\alpha. \end{align}

Thus, we set

(22)
\begin{align} R(r) = \left(\frac{r}{a}\right)^m, \end{align}

where the factor of $a$ has been put in as a convenience so that $R(a)=1\quad\forall\,\, m$.

Note that we have not yet restricted the value of $m$. We do so now by observing that we must have $\Theta(r)=\Theta(r+2\pi)$, and so $m$ must be an integer. Thus, we have that the most general form for $V$ is,

(23)
\begin{align} V(r,\theta) = \sum_{m=0}^\infty \left(\frac{r}{a}\right)^m \left[A_m\cos(m\theta) + B_m\sin(m\theta)\right]. \end{align}

Q.E.D.

B. [15 points] Find the coefficients A_m and B_m.

We use the orthogonality condition:

(24)
\begin{align} \int_0^{2\pi} \sin(m\theta)\sin(n\theta) = \delta_{mn} \pi \end{align}

Thus,

(25)
\begin{aligned} A_m &= \frac{1}{\pi}\left[\int_0^{\pi}V_1\sin(m\theta)+\int_{\pi}^{2\pi}V_2\sin(m\theta)\right] \\ &= \frac{V_1\cos(m\theta)\big|^\pi_0 + V_2\cos(m\theta)\big|^{2\pi}_\pi}{m\pi} \\ &= \frac{(V_2-V_1)[1-(-1)^m]}{m\pi} \\ \end{aligned}

and

(26)
\begin{aligned} B_m &= \frac{1}{\pi}\left[\int_0^{\pi}V_1\cos(m\theta)+\int_{\pi}^{2\pi}V_2\cps(m\theta)\right] \\ &= \frac{V_1\sin(m\theta)\big|^\pi_0 + V_2\sin(m\theta)\big|^{2\pi}_\pi}{m\pi} \\ &= 0 \\ \end{aligned}

3. [30 points total] Electromagentic Waves in the Ionosphere

The ionosphere can be considered as an ionized medium containing N free electrons per unit volume. A lineraly polarized electromagnetic plane wave is propagating in the +z-direction in the small locally uniform magnetic field of the earth, $\textbf{B}=B_0\hat e_z$, that is aligned along the +z-axis. At $z=0$, the electric field associated with the plane wave is linearly polarized in the x-direction,

(27)
\begin{align} \textbf{E}(0,t) = E_0 e^{-i\omega t} \hat e_x \end{align}

A. [10 points] The electric field associated with a circularly polarized electromagnetic plane wave moving in the +z-direction is

(28)
\begin{align} \textbf{E}^{(\lambda)}(z,t) = E_\lambda e^{i(k_\lambda z-\omega t)} (\hat e_x + i\lambda \hat e_y) \end{align}

where $\lambda=\pm 1$ determines the polarization of the wave, and the wave-number $k_\lambda$, in general, depends upon $\lambda$. Assume that the velocity of an electron in the ionosphere through which a circularly polarized plane wave is traveling will have the form

(29)
\begin{align} \textbf{v}^{(\lambda)} = v_0^{(\lambda)} e^{i(k_\lambda z-\omega t)}(\hat e_x-i\lambda \hat e_y) \end{align}

Show that

(30)
\begin{align} v_0^{(\lambda)} = \frac{-i|e|E_\lambda}{m(\omega-\lambda\omega_0)}, \end{align}

where $m$ is the electron mass and $\omega_0=\frac{|e|V}{m}$.

Greg's Solution:

This is just an application of the Lorentz Force Law:

(31)
\begin{aligned} m\dot{ \textbf{v} }&= e \textbf{E} + q\textbf{v} \times \textbf{B} \\ -i\omega m v_0 e^{i(k_\lambda z-\omega t)} (\hat e_x+i\lambda \hat e_y) &= - |e| E_\lambda e^{i(k_\lambda z-\omega t)} (\hat e_x+i\lambda \hat e_y) - |e| (i\lambda)B v_0 e^{i(k_\lambda z-\omega t)} (\hat e_x+i\lambda \hat e_y) \\ \Rightarrow v_0 &= \frac{-i |e| E_\lambda}{m\left(\omega-\lambda \omega_0\right)}, \\ \end{aligned}

where $\omega_0:=\frac{|e|B}{m}$. (Note that we employed the trick that $(\hat e_x+i\lambda\hat e_y) \times \hat e_z = (-\hat e_y +i\lambda \hat e_x) \equiv i\lambda (\hat e_x + i\lambda \hat e_y$.)

B. Show that the refractive index for each circular polarization state may be written as

(32)
\begin{align} n_\lambda = \sqrt{1-\frac{\omega_p^2}{\omega(\omega-\lambda\omega_0)}}, \end{align}

and define $\omega_p$.

Greg's Solution:

Each oscillating electron may be thought of as a dipole with moment

(33)
\begin{align} p=-\epsilon_0 |e||\textbf{x}|= \frac{-\epsilon_0 |e|^2 E_\lambda}{m\omega (\omega-\lambda\omega_0)} \end{align}

Thus, given that there are N free electrons per unit volume, the polarization induced by the electromagnetic wave is

(34)
\begin{align} P=Np = \frac{-N \epsilon_0 |e|^2 E_\lambda}{m\omega (\omega-\lambda\omega_0)}. \end{align}

We now observe that,

(35)
\begin{aligned} D &= \epsilon_0 E_\lambda + P \\ &= \epsilon_0 E_\lambda \left(1-\frac{N |e|^2 E_\lambda}{m\omega(\omega-\lambda\omega_0)} \\ &= \epsilon_0 E_\lambda \left(1-\frac{\om_p^2}{\omega(\omega-\lambda\omega_0)} \\ &= \epsilon_0 E_\lambda n^2 \\ &= \epsilon E_\lambda, \end{aligned}

where

(36)
\begin{align} \begin{aligned} \epsilon &= n^2 \epsilon_0 \\ n &= \sqrt{1- \frac{\om_p^2}{\omega(\omega-\lambda\omega_0)}} \\ \om_p &= \frac{|e|B}{m}, \end{align}

the last being the plasma frequency of the medium.

C. [10 points] What is the angle between the plane of polarization of the electromagnetic wave at $z=z_0$ and $z=0$?

The wavelength of the wave will change when it enters the medium since the speed of light changes —

(37)
\begin{align} k' = \frac{\om}{nc} = \frac{k_\lambda}{n} \end{align}

Thus, the number of radians through which the wave progressed between $0$ and $z_0$ is given by

(38)
\begin{align} \theta = k' z_0, \end{align}

and this also gives us the amount by which the angle of the plane of polarization has processed. For convenience one can map this angle onto the interval $[-\pi/2,\pi/2]$ — note that a rotation of $\pi$ radians is equivalent to not changing the plane at all, but changing the phase by an angle of $\pi$.

page revision: 4, last edited: 30 Aug 2006 19:28