Question

Calculate the force of magnetic attraction between the northern and southern hemispheres of a uniformly charged spinning spherical shell, with radius R, angular velocity $\mathit{\omega }$, and surface charge density $\mathit{\sigma }$. [This is the same as Prob.$5.44$, but this time use the Maxwell stress tensor and Eq.8.21.]

Short answer

The force of magnetic attraction between the northern and southern hemispheres is $F=-\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{2}\right)}^2\widehat{z}$.

Step-by-step solution

Expression for Maxwell-stress tensor:

Write the expression of Maxwell-stress tensor:

$T_{ij}={\epsilon }_0\left(E_i{E}_j-\frac{1}{2}{\delta }_{ij}{E}^2\right)+\frac{1}{{\mu }_0}\left(B_i{B}_j-\frac{1}{2}{\delta }_{ij}{B}^2\right)$…… (1)

Here, $T_{ij}$is the magnitude of the force acting per unit area in the $i^{\text{th}}$direction of the surface, which is oriented in the $j^{\text{th}}$ direction, E is the electric field, and B is the magnetic field.

Determine the magnitude of the net force for the hemisphere:

Write the formula for the magnitude of an electromagnetic force on a charge in volume V.

$F=\oint \overrightarrow{T}\cdot da-{\mu }_0{\epsilon }_0\frac{d}{dt}\int S\cdot d\tau$

Using equation (1), the Maxwell-stress tensor equation becomes,

$\begin{array}{c}{\left(\overrightarrow{T}\cdot da\right)}_z=T_{zx}d{a}_x+T_{zy}d{a}_y+T_{zz}d{a}_z\\ {\left(\overrightarrow{T}\cdot da\right)}_z=\frac{1}{{\mu }_0}\left(B_z{B}_{x}d{a}_x+B_z{B}_{y}d{a}_y+B_z{B}_{z}d{a}_z-\frac{1}{2}{B}^{2}d{a}_z\right)\\ {\left(\overrightarrow{T}\cdot da\right)}_z=\frac{1}{{\mu }_0}\left[B_z\left(B\cdot da\right)-\frac{1}{2}{B}^{2}d{a}_z\right]\end{array}$….. (2)

Write the magnetic field inside the sphere.

$B=\frac{2}{3}{\mu }_0\sigma R\omega \widehat{z}$

Write the magnetic field outside the sphere.

$B=\frac{{\mu }_{0}m}{4\pi r^3}\left(2\cos \theta \widehat{r}+\sin \theta \widehat{\theta }\right)$….. (3)

Here, m is the magnetic momentum which is given by,

$m=\frac{4}{3}\pi R^3\left(\sigma \omega R\right)$

Here, R is the radius of the spherical shell, $\sigma$is the surface charge density and $\omega$is the angular velocity of the spinning shell.

For the hemisphere, the equation (3) becomes,

$\begin{array}{c}{B}_z=\frac{{\mu }_{0}m}{4\pi R^3}\left[2\cos \theta {\left(\widehat{r}\right)}_z+\sin \theta {\left(\widehat{\theta }\right)}_z\right]\\ B_z=\frac{{\mu }_{0}m}{4\pi R^3}\left[2{\cos }^2\theta -{\sin }^2\theta \right]\\ B_z=\frac{{\mu }_{0}m}{4\pi R^3}\left(3{\cos }^2\theta -1\right)\end{array}$

Write the areal vector.

$\begin{array}{c}da=R^2\sin \theta d\theta d\varphi \widehat{r}\\ d{a}_z=R^2\sin \theta d\theta d\varphi \cos \theta \end{array}$

Rewrite the equation as,

$B_z=\frac{{\mu }_{0}m}{4\pi R^3}\left(2\cos \theta \right)R^2\sin \theta d\theta d\varphi$

Solve further as,

$\begin{array}{c}{B}^2={\left(\frac{{\mu }_{0}m}{74\pi R^3}\right)}^2\left(4{\cos }^2\theta +{\sin }^2\theta \right)\\ B^2={\left(\frac{{\mu }_{0}m}{4\pi R^3}\right)}^2\left(3{\cos }^2\theta +1\right)\end{array}$

Substitute the known values in equation (2).

$\begin{array}{c}{\left(\overrightarrow{T}\cdot da\right)}_z=\frac{1}{{\mu }_0}{\left(\frac{{\mu }_{0}m}{4\pi R^3}\right)}^2\left[\begin{array}{l}\left(3{\cos }^2\theta -1\right)2\cos \theta R^2\sin \theta d\theta d\varphi \\ -\frac{1}{2}\left(3{\cos }^2\theta +1\right)R^2\sin \theta \cos \theta d\theta d\varphi \end{array}\right]\\ {\left(\overrightarrow{T}\cdot da\right)}_z={\mu }_0{\left(\frac{\sigma \omega R}{3}\right)}^2\left[\frac{1}{2}{R}^2\sin \theta \cos \theta d\theta d\varphi \right]\left(12{\cos }^2\theta -4-3{\cos }^2\theta -1\right)\\ {\left(\overrightarrow{T}\cdot da\right)}_z=\frac{{\mu }_0}{2}{\left(\frac{\sigma \omega R^2}{3}\right)}^2\left(9{\cos }^2\theta -5\right)\sin \theta \cos \theta d\varphi \end{array}$

Calculate the net force for the hemisphere.

$\begin{array}{c}{\left(F_{\text{hemi}}\right)}_z=\frac{{\mu }_0}{2}{\left(\frac{\sigma \omega R^2}{3}\right)}^{2}2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(9{\cos }^2\theta -5\right)\sin \theta d\theta \\ {\left(F_{\text{hemi}}\right)}_z={\mu }_0\pi {\left(\frac{\sigma \omega R^2}{3}\right)}^2{\left[-\frac{9}{4}{\cos }^4\theta +\frac{5}{2}{\cos }^2\theta \right]}_0^{\frac{\pi }{2}}\\ {\left(F_{\text{hemi}}\right)}_z={\mu }_0\pi {\left(\frac{\sigma \omega R^2}{3}\right)}^2\left[0+\frac{9}{4}-\frac{5}{2}\right]\\ {\left(F_{\text{hemi}}\right)}_z=-\frac{{\mu }_0\pi }{4}{\left(\frac{\sigma \omega R^2}{3}\right)}^2\end{array}$

Determine the magnitude of the net force for the disk:

Write the magnetic field inside the disk.

$B_z=\frac{2}{3}{\mu }_0\sigma R\omega$

Write the areal vector.

$\begin{array}{rcl}da& =& rdrd\varphi \hat{\varphi }\\ da& =& -rdrd\varphi \hat{z}\backslash \\ d{a}_z& =& -rdrd\varphi \end{array}$

Consider the magnetic field equation.

$B·d{a}_z=-\frac{2}{3}{\mu }_0\sigma R\omega rdrd\varphi$

Rewrite the magnetic field equation as,

$B^2={\left(\frac{2}{3}{\mu }_0\sigma R\omega \right)}^2$

Substitute the known values in equation (2).

$\begin{array}{rcl}{\left(\overrightarrow{T}·da\right)}_z& =& \frac{1}{{\mu }_0}{\left(\frac{2}{3}{\mu }_0\sigma R\omega \right)}^2\left[-rdrd\varphi +\frac{1}{2}rdrd\varphi \right]\\ {\left(\overrightarrow{T}·da\right)}_z& =& -\frac{1}{2{\mu }_0}{\left(\frac{2}{3}{\mu }_0\sigma R\omega \right)}^{2}rdrd\varphi \\ {\left(\overrightarrow{T}·da\right)}_z& =& -2{\mu }_0{\left(\frac{\sigma \omega R}{3}\right)}^{2}2\pi rdrd\varphi \end{array}$

Calculate the net force for a disk.

$$\begin{gathered} {\left(F_{\text{disk}}\right)}_z=-2{\mu }_0{\left(\frac{\sigma \omega R}{3}\right)}^{2}2\pi {\int }_0^{R}rdr \\ {\left(F_{\text{disk}}\right)}_z=-2\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{3}\right)}^2 \end{gathered}$$

Determine the force of magnetic attraction between the northern and southern hemispheres:

Write the force of magnetic attraction between the northern and southern hemispheres.

$F={\left(F_{\text{hemi}}\right)}_z+{\left(F_{\text{disk}}\right)}_z$

Substitute the known values in the above equation.

$$\begin{gathered} F=\left[-\frac{{\mu }_0\pi }{4}{\left(\frac{\sigma \omega R^2}{3}\right)}^2\right]+\left[-2\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{3}\right)}^2\right] \\ F=-\frac{{\mu }_0\pi }{4}{\left(\frac{\sigma \omega R^2}{3}\right)}^2-2\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{3}\right)}^2 \\ F=-\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{3}\right)}^2\left(2+\frac{1}{4}\right)\hat{z} \\ F=-\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{2}\right)}^2\hat{z} \end{gathered}$$

Final Solution:

Therefore, the force of magnetic attraction between the northern and southern hemispheres is $F=-\pi {\mu }_0{\left(\frac{\sigma \omega R^2}{2}\right)}^2\hat{z}$.