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EM M97 2

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An infinite conducting plane (taken here to be the x-y plane) is electrically neutral and carries a uniform surface current $t\leq0$ and $K = K 0$ for t>0 with $K 0$ equal to a constant.

a. Find the magnitude and direction of the electric and magnetic fields a height z above the plane.

b. Compute the total power radiated per unit area of the x-y plane

The magnetic field is given by:

$\mathbf{B}(\mathbf{x})=\frac{\mu_{0}}{4\pi}\int\mathbf{J}(\mathbf{x}^{\prime})\times\frac{\mathbf{x}-\mathbf{x}^{\prime}}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|^{3}}d^{3}x^{\prime}$

We have $\mathbf{J}(\mathbf{x})=K\hat{\mathbf{j}}\delta(z)$ so taking the cross product and the integral over z :

$\mathbf{B}(\mathbf{x})=\frac{\mu_{0}}{4\pi}\int_{z^{\prime}=0}K\hat{\mathbf{i}}\frac{z}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|^{3}}d^{2}x^{\prime}$

Performing this integral:

$\mathbf{B}(\mathbf{x})=\frac{\mu_{0}K}{2}\hat{\mathbf{i}}$

However, this can obviously only travel at the speed of light, so that:

$\begin{matrix}\mathbf{B}(|z|<ct) & = & \frac{\mu_0 K}{2}\hat{\mathbf{i}} \\ \mathbf{B}(|z|>ct) & = & 0 \end{matrix}$

Which gives us an electric field by Faraday's Law:

$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}=-\frac{\mu_{0}K}{2}\delta(z+ct)+\frac{\mu_{0}K}{2}\delta(z-ct)$

If we think of this as electromagnetic radiation, we wouldn't have E parallel to the direction of travel, and it must be perpendicular to B. Therefore, if we write:

$\mathbf{E}=E(\mathbf{x})\hat{\mathbf{y}}$

We find:

$\frac{\partial E(\mathbf{x})}{\partial z}=-\frac{\mu_{0}K}{2}\delta(z+ct)+\frac{\mu_{0}K}{2}\delta(z-ct)$ $\frac{\partial E(\mathbf{x})}{\partial x}=\frac{\partial E(\mathbf{x})}{\partial y}=0$

So that:

$\begin{matrix}\mathbf{E}(|z|<ct) & = & -\frac{\mu_0 K}{2}\hat{\mathbf{i}} \\ \mathbf{E}(|z|>ct) & = & 0 \end{matrix}$

We find the power:

$P=\frac{\partial}{\partial t}\int dV(E^{2}+B^{2})/8\pi=\frac{1}{4\pi}\int dV\left(E\frac{\partial E}{\partial t}+B\frac{\partial B}{\partial t}\right)$

Per unit area, using symmetry:

$\frac{P}{A}=\frac{1}{2\pi}\frac{\partial}{\partial t}\int_{0}^{\infty}dz\left(E^{2}+B^{2}\right)$

Plugging in, using $Θ(s)$ as the heavyside function:

$\frac{P}{A}=\frac{1}{2\pi}\frac{\partial}{\partial t}\int_{0}^{\infty}dz\left(\left(-\frac{\mu_{0}K}{2}\Theta(ct-z)\right)^{2}+\left(\frac{\mu_{0}K}{2}\Theta(ct-z)\right)^{2}\right)$ $\frac{P}{A}=\frac{1}{2\pi}\frac{\partial}{\partial t}\int_{0}^{ct}dz\left(\frac{\mu_{0}K}{2}\right)^{2}=\frac{c}{2\pi}\left(\frac{\mu_{0}K}{2}\right)^{2}$

The power radiated per unit area, as defined on page 347 of Griffiths, is simply the Poynting vector (Given in SI units, with |E| = c |B|):