2002 Autumn Electromagnetism
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You must answer Question 1 and EITHER answer Question 2 OR Question 3.

1. [50 points total] 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\tau'\,\frac{\rho(\textbf{r}',t_r}{|\textbf{r}-\textbf{r'}|},\quad \textbf{A}(\textbf{r},t) = \frac{\mu_0}{4\pi}\int\,d\tau'\,\frac{\textbf{j}(\textbf{r'},t_r)}{|\textbf{r}-\textbf{r'}|} \end{align}

A. [12 points] Carefully define the quantities $\mu_0,\epsilon_0,\rho,\textbf{j},\tau',\textbf{r},\textbf{r'},$ and $t_r$.

permeability of free space, a constant which gives the strength of the electric force
permittivity of free space, a constant which gives the strength of the magnetic force
charge density — the amount of charge per unit volume at a point in space
current density — the amount of current flowing through a point in space per unit area, where the area is perpendicular to the flow of charge
the volume differential
the location in space at which the fields are being computed
the variable being integrated over; a location in space where the charge/current density is being evaluated
the "retarded" time, defined to be $t_r:=t-\frac{|\textbf{r}-\textbf{r'}}{c}$, which compensates for the time needed for information to travel from a point in the source to the point at which the field is being evaluated

Problem description, continued

Two infinite, parallel, straight wires (wire 1 and wire 2) separated by a distance 3d, lie parallel to the z-axis. One carries a current I(t) in the +z-direction, while the other carries a current -I(t) in the +z-direction, where

\begin{align} I(t) := q_0 \delta(t), \end{align}

where $\delta(r)$ is the Dirac-Delta function. A point P lies in the plane of the wires and in the z=0 plane. It is a distance d from wire 1, and a distance 2d from the wire 2, as shown in Fig. 1.

B. [8 points] If an event occurs on wiree 1 at a distance z along the z-axis above the point P, how much time must pass before this event can influence physics at the point P? What is this time difference for an event on wire 2 with the same z-coordinate as the event on wire 1?

Greg's Solution:

In both cases, it will take the same amount of time for information to travel to P as it does for light to travel to P. That is, in the first case it will take $\frac{\sqrt{z^2+d^2}}{c}$ time, and in the second case it will take $\frac{\sqrt{z^2+4d^2}}{c}$ time.

C. [18 points] What are the vector potential $\textbf{A}(t)$ and scalar potential $V(t)$ at the point P due to I(t), as a function of time?

Greg's Solution:

Since we have not been told otherwise, we assume that the charge density over all space is zero, so therefore the electric potential is everywhere zero (employing the convention that we want the potential to vanish at infinity).

The current density has already been given. Assume that the point P lies in the xz plane and constitutes the origin of our coordinate system, so that the wires are located at $x=-d$ and $x=+2d$. Then the current density of the entire system is,

\begin{align} \textbf{j} = q_0 \hat e_z\delta(t)[\delta(x+d)\delta(y)-\delta(x-2d)\delta(y)] \end{align}

Thus, the vector potential is given by,

\begin{aligned} \textbf{A} &=\frac{\mu_0}{4\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \,dx\,dy\,dz \frac{ q_0 \hat e_z\delta\overbrace{\left(t-\frac{\sqrt{x^2+y^2+z^2}}{c}\right)}^{t_r}[\delta(x+d)\delta(y)-\delta(x-2d)\delta(y)]}{\sqrt{x^2+y^2+z^2}} \\ &=\frac{\mu_0}{4\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \,dx\,dz \frac{ q_0 \hat e_z\delta\left(t-\frac{\sqrt{x^2+z^2}}{c}\right)[\delta(x+d)-\delta(x-2d)]}{\sqrt{x^2+z^2}} \\ &=\frac{\mu_0}{4\pi}\int_{-\infty}^{\infty}\,dz \, q_0 \hat e_z\left[\frac{\delta\left(t-\frac{\sqrt{d^2+z^2}}{c}\right)}{\sqrt{d^2+z^2}} -\frac{\delta\left(t-\frac{\sqrt{4d^2+z^2}}{c}\right)}{\sqrt{4d^2+z^2}}\right]\\ \end{aligned}

We perform a change of variables: let

\begin{aligned} u&=\frac{\sqrt{d^2+z^2}}{c} \\ du&=\frac{z\,dz}{\sqrt{d^2+z^2} c} \Rightarrow dz= \frac{uc^2\,du}{\sqrt{u^2-d^2}} \\ \end{aligned}


\begin{aligned} v&=\frac{\sqrt{4d^2+z^2}}{c} \\ dv&=\frac{z\,dz}{\sqrt{4d^2+z^2} c} \Rightarrow dz= \frac{vc^2\,dv}{\sqrt{v^2-4d^2}} \\ \end{aligned}

Then the field integral becomes,

\begin{aligned} \textbf{A} &=\frac{c\mu_0 q_0 \hat e_z}{2\pi}\left[\int_{d/c}^{\infty}\frac{\delta(t-u)\,du}{\sqrt{c^2u^2-d^2}} -\int_{2d/c}^{\infty}\frac{\delta(t-v)\,du}{\sqrt{c^2v^2-4d^2}} \right] \\ &=\frac{c\mu_0 q_0 \hat e_z}{2\pi}\left[\frac{\Theta(t-d/c)}{\sqrt{c^2t^2-d^2}}- \frac{\Theta(t-2d/c)}{\sqrt{c^2t^2-4d^2}} \right] \\ \end{aligned}

D. [12 points] Determine the leading time-dependence of the electric field at the point P for $t\gg 2d/c$.

We have that,

\begin{aligned} \textbf{E}(t>2d/c) &= \frac{\partial \textbf{A}}{c\partial t} - \overbrace{\nabla \phi}^{=0\,\,\text{from 1.C}} \\ &= \frac{\mu_0 q_0 \hat e_z}{2\pi}\left[\frac{c^2t}{(c^2t^2-4d^2)^{3/2}}-\frac{c^2t}{(c^2t^2-d^2)^{3/2}}\right] \\ \end{aligned}

2. [50 points total] Magnetostatics

A. [6 points] Write Maxwell's equations for D, E, H, and B in the presence of a free charge density $\rho$ and a free current density $\textbf{j}$.

Greg's Solution:

\begin{aligned} \nabla\cdot \textbf{D} &= \rho\\ \nabla\times \textbf{E} &= -\frac{\partial \textbf{B}}{\partial t}\\ \nabla\cdot \textbf{B} &= 0\\ \nabla\times\textbf{H} &= \textbf{j}+\frac{\partial \textbf{D}}{t} \end{aligned}

B. [5 points] In terms of the magnetic field, B, the magnetization, M, and the permeability of free space, define the field H.

Greg's Solution:

\begin{align} \textbf{H} &:= \frac{\textbf{B}}{\mu_0} - \textbf{M} \end{align}

C. [14 points] Consider an infinite straight wire in free-space carrying current I, located at $x=y=0$ aligned in the z-direction. State Ampere's law and use it to find the magnetic field $B$ a distance $r$ away from the wire. Show that everywhere except at the wire, the field $\textbf{H}$ can be written in terms of a scalar potential $\textbf{H}=-\nabla\phi$, where (up to a constant)

\begin{align} \phi=\frac{I}{2\pi}\text{Im}[\log(x+iy)] \end{align}

Greg's Solution:

Ampere's law states that whenever one has a surface S in space, then

\begin{align} \oint_{\partial S} \textbf{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}, \end{align}

where $\partial S$ is the boundary of the surface, $d\vec{l}$ is a vector differential element of the boudary, and $I_{\text{enc}}$ is the amount of current passing through the surface.

We apply Ampere's law by drawing a circular loop around the wire with radius r; since the quantity $\textbf{B}\cdot\vec{l}$ is then cylindrically symmetric, we have that

\begin{align} \oint \textbf{B}\cdot d\vec{l} = 2\pi r B = \mu_0 I_{\text{enc}}\Rightarrow \textbf{B} = \frac{\mu_0 I}{2\pi r}\left(\frac{-y}{r},\frac{x}{r}\right) \end{align}

The curl of $\textbf{H}$ is only nonvanishing at the location of the wire. Thus, everywhere else in space one may represent $\textbf{H}$ as the gradient of a potential. In particular, we observe that if $\phi=\frac{I}{2\pi}\text{Im}[\log(x+iy)]$, then

\begin{aligned} -\nabla \phi &= \frac{I}{2\pi} \text{Im}\left(\frac{1}{x+iy},\frac{i}{x+iy}\right) \\ &= \frac{I}{2\pi} \text{Im}\left(\frac{x-iy}{x^2+y^2},\frac{ix+y}{x^2+y^2}\right) \\ &= \frac{I}{2\pir} \left(\frac{-y}{r},\frac{x}{r}\right), \\ &= \frac{\textbf{B}}{\mu_0} \\ &= \textbf{H} \end{aligned}

A long straight wire carrying current I is placed a distance d above a semi-infinite magnetic medium of permeability $\mu$.

D. [10 points] Given that there are no free charges or currents at the boundary between the magnetic medium and free-space, write down the relation between components of $\textbf{B}$ just above the boundary and the components of $\textbf{B}$ just below the boundary. Write this relation in terms of the scalar potential just above the boundary, $\phi_>$, and just below the boundary, $\phi_>$.

__Greg's Solution:_

(Without loss of generality, we assume that the plane bonding the semi-magnetic region is located at $y=0$.)

Since there are no free currents, $\nabla\times\textbf{H}=0$ and so there must be a scalar function $\phi$ such that $\textbf{H}=-\nabla\phi$; since this function must be differentiable, it must also be continuous, so that

\begin{align} \phi_< = \phi_> \end{align}

Furthermore, the absense of free currents means that the tangental component of $\textbf{H}$ must be continuous, so that

\begin{aligned} \textbf{H}_{<,\|} &= \textbf{H}_{>,\|} \\ \frac{\partial \phi_{<}}{\partial(x,z)}&=\frac{1}{\mu_0}\frac{\partial \phi_{>}}{\partial (x,z)} \\ \end{aligned}

Finally, $\textbf{B}$ must always be divergence-free, so we have that

\begin{aligned} B_{<,\perp} &= B_{>,\perp}\\ \mu H_{<,\perp} &= \mu_0 H_{>,\perp}\\ \mu \frac{\partial \phi_<}{\partial y} &= \mu_0 \frac{\partial \phi_>}{\partial y} \end{aligned}

E. [15 points] Find the force per unit length (including direction) on the wire using the method of images.

Greg's Solution:

For convenince, we assume that the wire is at $y=+d$ so that the boundary of the medium is at $y=0$.

We seek the magnitude of the field at the location of the wire. To do this, we will solve for the magnetic potential over all space using the boundary conditions given in the previous part, and the following Ansatzs:

\begin{aligned} \phi_> &= -\frac{I}{2\pi}\text{Im}\{\log[x+i(y-d)]\} - \frac{I'}{2\pi}\text{Im}\{\log[x+i(y+d)] \}\\ \phi_< &= -\frac{I''}{2\pi}\text{Im}\{\log[x+i(y-d)] \}\\ \end{aligned}

That is, we assume that in the free-space region there appears to be a mirror-image of the wire in the semi-magnetic region, while in the semi-magnetic region only the real wire is seen but it appears to be screened. We can find the constants $I'$ and $I''$ from the boundary conditions; first, from continuity of $\phi$,

\begin{aligned} \phi_<(y=0) &= \phi_>(y=0) \\ -\frac{I''}{2\pi}\text{Im}\{\log[x-id]\} &= -\frac{I}{2\pi}\text{Im}\{\log[x-id]\} - \frac{I'}{2\pi}\text{Im}\{\log[x+id]\} \\ I'' &= I + I' \frac{\text{Im}\{\log[x+id]\}}{\text{Im}\{\log[x-id]\}} \\ I'' &= I + I' \frac{\text{Im}\{\log[x+id]\}}{\text{Im}\{\log[(x+id)^*]\}} \\ I'' &= I + I' \frac{\text{Im}\{\log[x+id]\}}{\text{Im}\{\log[x+id]^*\}} \\ I'' &= I - I'\\ \end{aligned}

From the condition that $\textbf{H}_{\|}$ is continuous, we have that

\begin{aligned} \frac{\partial \phi_<}{\partial x}&=\frac{\partial \phi_>}{\partial x} \\ \frac{I'' d}{2\pi (x^2+d^2)}&= \frac{1}{\mu_0}\left(\frac{I d}{2\pi(x^2+d^2)} - \frac{I' d}{2\pi(x^2+d^2)}\right) \\ I'' &= I'-I, \\ \end{aligned}

but we see that this turns out to give us redundant information. The reason for this is that we were clever enough to have put the image wire at the same distance $d$ from the border as the real wire; if we had the distance be another constant which needed to be found, then this redundancy would have disappeared and we would have had another independent equation. Furthermore, if our choice of $d$ had been incorrect, then we would have seen an inconsistency in the equations

Next, we observe that since $\phi$ is independent of z, the condition that $\frac{1}{\mu}\frac{\partial \phi_<}{\partial z}=\frac{1}{\mu_0}\frac{\partial \phi_>}{\partial z}$ is automatically satisfied.

Finally, we have the condition that,

\begin{aligned} \mu \frac{\partial \phi_<}{\partial y}(x=0) &= \mu_0\frac{\partial \phi_>}{\partial y}(x=0) \\ \mu \frac{I'' x}{2\pi (d^2+y^2)}&= \mu_0\left(\frac{I x}{2\pi(d^2+y^2)} +\frac{I' x}{2\pi(d^2+y^2)}\right) \\ \mu I''&= \mu_0 (I + I')) \\ \end{aligned}

From our two equations, we conclude that

\begin{aligned} I' &= \frac{\mu-\mu_0}{\mu+\mu_0}\\ I'' &= \frac{2\mu_0}{\mu+\mu_0} \end{aligned}

Thus, we see that our wire feels the magnetic field from a parallel image wire with current $I'=\frac{\mu-\mu_0}{\mu+\mu_0}$ at a distance $2d$ away,

\begin{align} \textbf{B} = \frac{-I'\mu_0 \hat e_x}{2\pi (2d)}, \end{align}

and so the force acting per unit length on the wire is given by,

\begin{aligned} \frac{\textbf{F}}{L} &= \textbf{I}\times\textbf{B} \\ &= -(\hat e_z\times \hat e_x) \frac{II'\mu_0}{4\pi d} \\ &= -\hat e_y \frac{I^2\mu_0}{4\pi d} \frac{\mu-\mu_0}{\mu+\mu_0} \\ \end{aligned}

3. [50 points total] Reflection, Refraction, and Brewster's Angle

A monochromatic electromagnetic plane-wave polarized in the plane of incidence is incident upon a boundary between two linear, homogenous media at an angle $\theta_I$ to the noral. Medium 1 has refractive index $n_1$ and medium 2 has refractive index $n_2$. Both media have permeabilities equal to that of the vacuum, $\mu_1=\mu_2=\mu_0$.

A. [7 points] Write down the boundary conditions that relate arbitrary electric and magnetic fields on either side of the boundary.

Greg's Solution:

In general, for the magnetic field we have that

\begin{aligned} B_{1,\perp}\big|_{z=0} &= B_{2,\perp}\big|_{z=0} \\ \frac{1}{\mu_1}\textbf{B}_{1,\|}\bigg|_{z=0} &= \frac{1}{\mu_2}\textbf{B}_{2,\|}\bigg|_{z=0} \Leftrightarrow \textbf{H}_{1,\|}\bigg|_{z=0} &= \textbf{H}_{2,\|}\bigg|_{z=0} \\ \end{aligned}

However, since $\mu_1=\mu_2$, we have that the magnetic field must be completely continuous across the boundary,

\begin{align} \textbf{B}_1\bigg|_{z=0} = \textbf{B}_2\bigg|_{z=0} \end{align}

In general, for the electric field, we have that

\begin{aligned} \textbf{E}_{1,\|}\big|_{z=0} &= \textbf{E}_{2,\|}\big|_{z=0} \\ \epsilon_1 E_{1,\perp}\bigg|_{z=0} &= \epsilon_2 E_{2,\perp}\bigg|_{z=0} \Leftrightarrow D_{1,\perp}\bigg|_{z=0} &= D_{2,\perp}\bigg|_{z=0} \\ \end{aligned}

Since $e_i=n_i^2 e_0$, we can rewrite the above in terms of the indices of refraction in the form,

\begin{aligned} \textbf{E}_{1,\|}\big|_{z=0} &= \textbf{E}_{2,\|}\big|_{z=0} \\ n_1^2 E_{1,\perp}\bigg|_{z=0} &= n_2^2 E_{2,\perp}\bigg|_{z=0} \\ \end{aligned}

B. [8 points] Write down general expressions for the incident, reflected and transmitted electric and magnetic fields. Indicate all spatial and time dependencies.

Greg's Solution:

The electric and magnetic fields all take the same general form: (assume that at the end of the day we will take the real parts of all quantities)

\begin{aligned} \textbf{E}_I(\textbf{r},t) &= E_I (\hat y \times \hat k_I) e^{i(\textbf{r}\cdot\textbf{k}_I-\om t)} \\ \textbf{E}_R(\textbf{r},t) &= E_R (\hat y \times \hat k_R) e^{i(\textbf{r}\cdot\textbf{k}_R-\om t)} \\ \textbf{E}_T(\textbf{r},t) &= E_T (\hat y \times \hat k_T) e^{i(\textbf{r}\cdot\textbf{k}_T-\om t)} \\ \end{aligned}
\begin{aligned} \textbf{B}_I(\textbf{r},t) &= B_I (\hat y) e^{i(\textbf{r}\cdot\textbf{k}_I-\om t)} \\ \textbf{B}_R(\textbf{r},t) &= B_R (\hat y) e^{i(\textbf{r}\cdot\textbf{k}_R-\om t)} \\ \textbf{B}_T(\textbf{r},t) &= B_T (\hat y) e^{i(\textbf{r}\cdot\textbf{k}_T-\om t)} \\ \end{aligned}

Note that the angular frequency of the wave is the same everywhere in space, so therefore the wave numbers are related by,

\begin{align} k_I = k_R = \frac{n_1}{n_2}k_T \end{align}

C. [8 points] Derive relations between the reflected scattering angle $\theta_R$ and the incident angle $\theta_I$, and between the transmitted angle $\theta_T$ and the incident angle $\theta_I$.

The phase of the radiation must be continuous across the boundary; thus, at $t=0$ and $z=0$ we must have for all $x$ that

\begin{align} x k_I \sin\theta_I = x k_R \sin\theta_R = x k_T \sin\theta_T \end{align}

Since $k_I=k_R$, we immediately obtain $\theta_I=\theta_R$. Since $\frac{k_I}{n_1} = \frac{k_T}{n_2}$, we immediately obtain Snell's Law: $n_1 \sin\theta_I n_1 = n_2 \sin\theta_T$.

D. [9 points] Relate the reflected electric field to the incident electric field.

We use the boundary conditions given in part A; from the condition that $D_\perp$ must be continuous, we obtain

\begin{aligned} n_1^2\hat z \cdot (\textbf{E}_I+\textbf{E}_R) &= n_2^2\hat z \cdot \textbf{E}_T \\ n_1^2 \sin\theta_I (E_I+E_R) &= n_2^2\sin\theta_T E_T \\ n_1 (E_I+E_R) &= n_2 E_T \\ \end{aligned}

From the condition that $\textbf{E}_\|$ must be continuous, we obtain

\begin{aligned} \hat x \cdot (\textbf{E}_I+\textbf{E}_R) &= \hat x \cdot \textbf{E}_T \\ \cos\theta_I (E_I-E_R) &= \cos\theta_T E_T\\ \end{aligned}

Putting these equations together with the fact that $\cos\theta_T = \sqrt{1-\sin\theta_T^2} = \sqrt{1-\frac{n_1}{n_2}\sin\theta_I^2}$, we obtain the relation

\begin{aligned} \frac{n_1}{n_2}(E_I+E_R) &= \frac{\cos\theta_I}{\sqrt{1-\frac{n_1}{n_2}\sin\theta_I^2}}(E_I-E_R)\\ E_R &= E_I \frac{\frac{n_1}{n_2}-\frac{\cos\theta_I}{\sqrt{1-\frac{n_1}{n_2}\sin\theta_I^2}}}{\frac{n_1}{n_2}+\frac{\cos\theta_I}{\sqrt{1-\frac{n_1}{n_2}\sin\theta_I^2}}} \\ \end{aligned}

E. [9 points] Find an expression for Brewster's angle, $\theta_B$. What is $\theta_B$ for media with $n_1=2$?

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