91Ó°ÊÓ

The given data can be listed below as:

- The charge density of the solid sphere is $\mathrm{ÒÏ}$.

- The radius of the sphere is $R$.

- The angular velocity of the sphere is $\mathrm{Ó\neg }$.

The magnetic field is described as a region that surrounds a moving charge which helps the charge to exert magnetism force on another object. The magnetic field is also helps to distribute the magnetic forces around a particular magnetic material.

The equation of the example 5.11 is expressed as:

$\begin{array}{rcl}A\left(r,\mathrm{θ},\mathrm{Ï•}\right)& =& \frac{{\mathrm{μ}}_{0}R\mathrm{Ó\neg }\mathrm{σ}}{3}r\sin \mathrm{θ}\widehat{\mathrm{Ï•}}\left(râ\copyright ½R\right)\\ & =& \frac{{\mathrm{μ}}_0{R}^4\mathrm{Ó\neg }\mathrm{σ}}{3}\frac{\sin \mathrm{θ}}{r^2}\widehat{\mathrm{Ï•}}\left(râ\copyright ¾R\right)\\ & & \end{array}$

Here, $A(r,\mathrm{θ},\mathrm{Ï•})$is described as the vector potential of the cylindrical coordinates, ${\mathrm{μ}}_0$is the permeability,$R$is the radius of the sphere, $\mathrm{Ó\neg }$is the angular velocity, $\mathrm{σ}$is the elongation, $r$is the change in the radius, $\mathrm{θ}$is the angle subtended by the sphere and $\widehat{\mathrm{Ï•}}$is the unit vector.

Substitute $R→\stackrel{¯}{r}$and $\mathrm{σ}→\mathrm{ÒÏ}d\stackrel{¯}{r}$in the above equation.

localid="1658743919596" $\begin{array}{rcl}A& =& \frac{{\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}}{3}\frac{\sin \mathrm{θ}}{r^2}\widehat{\mathrm{Ï•}}\underset{0}{\overset{r}{∫}}{\stackrel{¯}{r}}^{4}d\stackrel{¯}{r}+\frac{{\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}}{3}r\sin \mathrm{θ}\widehat{\mathrm{Ï•}}\underset{r}{\overset{R}{∫}}\stackrel{¯}{r}d\stackrel{¯}{r}\\ & =& \frac{{\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}}{3}\sin \mathrm{θ}\left[\frac{1}{r^2}\left(\frac{r^5}{5}\right)+\frac{r}{2}\left(R^2-r^2\right)\right]\widehat{\mathrm{Ï•}}\\ & =& \frac{{\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}}{2}r\sin \mathrm{θ}\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\widehat{\mathrm{Ï•}}\\ & & \end{array}$

The equation of the magnetic field is expressed as:

$B=∇×A$

Here, $B$is the magnetic field and $∇$is the curl.

$\frac{{\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}}{2}r\sin \mathrm{θ}\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\widehat{\mathrm{Ï•}}$for $A$in the above equation.

Substitute

$B=\frac{{\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}}{2}\left\{\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left[\sin \mathrm{θ}r\sin \mathrm{θ}\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\right]\hat{r}-\frac{1}{r}\frac{∂}{∂r}\left[r^2\sin \mathrm{θ}\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\right]\widehat{\mathrm{θ}}\right\}$

$={\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}\left[\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\cos \mathrm{θ}\hat{r}-\left(\frac{R^2}{3}-\frac{2r^2}{5}\right)\sin \mathrm{θ}\widehat{\mathrm{θ}}\right]$

Thus, the magnetic field inside a solid sphere is ${\mathrm{μ}}_0\mathrm{Ó\neg }\mathrm{ÒÏ}\left[\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\cos \mathrm{θ}\hat{r}-\left(\frac{R^2}{3}-\frac{2r^2}{5}\right)\sin \mathrm{θ}\widehat{\mathrm{θ}}\right]$.