91Ó°ÊÓ

Write the expression for the induced emf in the ring.

$\mathit{Î\mu }\mathbf{=}{\mathit{l}}_{\mathbf{r}}\mathit{R}$…… (1)

Here, $l_r$is the current and R is the resistance.

(a)

Write the expression for the induced emf.

$\mathrm{Î\mu }=-\frac{d\mathrm{Ï•}}{dt}$…… (2)

Here, $\mathrm{Ï•}$is the magnetic flux.

Write the expression for the magnetic flux.

$$\begin{gathered} \mathrm{Ï•}=BA \\ \mathrm{Ï•}=\mathrm{Ï€}{a}^{2}B \end{gathered}$$

Here, B is the magnetic field, and a is the radius.

$B={\mathrm{μ}}_{0}n{l}_s$

Here, ${\mathrm{μ}}_0$is the permeability of free space, n is the number of turns and $l_s$is the current.

Substitute all the known values in equation (2).

$$\begin{gathered} \mathrm{Î\mu }=-\frac{d}{dt}\left(\mathrm{Ï€}{a}^{2}B\right) \\ \mathrm{Î\mu }=-\frac{d}{dt}\left[\left(\mathrm{Ï€}{a}^2\right)\left({\mathrm{μ}}_{0}n{l}_s\right)\right] \\ \mathrm{Î\mu }=-{\mathrm{μ}}_{0}n\mathrm{Ï€}{a}^2\left(\frac{d{l}_s}{dt}\right) \end{gathered}$$

Substitute the known values in equation (1).

localid="1657521563292" $$\begin{gathered} -{\mathrm{μ}}_{0}n\mathrm{π}{a}^2\left(\frac{d{l}_s}{dt}\right)=l,R \\ {l}_r=\frac{-{\mathrm{μ}}_{0}n\mathrm{π}{a}^2{\displaystyle \frac{d{l}_s}{dt}}}{R} \\ {l}_r=-\frac{1}{R}\left({\mathrm{μ}}_0\mathrm{π}{a}^{2}n\right)\frac{d{l}_s}{dt} \end{gathered}$$

Therefore, the value of $l_r$in terms of $\frac{d{l}_s}{dt}$is$l_r=-\frac{1}{R}\left({\mathrm{μ}}_0\mathrm{π}{A}^{2}n\right)\frac{d{l}_s}{dt}$.

(b)

Write the expression for the Poynting vector.

$S=\frac{1}{{\mathrm{μ}}_0}(E×B)$…… (3)

Here, E is the electric field, and B is the magnetic field.

Using Gauss law, write the expression for an electric field.

$âˆ\circledR E.dl=-\frac{d\mathrm{Ï•}}{dt}$

Substitute the known values in the above expression.

$$\begin{gathered} E\left(2\mathrm{π}a\right)=-{\mathrm{μ}}_{0}n\mathrm{π}{a}^2\left(\frac{d{l}_s}{dt}\right) \\ E=-\frac{1}{2}{\mathrm{μ}}_{0}an\left(\frac{d{l}_s}{dt}\right)\widehat{\mathrm{ϕ}} \end{gathered}$$

Write the expression for a magnetic field.

$B=\frac{{\mathrm{μ}}_{0l_r}}{2}\frac{b^2}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\hat{Z}$

Substitute all the known values in equation (3).

$$\begin{gathered} S=\frac{1}{{\mathrm{μ}}_0}-\frac{1}{2}{\mathrm{μ}}_{0}an\left(\frac{d{l}_s}{dt}\right)\widehat{\mathrm{ϕ}}×\frac{{\mathrm{μ}}_0{l}_r}{2}\frac{b^2}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}} \\ S=-\frac{1}{4}{\mathrm{μ}}_0{l}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\hat{S} \end{gathered}$$

Write the expression for the delivered power.

$$\begin{gathered} P=∫S.da \\ P={∫}_{-00}^{00}S\left(2\mathrm{π}a\right)dz \end{gathered}$$

Substitute the known values in the above expression.

$$\begin{gathered} P=∫-\frac{1}{4}{\mathrm{μ}}_0{L}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2+)}^{{\displaystyle \frac{3}{2}}}}\widehat{S\left(2\mathrm{π}a\right)dz} \\ P=-\frac{1}{2}\mathrm{π}{\mathrm{μ}}_0{a}^2{b}^{2}n{l}_r\frac{d{l}_s}{dt}{∫}_{-00}^{00}\frac{1}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}dz \end{gathered}$$

Take the surface integral of the above equation.

localid="1657521729321" $$\begin{gathered} P=-\frac{1}{2}\mathrm{π}{\mathrm{μ}}_0{a}^2{b}^{2}n{l}_r\frac{d{l}_s}{dt}{∫}_{-∞}^{∞}d\frac{1}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}} \\ P=-\frac{1}{2}\mathrm{π}{\mathrm{μ}}_0{a}^2{b}^{2}n{l}_r{\left[\frac{d{l}_s}{dt}\frac{z}{b^2\sqrt{z^2}+b^2}\right]}_{-∞}^{∞} \\ P=-\frac{1}{2}\mathrm{π}{\mathrm{μ}}_0{a}^2{b}^{2}n{l}_r\frac{d{l}_s}{dt}\left(\frac{2}{b^2}\right) \end{gathered}$$

On further solving,

$P=-\left(\mathrm{π}{\mathrm{μ}}_0{a}^{2}n\frac{d{l}_s}{dt}\right)l_r$

Here, $R{l}_r=-{\mathrm{μ}}_0\mathrm{π}{a}^{2}n\frac{d{l}_s}{dt}$.

Hence,

$P=L_r^2$

Therefore, the Poynting vector is $S=-\frac{1}{4}{\mathrm{μ}}_0{l}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\hat{S}$, and the total power is $p=-l_r^{2}R$.