91Ó°ÊÓ

Write the expression of Maxwell-stress tensor:

$T_{ij}={\mathrm{Î\mu }}_0\left(E_i{E}_j-\frac{1}{2}{\mathrm{δ}}_{ij}{E}^2\right)+\frac{1}{{\mathrm{μ}}_0}\left(B_i{B}_j-\frac{1}{2}{\mathrm{δ}}_{ij}{B}^2\right)$…… (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.

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

$F=âˆ\circledR \stackrel{→}{T}â‹\dots da−{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{d}{dt}∫Sâ‹\dots d\mathrm{Ï„}$

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

$\begin{array}{c}{\left(\stackrel{→}{T}â‹\dots da\right)}_z=T_{zx}d{a}_x+T_{zy}d{a}_y+T_{zz}d{a}_z\\ {\left(\stackrel{→}{T}â‹\dots da\right)}_z=\frac{1}{{\mathrm{μ}}_0}\left(B_z{B}_{x}d{a}_x+B_z{B}_{y}d{a}_y+B_z{B}_{z}d{a}_z−\frac{1}{2}{B}^{2}d{a}_z\right)\\ {\left(\stackrel{→}{T}â‹\dots da\right)}_z=\frac{1}{{\mathrm{μ}}_0}\left[B_z\left(Bâ‹\dots da\right)−\frac{1}{2}{B}^{2}d{a}_z\right]\end{array}$….. (2)

Write the magnetic field inside the sphere.

$B=\frac{2}{3}{\mathrm{μ}}_0\mathrm{σ}R\mathrm{Ó\neg }\widehat{z}$

Write the magnetic field outside the sphere.

$B=\frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{r}^3}\left(2\cos \mathrm{θ}\widehat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)$….. (3)

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

$m=\frac{4}{3}\mathrm{Ï€}{R}^3\left(\mathrm{σ}\mathrm{Ó\neg }R\right)$

Here, R is the radius of the spherical shell, $\mathrm{σ}$is the surface charge density and $\mathrm{Ó\neg }$is the angular velocity of the spinning shell.

For the hemisphere, the equation (3) becomes,

$\begin{array}{c}{B}_z=\frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{R}^3}\left[2\cos \mathrm{θ}{\left(\widehat{r}\right)}_z+\sin \mathrm{θ}{\left(\widehat{\mathrm{θ}}\right)}_z\right]\\ B_z=\frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{R}^3}\left[2{\cos }^2\mathrm{θ}−{\sin }^2\mathrm{θ}\right]\\ B_z=\frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{R}^3}\left(3{\cos }^2\mathrm{θ}−1\right)\end{array}$

Write the areal vector.

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

Rewrite the equation as,

$B_z=\frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{R}^3}\left(2\cos \mathrm{θ}\right)R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}$

Solve further as,

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

Substitute the known values in equation (2).

$\begin{array}{c}{\left(\stackrel{→}{T}â‹\dots da\right)}_z=\frac{1}{{\mathrm{μ}}_0}{\left(\frac{{\mathrm{μ}}_{0}m}{4\mathrm{Ï€}{R}^3}\right)}^2\left[\begin{array}{l}\left(3{\cos }^2\mathrm{θ}−1\right)2\cos \mathrm{θ}{R}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\\ −\frac{1}{2}\left(3{\cos }^2\mathrm{θ}+1\right)R^2\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\end{array}\right]\\ {\left(\stackrel{→}{T}â‹\dots da\right)}_z={\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }R}{3}\right)}^2\left[\frac{1}{2}{R}^2\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\right]\left(12{\cos }^2\mathrm{θ}−4−3{\cos }^2\mathrm{θ}−1\right)\\ {\left(\stackrel{→}{T}â‹\dots da\right)}_z=\frac{{\mathrm{μ}}_0}{2}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2\left(9{\cos }^2\mathrm{θ}−5\right)\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{Ï•}\end{array}$

Calculate the net force for the hemisphere.

$\begin{array}{c}{\left(F_{\text{hemi}}\right)}_z=\frac{{\mathrm{μ}}_0}{2}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^{2}2\mathrm{Ï€}\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}\left(9{\cos }^2\mathrm{θ}−5\right)\sin \mathrm{θ}d\mathrm{θ}\\ {\left(F_{\text{hemi}}\right)}_z={\mathrm{μ}}_0\mathrm{Ï€}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2{\left[−\frac{9}{4}{\cos }^4\mathrm{θ}+\frac{5}{2}{\cos }^2\mathrm{θ}\right]}_0^{\frac{\mathrm{Ï€}}{2}}\\ {\left(F_{\text{hemi}}\right)}_z={\mathrm{μ}}_0\mathrm{Ï€}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2\left[0+\frac{9}{4}−\frac{5}{2}\right]\\ {\left(F_{\text{hemi}}\right)}_z=−\frac{{\mathrm{μ}}_0\mathrm{Ï€}}{4}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2\end{array}$

Write the magnetic field inside the disk.

$B_z=\frac{2}{3}{\mathrm{μ}}_0\mathrm{σ}R\mathrm{Ó\neg }$

Write the areal vector.

$\begin{array}{rcl}da& =& rdrd\mathrm{Ï•}\widehat{\mathrm{Ï•}}\\ da& =& -rdrd\mathrm{Ï•}\hat{z}\backslash \\ d{a}_z& =& -rdrd\mathrm{Ï•}\end{array}$

Consider the magnetic field equation.

$B·d{a}_z=-\frac{2}{3}{\mathrm{μ}}_0\mathrm{σ}R\mathrm{Ó\neg }rdrd\mathrm{Ï•}$

Rewrite the magnetic field equation as,

$B^2={\left(\frac{2}{3}{\mathrm{μ}}_0\mathrm{σ}R\mathrm{Ó\neg }\right)}^2$

Substitute the known values in equation (2).

$\begin{array}{rcl}{\left(\stackrel{→}{T}·da\right)}_z& =& \frac{1}{{\mathrm{μ}}_0}{\left(\frac{2}{3}{\mathrm{μ}}_0\mathrm{σ}R\mathrm{Ó\neg }\right)}^2\left[-rdrd\mathrm{Ï•}+\frac{1}{2}rdrd\mathrm{Ï•}\right]\\ {\left(\stackrel{→}{T}·da\right)}_z& =& -\frac{1}{2{\mathrm{μ}}_0}{\left(\frac{2}{3}{\mathrm{μ}}_0\mathrm{σ}R\mathrm{Ó\neg }\right)}^{2}rdrd\mathrm{Ï•}\\ {\left(\stackrel{→}{T}·da\right)}_z& =& -2{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }R}{3}\right)}^{2}2\mathrm{Ï€}rdrd\mathrm{Ï•}\end{array}$

Calculate the net force for a disk.

$$\begin{gathered} {\left(F_{\text{disk}}\right)}_z=-2{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }R}{3}\right)}^{2}2\mathrm{Ï€}{∫}_0^{R}rdr \\ {\left(F_{\text{disk}}\right)}_z=-2\mathrm{Ï€}{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2 \end{gathered}$$

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.

$$\begin{gathered} F=\left[-\frac{{\mathrm{μ}}_0\mathrm{Ï€}}{4}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2\right]+\left[-2\mathrm{Ï€}{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2\right] \\ F=-\frac{{\mathrm{μ}}_0\mathrm{Ï€}}{4}{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2-2\mathrm{Ï€}{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2 \\ F=-\mathrm{Ï€}{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{3}\right)}^2\left(2+\frac{1}{4}\right)\hat{z} \\ F=-\mathrm{Ï€}{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{2}\right)}^2\hat{z} \end{gathered}$$

Therefore, the force of magnetic attraction between the northern and southern hemispheres is $F=-\mathrm{Ï€}{\mathrm{μ}}_0{\left(\frac{\mathrm{σ}\mathrm{Ó\neg }{R}^2}{2}\right)}^2\hat{z}$.