Subfield: Electromagnetism

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Consider a spherical shell of thickness $\delta=0.5 \mathrm{~cm}$ and radius $R=5 \mathrm{~cm}$ made of an Ohmic material with resistivity $\rho=10^{-7} \Omega \mathrm{m}$. A spherical laser source with a tuned frequency of $f_{0}$ and intensity $I_{0}=10^{5} \mathrm{~W} / \mathrm{m}^{2}$ is placed at the center of the shell and is turned on. Working in the limit $\delta \ll \frac{c}{f_{0}} \ll R$, approximate the initial average power dissipated by the shell. Neglect inductance.

Solution:
Note that the proper treatment of this problem requires more advanced electromagnetism. The following might seem sketchy, but this will do. We first consider the case of a thin plane made of an ideal conductor placed at $z=0$. A monochromatic electromagnetic wave with an $E$ field: $\vec{E}_{i}(t, z)=E_{0} \cos (k z-\omega t) \hat{x}$ is incident on the plane. The corresponding $B$ field is of course $\vec{B}_{i}(t, z)=\frac{E_{0}}{c} \cos (k z-\omega t) \hat{y}$. We expect a current density $\hat{K}(t)=K(t) \hat{x}$ to form in the direction of the $E$ field, as the electric field accelerates the charges on the sheet $-K(t)$ will oscillate with the $E$ field. We know that perfect conductors completely reflect electromagnetic waves, so we expect no EM field in the $z>0$ region. The total EM field is a combination of the incident wave and the wave generated by the oscillating surface charges, so we are to believe that the latter provides just the right EM wave to cancel out $\vec{B}_{i}$ and $\vec{E}_{i}$ in the $z>0$ region. Consider an instance where the EM waves have an orientation shown in the figure below. It's easy to verify that the magnetic field generated by a uniform sheet of current pointing out of the page is given as shown. Since the electric wave generated by $\vec{K}(t)$ must cancel the $\vec{E}_{i}$ pointing out of page in $z>0$, we get a poynting vector (for the wave generated by $\vec{K}$ ) in the $\hat{z}$ direction in $z>0$. By symmetry, this means that the poynting vector points in the $-\hat{z}$ direction in $z<0$. Such simple symmetry arguments fully determine the wave generated by an oscillating sheet of charge: $\left|\vec{E}_{k}(t, z)\right|=\left|\vec{E}_{i}(t, z)\right|$, $\left|\vec{B}_{k}(t, z)\right|=\left|\vec{B}_{i}(t, z)\right|$ with the orientations shown below.
<img_4292>

We therefore get that the total magnetic field in the $z<0$ region is $\vec{B}=2 \vec{B}_{i}(t, z)$. Let us now take an Ampere loop as shown in the figure below:

<img_4285>



We have:
$$
T_{\epsilon} \equiv \oint_{\Gamma_{\epsilon}} \vec{B}(t,-\epsilon / 2) \cdot d \vec{l}=\mu_{0} I_{e n c}+\frac{1}{c^{2}} \frac{\partial}{\partial t} \int_{S_{\epsilon}} \vec{E}(t, z) \cdot d \vec{A}
$$
where $S_{\epsilon}$ is the surface bounded by $\Gamma_{\epsilon}$. Since $\vec{E}=0$ everywhere and $I_{e n c}=K(t) l$,
$$
\lim _{\epsilon \rightarrow 0} T_{\epsilon}=\frac{2 E_{0} l}{c} \cos (\omega t)=\mu_{0} K(t) l
$$
Hence, we have $\vec{K}(t)=\frac{2 E_{0}}{\mu_{0} c} \cos (\omega t) \hat{x}$. The key find in this long introduction is that an oscillating sheet of current $\vec{K}(t)=K_{0} \cos (\omega t) \hat{x}$ generates an EM wave with an electric field $\vec{E}_{K}(t, 0)=-\frac{\mu_{0} c}{2} \vec{K}(t)$ near the sheet (at $z=0$ ). We finally return to the original problem.

All laser beams have poynting vectors pointing radially outwards, so the $E_{i}$ field must point in the tangential direction to the sphere at all normal incidence. Say the $E_{i}$ field points in the $\hat{\theta}$ direction, so that the current in the shell flows from the north pole to the south pole. The condition $\delta \ll \frac{c}{f_{0}} \ll R$ allows us to neglect attenuation due to skin effects - essentially, the electric field, thus the surface current, is approximately uniform in the shell. The condition $\frac{c}{f_{0}} \ll R$ allows us to treat the incidence of the laser beams as an EM wave hitting an Ohmic plane of thickness $\delta$, so we consider that problem first. Denote the total electric field inside the the plane at time $t$ as $\vec{E}(t)$. The current density is given by Ohm's law: $\vec{J}(t)=\vec{E}(t) / \rho$. The corresponding surface current is of course $\vec{K}(t)=\delta \vec{E} / \rho(t)$. Carefully note the distinction between $\vec{E}$ and $\vec{E}_{i}$. Since the source wave $\vec{E}_{i}(t)$ is sinusoidal, we expect $\vec{E}(t)$ to be sinusoidal as well so that $\vec{E}_{K}(t)=-\frac{\mu_{0} c}{2} \vec{K}(t)$ applies. The total electric field inside the sheet is a sum of $\vec{E}_{K}$ and $\vec{E}_{i}$, thus:
$$
\vec{E}(t)=\vec{E}_{i}(t)-\frac{\mu_{0} c \delta}{2 \rho} \vec{E}(t)
$$
Rearranging for $\vec{E}(t)$, we get:
$$
\vec{E}(t)=\frac{\vec{E}_{i}(t)}{1+\mu_{0} c \delta / 2 \rho}
$$
On the sphere, $\vec{E}(t)$ points in the $\hat{\theta}$ direction and has the same magnitude everywhere on the sphere. The relevant cross-section will be the lateral surface of a truncated cone, as shown below:

<img_4286>



The current through this surface is therefore:
$$
I(\theta, t)=\frac{\pi \delta(2 R+\delta)\left|\vec{E}_{i}(t)\right|}{\rho+\mu_{0} c \delta / 2} \sin \theta
$$
The power dissipated by a volume generated by $\theta \sim \theta+d \theta$ is
$$
d P(t)=I^{2}(\theta, t) \frac{\rho(R+\delta / 2) d \theta}{\pi \delta(2 R+\delta) \sin \theta}
$$
We integrate this expression from $\theta=0$ to $\theta=\pi$ :
$$
P(t)=\frac{\pi \delta \rho(2 R+\delta)^{2}}{\left(\rho+\mu_{0} c \delta / 2\right)^{2}}\left|\vec{E}_{i}(t)\right|^{2}
$$
Now, the incident electric field has the form $\left|\vec{E}_{i}(t)\right|^{2}=\frac{2 I_{0}}{\epsilon_{0} c} \cos ^{2}\left(2 \pi f_{0} t\right)$. The average of this function from $t=0$ to $t=\frac{1}{f_{0}}$ is $1 / 2$, so our final answer is:
$$
\bar{P}=\frac{\pi \delta \rho(2 R+\delta)^{2}}{\left(\rho+\mu_{0} c \delta / 2\right)^{2}} \frac{I_{0}}{\epsilon_{0} c}
$$
which turns out to be $2.39078 \times 10^{-15} \mathrm{~W}$

Final answer: $2.39078 \times 10^{-15}$ W