Question

Imagine an iron sphere of radius R that carries a charge Q and a uniform magnetization M=M$\hat{\mathit{z}}$. The sphere is initially at rest.

(a) Compute the angular momentum stored in the electromagnetic fields.

(b) Suppose the sphere is gradually (and uniformly) demagnetized (perhaps by heating it up past the Curie point). Use Faraday’s law to determine the induced electric field, find the torque this field exerts on the sphere, and calculate the total angular momentum imparted to the sphere in the course of the demagnetization.

(c) Suppose instead of demagnetizing the sphere we discharge it, by connecting a grounding wire to the north pole. Assume the current flows over the surface in such a way that the charge density remains uniform. Use the Lorentz force law to determine the torque on the sphere, and calculate the total angular momentum imparted to the sphere in the course of the discharge. (The magnetic field is discontinuous at the surface ….does this matter?) [Answer:$\frac{\mathbf{2}}{\mathbf{9}}{\mathit{\mu }}_{\mathbf{0}}\mathit{M}\mathit{Q}{\mathit{R}}^{\mathbf{2}}$ ]

Short answer

(a) The angular momentum stored in the electromagnetic fields is $L=\frac{2}{9}{\mu }_{0}MQ{R}^2\hat{z}$.

(b) The induced electric field is $E=-\frac{{\mu }_0}{3}\frac{dM}{dt}\left(r\sin \theta \right)\hat{\varphi }$, the torque exerted on the sphere is $N=-\frac{2{\mu }_0}{9}Q{R}^2\frac{dM}{dt}\hat{z}$and the total angular momentum imparted to the sphere in the course of the demagnetization is $L=\frac{2}{9}{\mu }_{0}MQ{R}^2\hat{z}$.

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

Step-by-step solution

Expression for the electric and magnetic field when r<Rand r>R :

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

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

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

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

Determine the angular momentum stored in the electromagnetic fields:

(a)

Write the expression for the angular momentum density.

$l=r\times g$…… (1)

Here, g is the momentum density.

Write the expression for the momentum density.

$g={\epsilon }_0\left(E\times B\right)$

Substitute the known values in the above expression.

$g=\epsilon \left(\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\hat{r}\times \frac{{\mu }_{0}m}{4\pi r^3}\left[2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right]\right)$

$g=\frac{{\mu }_0}{{\left(4\pi r\right)}^2}\frac{Qm}{r^5}\left(\hat{r}\times \hat{\theta }\right)\sin \theta$

$g=\frac{{\mu }_0}{{\left(4\pi \right)}^2}\frac{Qm}{r^5}s\mathrm{in}\theta \hat{\varphi }$

Substitute the known values in equation (1).

$l=r\left(\frac{{\mu }_0}{{\left(4\pi \right)}^2}\frac{Qm}{r^5}s\mathrm{in}\theta \hat{\varphi }\right)$

$l=\frac{{\mu }_0}{{\left(4\pi \right)}^2}\frac{mQ}{r^4}s\mathrm{in}\theta \hat{\varphi }\left(\hat{r}\times \hat{\varphi }\right)$

Since,

$\left(\hat{r}\times \hat{\varphi }\right)=-\hat{\theta }$

Calculate the angular momentum stored in the electromagnetic fields.

$L=\frac{{\mu }_{0}mQ}{{\left(4\pi \right)}^2}\hat{z}\int \frac{{\sin }^2}{r^4}\left(r^2\sin \theta drd\theta d\varphi \right)$ ........(2)

Since,

$$\begin{gathered} {\int }_0^{2\pi r}d\varphi =2\pi \\ {\int }_0^{\pi r}{\sin }^{3}d\varphi =\frac{4}{3} \\ {\int }_R^{\infty }\frac{1}{r^2}dr=\left(-\frac{1}{r}\right) \\ {\left(-\frac{1}{r}\right)}_R^{\infty }=\frac{1}{R} \end{gathered}$$

Substitute the values in equation (2).

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

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

Therefore, the angular momentum stored in the electromagnetic fields is

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

Determine the induced electric field, and the torque exerts on the sphere and the total angular momentum.

(b)

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

$$\begin{gathered} \oint E.dl=-\frac{d\varphi }{dt} \\ E=-\pi {\left(r\sin \theta \right)}^2\left(\frac{2}{3}{\mu }_0\frac{dM}{dt}\right) \\ E=-\frac{{\mu }_0}{3}\frac{dM}{dt}\left(r\sin \theta \right)\hat{\varphi } \end{gathered}$$

Write the expression for the torque on the patch.

$dN=r\times dF$ …… (3)

Here, is the force on a patch.

Write the expression for the force on the patch.

$dF=\sigma Eda$

Substitute the known values in the above expression.

$$\begin{gathered} dF=\left(\frac{Q}{4\pi R^2}\right)\left(-\frac{{\mu }_0}{3}\frac{dM}{dt}\left(r\sin \theta \right)\hat{\varphi }\right)da \\ dF=-\frac{{\mu }_0\sigma }{3}\frac{dM}{dt}\left(r\sin \theta \right)da\hat{\varphi } \end{gathered}$$

Substitute the known values in equation (3).

$\int dN=\int r\times \left(-\frac{{\mu }_0\sigma }{3}\frac{dM}{dt}\left(r\sin \theta \right)da\hat{\varphi }\right)$

$N=-\frac{{\mu }_0\sigma }{3}\frac{dM}{dt}\hat{z}\int r^2{\sin }^2\theta \left(r\sin \theta d\theta d\varphi \right).......\left(4\right)$

Since,

r=R

$$\begin{gathered} {\int }_0^{\pi }{\sin }^3\theta d\theta =\frac{4}{3} \\ {\int }_0^{2\pi }d\varphi =2\pi \end{gathered}$$

Substitute the values in equation (4).

$$\begin{gathered} N=-\frac{{\mu }_0\sigma }{3}\frac{dM}{dt}\hat{z}{R}^4\left(\frac{4}{3}\right)\left(2\pi \right) \\ N=-\frac{2{\mu }_0}{9}Q{R}^2\frac{dM}{dt}\hat{z} \end{gathered}$$

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

$L=\int Ndt$

Substitute the known values in the above expression.

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

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

Determine the torque on the sphere and the total angular momentum on the sphere:

(c)

Calculate the charge (South) of the ring.

$$\begin{gathered} q_s=\sigma \left(2\pi R^2\right){\int }_0^{\pi }\sin \theta \text{'}d\theta \text{'} \\ {q}_s=\frac{q}{2}{\left[\left(-\cos \theta \text{'}\right)\right]}_0^{\pi } \\ {q}_s=\frac{q}{2}\left(1+\cos \theta \right) \end{gathered}$$

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

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

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

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

The value of K is given as:

$$\begin{gathered} K\left(t\right)=\frac{1}{2\pi R\sin \theta }(-\hat{\theta }) \\ K\left(t\right)=\frac{1}{4\pi R}\frac{dq}{dt}\frac{\left(1+\cos \theta \right)}{\sin \theta }\hat{\theta } \end{gathered}$$

The value of B is given as:

$B=\left[\frac{2}{3}{\mu }_{0}M\hat{z}+\frac{{\mu }_0}{4\pi }\frac{{\displaystyle \frac{4}{3}}\pi R^{3}M}{R^3}\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right)\right]\frac{1}{2}=\frac{{\mu }_{0}M}{6}\left[2\hat{z}+\cos \theta \hat{r}+\sin \theta \hat{\theta }\right]$

Substitute the known values in equation (5).

$$\begin{gathered} dF=\frac{1}{4\pi R}\frac{dq}{dt}\frac{{\mu }_{0}M}{6}\frac{\left(1+\cos \theta \right)}{\sin \theta }\left[2\left(\hat{\theta }\times \hat{z}\right)+2\cos \theta \left(\hat{\theta }\times \hat{\hat{r}}\right)\right]] \\ dF=\frac{1}{4\pi R}\frac{dq}{dt}\frac{{\mu }_{0}M}{6}\frac{\left(1+\cos \theta \right)}{\sin \theta }\left[2\left(\hat{\theta }\times \hat{z}\right)+2\cos \theta \left(-\hat{\varphi }\right)\right] \end{gathered}$$

Calculate the torque on the sphere.

$$\begin{gathered} dN=R\hat{r}\times dF \\ dN=\frac{{\mu }_{0}M}{24\pi }\left(\frac{dq}{dt}\right)\frac{\left(1+\cos \right)}{\sin \theta }2\left[\hat{r}\times \left(\hat{\theta }\times \hat{z}\right)-\cos \theta \left(\hat{r}\times \hat{\varphi }\right)\right]R^2\sin \theta d\theta d\varphi \\ dN=\frac{{\mu }_{0}M}{24\pi }\left(\frac{dq}{dt}\right)\left(1+\cos \theta \right)R^2\left[\cos \theta \hat{\theta }+\cos \theta \hat{\theta }\right]d\theta d\varphi \\ dN=\frac{{\mu }_{0}M}{24\pi }\left(\frac{dq}{dt}\right)\left(1+\cos \theta \right)\cos \theta d\theta d\varphi \hat{\theta } \end{gathered}$$

Integrate the above equation.

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

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

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

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