8 Suppose the (electrically neutral) yz plane carries a time-dependent but uniform surface current K (t) Z.

(a) Find the electric and magnetic fields at a height x above the plane if

(i) a constant current is turned on at t = 0:

$\mathit{K}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\begin{array}{l}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{\le }\mathbf{0}\\ {\mathbf{K}}_{\mathbf{0}}\mathbf{,}\mathbf{t}\mathbf{>}\mathbf{0}\end{array}\mathbf{\right\}}$

(ii) a linearly increasing current is turned on at t = 0:

$\mathit{K}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\begin{array}{l}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{\le }\mathbf{0}\\ \mathbf{\alpha }\mathbf{t}\mathbf{,}\mathbf{t}\mathbf{>}\mathbf{0}\end{array}\mathbf{\right\}}$

(b) Show that the retarded vector potential can be written in the form, and from

$\mathit{A}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{c}}{\mathbf{2}}\widehat{\mathbf{z}}\underset{0}{\overset{\infty }{\int }}K(t-\frac{x}{c}-u)\mathit{d}\mathit{u}$

And from this determine E and B.

(c) Show that the total power radiated per unit area of surface is

$\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{c}}{\mathbf{2}}{\mathbf{\left[}\mathbf{K}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right]}}^{\mathbf{2}}$

Explain what you mean by "radiation," in this case, given that the source is not localized.22

When a statically charged particle is placed in space, the field will be produced due to charged particle, called an Electric field.

The field produced around a magnet and magnetism's effect is called the magnetic field.

(a)

We know that:

Modifying equation 11.16, we can write:

$A(x,t)=\frac{{\mu }_0}{4\pi }\text{}\int \frac{K\left(t_r\right)\text{}}{r}da\mathrm{...}\text{(i)}$

By solving simplifying equation (i), we can write:

$\begin{array}{c}A(x,t)=\frac{{\mu }_0\widehat{z}}{4\pi }\text{}\int \frac{K\left(t_r\right)\text{}}{\sqrt{r^2+x^2}}2\pi rdr\\ =\frac{{\mu }_0\widehat{z}}{2}\text{}\int \frac{K(t-\frac{\sqrt{r^2+x^2}}{c})\text{}}{\sqrt{r^2+x^2}}rdr\mathrm{...}\text{(ii)}\end{array}$

To maximize or minimize the function, we have to perform the derivative of equation (ii) and equate it with zero. Hence, we can find the condition of maximum r

$t-\frac{\sqrt{r^2+x^2}}{c}=0$

Therefore,

$r_{\max }=\sqrt{c^2{t}^2-x^2}\text{}$As we know that K(t) is 0 for t less than zero.

(b)

Let,

$u\equiv \frac{1}{c}(\sqrt{r^2+x^2}-x)\mathrm{...}\text{(iii)}$

Therefore, we can find its derivative as:

$\begin{array}{c}du=\frac{1}{c}\left(\frac{1}{2}\frac{1}{\sqrt{r^2+x^2}}2rdr\right)\\ =\frac{1}{c}\frac{r}{\sqrt{r^2+x^2}}dr\end{array}$

And we can also calculate,

$t-\frac{\sqrt{r^2+x^2}}{c}=t-\frac{x}{c}-u$

Because when r tends from 0 to infinity, u will also be tending from 0 to infinity.

Hence, we can modify equation (ii) as:

$A(x,t)=\frac{{\mu }_0\widehat{z}}{2}\text{}\underset{0}{\overset{\infty }{\int }}(\text{}t-\frac{x}{c}-u)du\mathrm{...}\text{(iv)}$

As we know, Electric field intensity (E) can be calculated as

$\begin{array}{c}E(x,t)=-\frac{\partial A}{\partial t}\\ =-\frac{{\mu }_0\widehat{z}}{2}\text{}\underset{0}{\overset{\infty }{\int }}\frac{\partial }{\partial t}K(\text{}t-\frac{x}{c}-u)du\end{array}$

Now, as we know that:

$\frac{\partial }{\partial t}K(\text{}t-\frac{x}{c}-u)=\frac{\partial }{\partial u}K(\text{}t-\frac{x}{c}-u)$

Therefore,

$\begin{array}{c}E(x,t)=-\frac{{\mu }_{0}c\widehat{z}}{2}\text{}\underset{0}{\overset{\infty }{\int }}\frac{\partial }{\partial u}K(\text{}t-\frac{x}{c}-u)du\\ =\frac{{\mu }_{0}c\widehat{z}}{2}K{\text{}(\text{}t-\frac{x}{c}-u)|}_0^{\infty }\\ =-\frac{{\mu }_{0}c}{2}\left[K\right(t-\frac{x}{c})-K(-\infty \left)\right]\widehat{z}\\ =-\frac{{\mu }_{0}c}{2}K(t-\frac{x}{c})\widehat{z}\end{array}$

As we know, Magnetic flux density (B) can be calculated as:

$\begin{array}{c}B(x,t)=\nabla \times A\\ =-\frac{\partial A_z}{\partial x}\widehat{y}\\ =-\frac{{\mu }_{0}c\widehat{y}}{2}\text{}\underset{0}{\overset{\infty }{\int }}\frac{\partial }{\partial x}K(\text{}t-\frac{x}{c}-u)du\end{array}$

Now, as we know that:

$\frac{\partial }{\partial x}K(\text{}t-\frac{x}{c}-u)=\frac{1}{c}\frac{\partial }{\partial u}K(\text{}t-\frac{x}{c}-u)$

Therefore,

$\begin{array}{c}B(x,t)=-\frac{{\mu }_0\widehat{y}}{2}\text{}\underset{0}{\overset{\infty }{\int }}\frac{\partial }{\partial u}K(\text{}t-\frac{x}{c}-u)du\\ =-\frac{{\mu }_0\widehat{y}}{2}{\left[K(\text{}t-\frac{x}{c}-u)\right]|}_0^{\infty }\\ =-\frac{{\mu }_0}{2}\left[K\right(t-\frac{x}{c})-K(-\infty \left)\right]\widehat{y}\\ =\frac{{\mu }_0}{2}K(t-\frac{x}{c})\widehat{y}\end{array}$

We know that to calculate the power radiated, we have to use thePoynting vector:

Therefore,

$\begin{array}{c}S=\frac{1}{{\mu }_0}(E\times B)\\ =\frac{1}{{\mu }_0}\left[-\frac{{\mu }_{0}c}{2}K(t-\frac{x}{c})\widehat{z}\times \frac{{\mu }_0}{2}K(t-\frac{x}{c})\widehat{y}\right]\\ =\frac{1}{{\mu }_0}\frac{{\mu }_{0}c}{2}\frac{{\mu }_0}{2}K(t-\frac{x}{c})[-\widehat{z}\times \widehat{y}]\\ =\frac{{\mu }_{0}c}{4}{\left[K(t-\frac{x}{c})\right]}^2\widehat{x}\end{array}$

Therefore, the power radiated we got by solving the Poynting vector is the power per unit area, which reaches a certain point x at a certain time t. It passes away from the surface at a time instant (t-x/c), and the amount of energy radiated towards the downward direction is the same.

Hence,

The time at which the total power left the surface is:

$\frac{{\mu }_{0}c}{2}{\left[K\right(t\left)\right]}^2$