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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:

$A(x,t)=\frac{{渭}_0}{4蟺}\text{鈥嬧赌�}鈭�\frac{K\left(t_r\right)\text{鈥�}}{r}da\mathrm{...}\text{(i)}$

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

$\begin{array}{c}A(x,t)=\frac{{渭}_0\widehat{z}}{4蟺}\text{鈥嬧赌�}鈭�\frac{K\left(t_r\right)\text{鈥�}}{\sqrt{r^2+x^2}}2蟺rdr\\ =\frac{{渭}_0\widehat{z}}{2}\text{鈥嬧赌�}鈭�\frac{K(t鈭�\frac{\sqrt{r^2+x^2}}{c})\text{鈥�}}{\sqrt{r^2+x^2}}rdr\mathrm{...}\text{(ii)}\end{array}$

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

$t鈭�\frac{\sqrt{r^2+x^2}}{c}=0$

Therefore,

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

(b)

Let,

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

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,

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

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

Hence, we can modify equation (ii) as:

$A(x,t)=\frac{{渭}_0\widehat{z}}{2}\text{鈥�}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}(\text{鈥�}t鈭�\frac{x}{c}鈭�u)du\mathrm{...}\text{(iv)}$

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

$\begin{array}{c}E(x,t)=鈭�\frac{鈭�A}{鈭�t}\\ =鈭�\frac{{渭}_0\widehat{z}}{2}\text{鈥�}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}\frac{鈭�}{鈭�t}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)du\end{array}$

Now, as we know that:

$\frac{鈭�}{鈭�t}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)=\frac{鈭�}{鈭�u}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)$

Therefore,

$\begin{array}{c}E(x,t)=鈭�\frac{{渭}_{0}c\widehat{z}}{2}\text{鈥�}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}\frac{鈭�}{鈭�u}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)du\\ =\frac{{渭}_{0}c\widehat{z}}{2}K{\text{鈥�}(\text{鈥�}t鈭�\frac{x}{c}鈭�u)|}_0^{\mathrm{鈭�}}\\ =鈭�\frac{{渭}_{0}c}{2}\left[K\right(t鈭�\frac{x}{c})鈭�K(鈭�\mathrm{鈭�}\left)\right]\widehat{z}\\ =鈭�\frac{{渭}_{0}c}{2}K(t鈭�\frac{x}{c})\widehat{z}\end{array}$

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

$\begin{array}{c}B(x,t)=鈭�脳A\\ =鈭�\frac{鈭�A_z}{鈭�x}\widehat{y}\\ =鈭�\frac{{渭}_{0}c\widehat{y}}{2}\text{鈥�}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}\frac{鈭�}{鈭�x}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)du\end{array}$

Now, as we know that:

$\frac{鈭�}{鈭�x}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)=\frac{1}{c}\frac{鈭�}{鈭�u}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)$

Therefore,

$\begin{array}{c}B(x,t)=鈭�\frac{{渭}_0\widehat{y}}{2}\text{鈥�}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}\frac{鈭�}{鈭�u}K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)du\\ =鈭�\frac{{渭}_0\widehat{y}}{2}{\left[K(\text{鈥�}t鈭�\frac{x}{c}鈭�u)\right]|}_0^{\mathrm{鈭�}}\\ =鈭�\frac{{渭}_0}{2}\left[K\right(t鈭�\frac{x}{c})鈭�K(鈭�\mathrm{鈭�}\left)\right]\widehat{y}\\ =\frac{{渭}_0}{2}K(t鈭�\frac{x}{c})\widehat{y}\end{array}$

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

Therefore,

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

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:

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