Question

Two concentric spherical shells carry uniformly distributed charges +Q(at radius a) and -Q (at radius ). They are immersed in a uniform magnetic field $\mathit{B}\mathbf{=}{\mathit{B}}_{\mathbf{0}}\hat{\mathbf{z}}$.

(a) Find the angular momentum of the fields (with respect to the center).

(b) Now the magnetic field is gradually turned off. Find the torque on each sphere, and the resulting angular momentum of the system.

Short answer

(a) The angular momentum of the fields is $L=\frac{1}{3}Q{B}_0\left(b^2-a^2\right)\hat{z}$.

(b) The net torque is $N=\frac{Q}{3}\frac{dB}{dt}\left(b^2-a^2\right)\hat{z}$and the angular momentum is $L=\frac{1}{3}Q{B}_0\left(b^2-a^2\right)\hat{z}$.

Step-by-step solution

Expression for the angular momentum of the fields:

Write the expression for the angular momentum of the fields.

$\mathit{L}\mathbf{=}\mathbf{\int }\left(r\times g\right)\mathit{d}\mathit{\tau }$ …… (1)

Here, g is the momentum density.

Determine the angular momentum of the fields:

(a)

Write the expression for the momentum density.

$g=\epsilon \left(E\times B\right)$ …… (2)

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

Write the expression for the electric field.

$E=\frac{Q}{4\pi \epsilon }\frac{1}{r^2}\hat{r}$

Substitute the values in equation (2).

localid="1657537044442" $$\begin{gathered} g={\epsilon }_0\left(\frac{Q}{4\pi {\epsilon }_0}\frac{1}{r^2}\hat{r}\times B\right) \\ g=\frac{Q{B}_0}{4\pi r^2}\left(\hat{r}\times \hat{z}\right) \end{gathered}$$

Substitute the known values in equation (1).

$$\begin{gathered} L=\int \left(r\times \frac{QB}{4\pi r^2}\left(\hat{r}\times \hat{z}\right)\right)d\tau \\ L=\frac{Q{B}_0}{4\pi r}\int \frac{1}{r^2}\left[r\times \left(\hat{r}\times \hat{z}\right)\right]r^2\sin \theta drd\theta d \\ L=\frac{Q{B}_0}{4\pi r}\int r\left[\hat{r}\left[\hat{r}\times \hat{z}\right]\right]\sin \theta drd\theta d......\left(3\right) \end{gathered}$$

Since,

$$\begin{gathered} \hat{r}\times \left(\hat{r}\times \hat{z}\right)=\hat{r}\left(\hat{r}.\hat{z}\right)-\hat{z}\left(\hat{r}.\hat{r}\right) \\ \hat{r}\times \left(\hat{r}\times \hat{z}\right)=\hat{r}\cos \theta \\ \hat{r}\times \left(\hat{r}\times \hat{z}\right)=\hat{z} \end{gathered}$$

As L has to be along the z-direction, pick the z component of $\hat{r}$. Hence,

localid="1657537019616" $$\begin{gathered} {\left[\hat{r}\times \left(\hat{r}\times \hat{z}\right)\right]}_z={\cos }^2\theta -1 \\ \left[\hat{r}\times {\left(\hat{r}\times \hat{z}\right)}_z\right]=-{\sin }^2\theta \end{gathered}$$

From equation (3),

localid="1657534367950" $$\begin{gathered} L=-\frac{Q{B}_0}{4\pi r}\int r{\sin }^3\theta drd\theta d\varphi \\ L=-\frac{Q{B}_0}{4\pi r}\left(2\pi \right){\int }_0^{\pi }{\sin }^3\theta d\theta {\int }_a^{b}rdr \\ L=-\frac{Q{B}_0}{2}\left(\frac{4}{3}\right)\left(\frac{b^2-a^2}{2}\right) \\ L=-\frac{1}{3}Q{B}_0\left(b^2-a^2\right)\hat{z} \end{gathered}$$

Therefore, the angular momentum of the fields is $L=-\frac{1}{3}Q{B}_0\left(b^2-a^2\right)\hat{z}$.

Determine the torque on each sphere and resulting angular momentum of the system:

(b)

Write the expression for the torque on the patch.

$N=s\times dF$ …… (4)

Here, dF is the force on a patch.

Write the expression for the force on a patch.

$dF=E\sigma da$ …… (5)

Write the expression for the electric field.

$E=-\frac{s}{2}\frac{dB}{dt}\hat{\varphi }$

Substitute the known values in equation (5).

$$\begin{gathered} dF=\left(-\frac{s}{2}\frac{dB}{dt}\hat{\varphi }\right)\sigma da\left(\begin{array}{c}\begin{array}{c}\therefore s=a\sin \theta \hat{s}\\ da=a^2\sin \theta d\theta d\varphi \end{array}\end{array}\right) \\ dF=\left(-\frac{s}{2}\frac{dB}{dt}\hat{\varphi }\frac{dB}{dt}\hat{\varphi }\right)\sigma \left(a^2\sin \theta d\theta d\varphi \right) \end{gathered}$$

Substitute the known values in equation (4) to calculate the net torque on the sphere at radius a.

$$\begin{gathered} N=s\times \left(-\frac{a\sin \theta \hat{s}}{2}\frac{dB}{dt}\hat{\varphi }\right)\sigma \left(a^2\sin \theta d\theta d\varphi \right) \\ {N}_a=--\frac{(a\sin \theta )}{2}\frac{dB}{dt}\sigma a^3{\sin }^2\theta \left(\hat{s}\times \hat{\varphi }\right)\left(d\theta d\varphi \right) \\ {N}_a=--\frac{(a^4{\sin }^3\theta )}{2}\frac{dB}{dt}\left(\frac{Q}{4\pi a^2}\right)\left(\hat{s}\times \hat{\varphi }\right)\left(d\theta d\varphi \right) \\ {N}_a=-\frac{Q{a}^2}{8\pi }\frac{dB}{dt}\hat{z}\left(2\pi \right){\int }_0^{\pi }{\sin }^3\theta d\theta {\int }_0^{2\pi }d\varphi \end{gathered}$$

On further solving,

$$\begin{gathered} N_a=-\frac{Q{a}^2}{8}\frac{dB}{dt}\left(\hat{z}\right)\left(2\pi \right)\left(\frac{4}{3}\right) \\ {N}_a=-\frac{Q{a}^2}{3}\frac{dB}{dt}\hat{z} \end{gathered}$$

Calculate the net torque on the sphere at radius b.

$N_b=\frac{Q{b}^2}{3}\frac{dB}{dt}\hat{z}$

Hence, the total torque on each sphere will be,

localid="1657536473852" $$\begin{gathered} N=N_b-N_a \\ N=\left(\frac{Q{b}^2}{3}\frac{dB}{dt}\hat{z}\right)-\left(\frac{Q{a}^2}{3}\frac{dB}{dt}\hat{z}\right) \\ N=\frac{Q}{3}\frac{dB}{dt}\left(b^2-a^2\right)\hat{z} \end{gathered}$$

Calculate the angular momentum delivered to the spheres.

$$\begin{gathered} L=\int Ndt \\ L=\int \frac{Q}{3}\frac{dB}{dt}\left(b^2-a^2\right)\hat{z}dt \\ L=\frac{Q}{3}\left(b^2-a^2\right)\hat{z}{\int }_{B_0}^0 \\ L=-\frac{1}{3}Q{B}_0\left(b^2-a^2\right)\hat{z} \end{gathered}$$

Therefore, the net torque is $N=\frac{Q}{3}\frac{dB}{dt}\left(b^2-a^2\right)\hat{z}$and the angular momentum is $L=-\frac{1}{3}Q{B}_0\left(b^2-a^2\right)\hat{z}$.