91Ó°ÊÓ

Write the expression for an electric field when $r<R$and $r>R$.

$E=\left\{\begin{array}{ll}0,& \left(r<R\right)\\ \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r^2}\hat{r}& \left(r>R\right)\end{array}\right.$

Write the expression for a magnetic field when $r<R$and role="math" localid="1657523196296" $r>R$.

$$\begin{gathered} B=\left\{\begin{array}{ll}\frac{2}{3}{\mathrm{μ}}_0\mathrm{M}\hat{\mathrm{z}}& \left(r<R\right)\\ \frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{r}^3}\left[2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right],& \left(r>R\right)\end{array}\right. \\ \mathrm{Here},\mathrm{m}=\frac{4}{3}\mathrm{π}{R}^{\mathit{3}}M \end{gathered}$$

(a)

Write the expression for the angular momentum density.

$l=r×g$…… (1)

Here, g is the momentum density.

Write the expression for the momentum density.

$g={\mathrm{Î\mu }}_0\left(E×B\right)$

Substitute the known values in the above expression.

$g=\mathrm{Î\mu }\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r^2}\hat{r}×\frac{{\mathrm{μ}}_{0}m}{4\mathrm{Ï€}{r}^3}\left[2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right]\right)$

$g=\frac{{\mathrm{μ}}_0}{{\left(4\mathrm{π}r\right)}^2}\frac{Qm}{r^5}\left(\hat{r}×\widehat{\mathrm{θ}}\right)\sin \mathrm{θ}$

$g=\frac{{\mathrm{μ}}_0}{{\left(4\mathrm{π}\right)}^2}\frac{Qm}{r^5}s\mathrm{in}\mathrm{θ}\widehat{\mathrm{ϕ}}$

Substitute the known values in equation (1).

$l=r\left(\frac{{\mathrm{μ}}_0}{{\left(4\mathrm{π}\right)}^2}\frac{Qm}{r^5}s\mathrm{in}\mathrm{θ}\widehat{\mathrm{ϕ}}\right)$

$l=\frac{{\mathrm{μ}}_0}{{\left(4\mathrm{π}\right)}^2}\frac{mQ}{r^4}s\mathrm{in}\mathrm{θ}\widehat{\mathrm{ϕ}}\left(\hat{r}×\widehat{\mathrm{ϕ}}\right)$

Since,

$\left(\hat{r}×\widehat{\mathrm{ϕ}}\right)=-\widehat{\mathrm{θ}}$

Calculate the angular momentum stored in the electromagnetic fields.

$L=\frac{{\mathrm{μ}}_{0}mQ}{{\left(4\mathrm{π}\right)}^2}\hat{z}∫\frac{{\sin }^2}{r^4}\left(r^2\sin \mathrm{θ}drd\mathrm{θ}d\mathrm{ϕ}\right)$ ........(2)

Since,

$$\begin{gathered} {∫}_0^{2\mathrm{π}r}d\mathrm{ϕ}=2\mathrm{π} \\ {∫}_0^{\mathrm{π}r}{\sin }^{3}d\mathrm{ϕ}=\frac{4}{3} \\ {∫}_R^{∞}\frac{1}{r^2}dr=\left(-\frac{1}{r}\right) \\ {\left(-\frac{1}{r}\right)}_R^{∞}=\frac{1}{R} \end{gathered}$$

Substitute the values in equation (2).

$L=\frac{{\mathrm{μ}}_{0}mQ}{{\left(4\mathrm{π}\right)}^2}\hat{z}\left(2\mathrm{π}\right)\left(\frac{4}{3}\right)\left(\frac{1}{R}\right)$

$L=\frac{2}{9}{\mathrm{μ}}_{0}MQ{R}^2\hat{z}$

Therefore, the angular momentum stored in the electromagnetic fields is

$L=\frac{2}{9}{\mathrm{μ}}_{0}MQ{R}^2\hat{z}$

(b)

Apply Faraday’s law to the ring to calculate the induced electric field.

$$\begin{gathered} âˆ\circledR E.dl=-\frac{d\mathrm{Ï•}}{dt} \\ E=-\mathrm{Ï€}{\left(r\sin \mathrm{θ}\right)}^2\left(\frac{2}{3}{\mathrm{μ}}_0\frac{dM}{dt}\right) \\ E=-\frac{{\mathrm{μ}}_0}{3}\frac{dM}{dt}\left(r\sin \mathrm{θ}\right)\widehat{\mathrm{Ï•}} \end{gathered}$$

Write the expression for the torque on the patch.

$dN=r×dF$ …… (3)

Here, is the force on a patch.

Write the expression for the force on the patch.

$dF=\mathrm{σ}Eda$

Substitute the known values in the above expression.

$$\begin{gathered} dF=\left(\frac{Q}{4\mathrm{π}{R}^2}\right)\left(-\frac{{\mathrm{μ}}_0}{3}\frac{dM}{dt}\left(r\sin \mathrm{θ}\right)\widehat{\mathrm{ϕ}}\right)da \\ dF=-\frac{{\mathrm{μ}}_0\mathrm{σ}}{3}\frac{dM}{dt}\left(r\sin \mathrm{θ}\right)da\widehat{\mathrm{ϕ}} \end{gathered}$$

Substitute the known values in equation (3).

$∫dN=∫r×\left(-\frac{{\mathrm{μ}}_0\mathrm{σ}}{3}\frac{dM}{dt}\left(r\sin \mathrm{θ}\right)da\widehat{\mathrm{ϕ}}\right)$

$N=-\frac{{\mathrm{μ}}_0\mathrm{σ}}{3}\frac{dM}{dt}\hat{z}∫r^2{\sin }^2\mathrm{θ}\left(r\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}\right).......\left(4\right)$

Since,

r=R

$$\begin{gathered} {∫}_0^{\mathrm{π}}{\sin }^3\mathrm{θ}d\mathrm{θ}=\frac{4}{3} \\ {∫}_0^{2\mathrm{π}}d\mathrm{ϕ}=2\mathrm{π} \end{gathered}$$

Substitute the values in equation (4).

$$\begin{gathered} N=-\frac{{\mathrm{μ}}_0\mathrm{σ}}{3}\frac{dM}{dt}\hat{z}{R}^4\left(\frac{4}{3}\right)\left(2\mathrm{π}\right) \\ N=-\frac{2{\mathrm{μ}}_0}{9}Q{R}^2\frac{dM}{dt}\hat{z} \end{gathered}$$

Write the expression for the total angular momentum imparted to the sphere in the course of the demagnetization.

$L=∫Ndt$

Substitute the known values in the above expression.

$$\begin{gathered} L=∫\left(-\frac{2{\mathrm{μ}}_0}{9}Q{R}^2\frac{dM}{dt}\hat{z}\right)dt \\ L=-\frac{2{\mathrm{μ}}_0}{9}Q{R}^2\hat{z}{∫}_M^{0}dM \\ L=\frac{2{\mathrm{μ}}_0}{9}MQ{R}^2\hat{z} \end{gathered}$$

Therefore, the induced electric field is $E=-\frac{{\mathrm{μ}}_0}{3}\frac{dM}{dt}\left(r\sin \mathrm{θ}\right)\widehat{\mathrm{ϕ}}$, the torque exerted on the sphere is $N=-\frac{2{\mathrm{μ}}_0}{9}Q{R}^2\frac{dM}{dt}\hat{z}$and the total angular momentum imparted to the sphere in the course of the demagnetization is $L=\frac{2{\mathrm{μ}}_0}{9}MQ{R}^2\hat{z}$.

(c)

Calculate the charge (South) of the ring.

$$\begin{gathered} q_s=\mathrm{σ}\left(2\mathrm{π}{R}^2\right){∫}_0^{\mathrm{π}}\sin \mathrm{θ}\text{'}d\mathrm{θ}\text{'} \\ {q}_s=\frac{q}{2}{\left[\left(-\cos \mathrm{θ}\text{'}\right)\right]}_0^{\mathrm{π}} \\ {q}_s=\frac{q}{2}\left(1+\cos \mathrm{θ}\right) \end{gathered}$$

Calculate the total current crossing the ring (flowing north).

$l\left(t\right)=-\frac{1}{2}\frac{dq}{dt}\left(1+\cos \mathrm{θ}\right)$

Write the expression for the force on a patch of area da.

$dF=\left(K×B\right)da.....\left(5\right)$

The value of K is given as:

$$\begin{gathered} K\left(t\right)=\frac{1}{2\mathrm{π}R\sin \mathrm{θ}}(-\widehat{\mathrm{θ}}) \\ K\left(t\right)=\frac{1}{4\mathrm{π}R}\frac{dq}{dt}\frac{\left(1+\cos \mathrm{θ}\right)}{\sin \mathrm{θ}}\widehat{\mathrm{θ}} \end{gathered}$$

The value of B is given as:

$B=\left[\frac{2}{3}{\mathrm{μ}}_{0}M\hat{z}+\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{{\displaystyle \frac{4}{3}}\mathrm{π}{R}^{3}M}{R^3}\left(2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)\right]\frac{1}{2}=\frac{{\mathrm{μ}}_{0}M}{6}\left[2\hat{z}+\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right]$

Substitute the known values in equation (5).

$$\begin{gathered} dF=\frac{1}{4\mathrm{π}R}\frac{dq}{dt}\frac{{\mathrm{μ}}_{0}M}{6}\frac{\left(1+\cos \mathrm{θ}\right)}{\sin \mathrm{θ}}\left[2\left(\widehat{\mathrm{θ}}×\hat{z}\right)+2\cos \mathrm{θ}\left(\widehat{\mathrm{θ}}×\hat{\hat{r}}\right)\right]] \\ dF=\frac{1}{4\mathrm{π}R}\frac{dq}{dt}\frac{{\mathrm{μ}}_{0}M}{6}\frac{\left(1+\cos \mathrm{θ}\right)}{\sin \mathrm{θ}}\left[2\left(\widehat{\mathrm{θ}}×\hat{z}\right)+2\cos \mathrm{θ}\left(-\widehat{\mathrm{ϕ}}\right)\right] \end{gathered}$$

Calculate the torque on the sphere.

$$\begin{gathered} dN=R\hat{r}×dF \\ dN=\frac{{\mathrm{μ}}_{0}M}{24\mathrm{π}}\left(\frac{dq}{dt}\right)\frac{\left(1+\cos \right)}{\sin \mathrm{θ}}2\left[\hat{r}×\left(\widehat{\mathrm{θ}}×\hat{z}\right)-\cos \mathrm{θ}\left(\hat{r}×\widehat{\mathrm{ϕ}}\right)\right]R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ} \\ dN=\frac{{\mathrm{μ}}_{0}M}{24\mathrm{π}}\left(\frac{dq}{dt}\right)\left(1+\cos \mathrm{θ}\right)R^2\left[\cos \mathrm{θ}\widehat{\mathrm{θ}}+\cos \mathrm{θ}\widehat{\mathrm{θ}}\right]d\mathrm{θ}d\mathrm{ϕ} \\ dN=\frac{{\mathrm{μ}}_{0}M}{24\mathrm{π}}\left(\frac{dq}{dt}\right)\left(1+\cos \mathrm{θ}\right)\cos \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}\widehat{\mathrm{θ}} \end{gathered}$$

Integrate the above equation.

$$\begin{gathered} N_z=\frac{{\mathrm{μ}}_{0}M{R}^2}{6\mathrm{π}}\left(\frac{dq}{dt}\right)\left(2\mathrm{π}\right){∫}_0^{\mathrm{π}}\left(1+\cos \mathrm{θ}\right)\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{θ} \\ {N}_z=\frac{{\mathrm{μ}}_{0}M{R}^2}{6\mathrm{π}}\left(\frac{dq}{dt}\right)\left(\frac{2}{3}\right) \\ {N}_z=-\frac{2{\mathrm{μ}}_0}{9}M{R}^2\frac{dq}{dt}\hat{z} \end{gathered}$$

Calculate the total angular momentum imparted to the sphere in the course of the discharge.

$$\begin{gathered} L=∫Ndt \\ L=∫-\frac{2{\mathrm{μ}}_0}{9}M{R}^2\frac{dq}{dt}\hat{z}dt \\ L=-\frac{2{\mathrm{μ}}_0}{9}M{R}^2\hat{z}{∫}_Q^{0}dq \\ L=-\frac{2{\mathrm{μ}}_0}{9}M{R}^{2}Q\hat{z} \end{gathered}$$

Therefore, the torque on the sphere is $N=-\frac{2{\mathrm{μ}}_0}{9}M{R}^2\frac{dq}{dt}\hat{z}$and the total angular momentum imparted to the sphere in the course of the discharge is $L=-\frac{2{\mathrm{μ}}_0}{9}M{R}^{2}Q\hat{z}$.