Question

Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell.

Short answer

The magnetic force of attraction is $\frac{\pi }{4}{\mu }_0{\omega }^2{\delta }^2{R}^4$.

Step-by-step solution

Significance of the magnetic force

The magnetic force mainly arises amongst the particles having electrically charged and in also motion. However, the magnetic force is mainly the repulsion or attraction amongst the charged particles.

Determination of the magnetic attraction force

The equation of the surface element’s force is expressed as:

$B=\frac{1}{2}(B_{in}+B_{out})$

Here, is the surface element’s force, $B_{in}$and $B_{out}$are force inside and outside of the spinning shell.

The equation of the magnetic scalar potential’s curl of the inside of the shell is expressed as:

$A_{out}=\frac{{\mu }_0{R}^4\omega \delta }{3}\frac{\sin \theta }{r^2}\hat{\varnothing }$

Here, $A_{in}$is the magnetic scalar potential’s curl of the inside of the shell, $R$is the radius, ${\mu }_0$is the permeability constant, $\omega$is the angular velocity, $\delta$is the variability measurement, is the distance from the field, is the angle subtended and localid="1657519378591" $\hat{\varphi }$is the cap of the angle.

The cross product of $\nabla$with the $A_{in}$is expressed as:

$$\begin{gathered} \nabla \times A_{in}=B_{in} \\ =\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\frac{{\mu }_{0}R\omega \delta }{3}r{\sin }^2\theta \right)\hat{r}-\frac{1}{r}\frac{\partial }{\partial \theta }\left(r\frac{{\mu }_{0}R\omega \delta }{3}r{\sin }^2\theta \right)\hat{\theta } \end{gathered}$$

The cross product of $\nabla$with the $A_{out}$is expressed as:

$$\begin{gathered} \nabla \times A_{out}=B_{out} \\ =\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\frac{{\mu }_0{R}^4\omega \delta }{3r^2}{\sin }^2\theta \right)\hat{r}-\frac{1}{r}\frac{\partial }{\partial \theta }\left(r\frac{{\mu }_0{R}^4\omega \delta }{3r^2}{\sin }^2\theta \right)\hat{\theta } \\ =\frac{2}{3}{\mu }_0{R}^4\delta \omega \frac{1}{r^3}(2\cos \theta \hat{r}+\sin \theta \hat{\theta }) \end{gathered}$$

The equation of the average surface of the sphere is expressed as:

$B_{avg}=\frac{1}{2}{\left[(B_{in}+B_{out})\right]}_{r-R}$

The above equation can be reduced as:

$$\begin{gathered} B_{avg}=\frac{1}{2}\left[\frac{1}{3}{\mu }_0{R}^4\delta \omega \frac{1}{R^3}(2\cos \theta \hat{r}+\sin \theta \hat{\theta })+\frac{2}{3}{\mu }_{0}R\delta \omega (\cos \theta \hat{r}-\sin \theta \hat{\theta }\right] \\ =\frac{1}{6}R\omega \delta \mu (4\cos \theta \hat{r}-\sin \theta \hat{\theta }) \end{gathered}$$

The equation of the force on the differential surface element is expressed as:

$$\begin{gathered} dF=dqv\times B_{avg} \\ =\delta dSv\times B_{avg} \end{gathered}$$ …(i)

Here, dqv is the differential surface element.

The equation of the velocity of the shell is expressed as:

$v=R\sin \theta \omega \hat{\varphi }$

The differentiation of the above equation is expressed as:

$dS=R^2\sin \theta d\varphi d\theta$

Substituting $R\sin \theta \omega \hat{\varphi }$for $v$and $R^2\sin \theta d\varphi d\theta$for $dS$in the equation (i).

$$\begin{gathered} dF=\frac{1}{6}{\mu }_0{\delta }^2{\omega }^2{R}^4{\sin }^2\theta d\theta d\varphi \hat{\varphi }\times (4\cos \theta \hat{r}-\sin \theta \hat{\theta }) \\ =\frac{1}{6}{\mu }_0{\delta }^2{\omega }^2{R}^4{\sin }^2\theta (4\cos \theta \hat{\theta }-\sin \theta \hat{r})d\theta d\varphi \end{gathered}$$

The equation of the z component of the force of the shell is expressed as:

$d{F}_z=dF‐\hat{z}$

…(ii)

Here, $dF$is the differential force and $\hat{z}$is the position vector.

The equation of the angle subtended by the force is expressed as:

$\hat{\theta }=\cos \theta \hat{s}-\sin \theta \hat{z}$

The equation of the position vector in the $z$direction is expressed as:

localid="1658556502484" $\hat{\mathit{r}}=\mathbf{sin}\theta \hat{s}+\mathbf{cos}\theta \hat{\mathbf{z}}$

Substituting the values in the equation (ii).

$$\begin{gathered} dF=\frac{1}{6}{\mu }_0{\omega }^2{\delta }^2{R}^4{\sin }^2(4\cos \theta \sin \theta -\cos \theta \sin \theta )d\theta \varphi \\ =\frac{1}{2}{\mu }_0{\omega }^2{\delta }^2{R}^4{\sin }^3\theta \cos \theta d\theta d\varphi \end{gathered}$$

Double integrating the above equation with the integral limits can be expressed as:

$$\begin{gathered} F=\frac{1}{2}{\mu }_0{\delta }^2{\omega }^2{R}^4{\int }_0^{2\pi }{\int }_0^{\pi /2}{\sin }^3\theta \cos d\theta d\varphi \\ =\pi {\mu }_0{\omega }^2{\delta }^2{R}^4{\int }_0^{\mathrm{\pi }/2}{\sin }^3\mathrm{\theta cos}\mathrm{d\theta } \end{gathered}$$

…(iii)

The change in the variables can be expressed as:

$$\begin{gathered} x=\sin \theta \\ dx=\cos \theta d\theta \end{gathered}$$

Substitute the value of the integral in the equation (iii).

$$\begin{gathered} F=\pi {\mu }_0{\omega }^2{\delta }^2{R}^4{\int }_0^1{x}^{3}dx \\ =\pi {\mu }_0{\omega }^2{\delta }^2{R}^4\left(\frac{{\mathrm{x}}^4}{4}\right) \\ =\pi {\mu }_0{\omega }^2{\delta }^2{R}^4\left(\frac{1}{4}\right) \\ =\frac{\pi }{4}\pi {\mu }_0{\omega }^2{\delta }^2{R}^4 \end{gathered}$$

Thus, the magnetic force of attraction is $\frac{\mathrm{\pi }}{4}{\mu }_0{\omega }^2{\delta }^2{R}^4$.