91Ó°ÊÓ

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:

$A(x,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\text{​â¶Ä‹}∫\frac{K\left(t_r\right)\text{​}}{r}da\mathrm{...}\text{(i)}$

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

$\begin{array}{c}A(x,t)=\frac{{\mathrm{μ}}_0\widehat{z}}{4\mathrm{Ï€}}\text{​â¶Ä‹}∫\frac{K\left(t_r\right)\text{​}}{\sqrt{r^2+x^2}}2\mathrm{Ï€}rdr\\ =\frac{{\mathrm{μ}}_0\widehat{z}}{2}\text{​â¶Ä‹}∫\frac{K(t−\frac{\sqrt{r^2+x^2}}{c})\text{​}}{\sqrt{r^2+x^2}}rdr\mathrm{...}\text{(ii)}\end{array}$

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,

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

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:

$A(x,t)=\frac{{\mathrm{μ}}_0\widehat{z}}{2}\text{​}\underset{0}{\overset{\mathrm{∞}}{∫}}(\text{​}t−\frac{x}{c}−u)du\mathrm{...}\text{(iv)}$

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

$\begin{array}{c}E(x,t)=−\frac{∂A}{∂t}\\ =−\frac{{\mathrm{μ}}_0\widehat{z}}{2}\text{​}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{∂}{∂t}K(\text{​}t−\frac{x}{c}−u)du\end{array}$

Now, as we know that:

$\frac{∂}{∂t}K(\text{​}t−\frac{x}{c}−u)=\frac{∂}{∂u}K(\text{​}t−\frac{x}{c}−u)$

Therefore,

$\begin{array}{c}E(x,t)=−\frac{{\mathrm{μ}}_{0}c\widehat{z}}{2}\text{​}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{∂}{∂u}K(\text{​}t−\frac{x}{c}−u)du\\ =\frac{{\mathrm{μ}}_{0}c\widehat{z}}{2}K{\text{​}(\text{​}t−\frac{x}{c}−u)|}_0^{\mathrm{∞}}\\ =−\frac{{\mathrm{μ}}_{0}c}{2}\left[K\right(t−\frac{x}{c})−K(−\mathrm{∞}\left)\right]\widehat{z}\\ =−\frac{{\mathrm{μ}}_{0}c}{2}K(t−\frac{x}{c})\widehat{z}\end{array}$

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

$\begin{array}{c}B(x,t)=∇×A\\ =−\frac{∂A_z}{∂x}\widehat{y}\\ =−\frac{{\mathrm{μ}}_{0}c\widehat{y}}{2}\text{​}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{∂}{∂x}K(\text{​}t−\frac{x}{c}−u)du\end{array}$

Now, as we know that:

$\frac{∂}{∂x}K(\text{​}t−\frac{x}{c}−u)=\frac{1}{c}\frac{∂}{∂u}K(\text{​}t−\frac{x}{c}−u)$

Therefore,

$\begin{array}{c}B(x,t)=−\frac{{\mathrm{μ}}_0\widehat{y}}{2}\text{​}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{∂}{∂u}K(\text{​}t−\frac{x}{c}−u)du\\ =−\frac{{\mathrm{μ}}_0\widehat{y}}{2}{\left[K(\text{​}t−\frac{x}{c}−u)\right]|}_0^{\mathrm{∞}}\\ =−\frac{{\mathrm{μ}}_0}{2}\left[K\right(t−\frac{x}{c})−K(−\mathrm{∞}\left)\right]\widehat{y}\\ =\frac{{\mathrm{μ}}_0}{2}K(t−\frac{x}{c})\widehat{y}\end{array}$

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

Therefore,

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

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:

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