Question

Use the results of Ex. $5.11$to find the magnetic field inside a solid sphere, of uniform charge density $\rho$and radius $R$, that is rotating at a constant angular velocity $\backslash omega.$

Short answer

The magnetic field inside a solid sphere is .

Step-by-step solution

Identification of given data

The given data can be listed below as:

- The charge density of the solid sphere is $\rho$.

- The radius of the sphere is $R$.

- The angular velocity of the sphere is $\omega$.

Significance of the magnetic field

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.

Determination of the magnetic field inside a solid sphere

The equation of the example 5.11 is expressed as:

$\begin{array}{rcl}A\left(r,\theta ,\varphi \right)& =& \frac{{\mu }_{0}R\omega \sigma }{3}r\sin \theta \hat{\varphi }\left(r⩽R\right)\\ & =& \frac{{\mu }_0{R}^4\omega \sigma }{3}\frac{\sin \theta }{r^2}\hat{\varphi }\left(r⩾R\right)\\ & & \end{array}$

Here, $A(r,\theta ,\varphi )$is described as the vector potential of the cylindrical coordinates, ${\mu }_0$is the permeability,$R$is the radius of the sphere, $\omega$is the angular velocity, $\sigma$is the elongation, $r$is the change in the radius, $\theta$is the angle subtended by the sphere and $\hat{\varphi }$is the unit vector.

Substitute $R\to \overline{r}$and $\sigma \to \rho d\overline{r}$in the above equation.

localid="1658743919596" $\begin{array}{rcl}A& =& \frac{{\mu }_0\omega \rho }{3}\frac{\sin \theta }{r^2}\hat{\varphi }\underset{0}{\overset{r}{\int }}{\overline{r}}^{4}d\overline{r}+\frac{{\mu }_0\omega \rho }{3}r\sin \theta \hat{\varphi }\underset{r}{\overset{R}{\int }}\overline{r}d\overline{r}\\ & =& \frac{{\mu }_0\omega \rho }{3}\sin \theta \left[\frac{1}{r^2}\left(\frac{r^5}{5}\right)+\frac{r}{2}\left(R^2-r^2\right)\right]\hat{\varphi }\\ & =& \frac{{\mu }_0\omega \rho }{2}r\sin \theta \left(\frac{R^2}{3}-\frac{r^2}{5}\right)\hat{\varphi }\\ & & \end{array}$

The equation of the magnetic field is expressed as:

$B=\nabla \times A$

Here, $B$is the magnetic field and $\nabla$is the curl.

$\frac{{\mu }_0\omega \rho }{2}r\sin \theta \left(\frac{R^2}{3}-\frac{r^2}{5}\right)\hat{\varphi }$for $A$in the above equation.

Substitute

$B=\frac{{\mu }_0\omega \rho }{2}\left\{\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left[\sin \theta r\sin \theta \left(\frac{R^2}{3}-\frac{r^2}{5}\right)\right]\hat{r}-\frac{1}{r}\frac{\partial }{\partial r}\left[r^2\sin \theta \left(\frac{R^2}{3}-\frac{r^2}{5}\right)\right]\hat{\theta }\right\}$

$={\mu }_0\omega \rho \left[\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\cos \theta \hat{r}-\left(\frac{R^2}{3}-\frac{2r^2}{5}\right)\sin \theta \hat{\theta }\right]$

Thus, the magnetic field inside a solid sphere is ${\mu }_0\omega \rho \left[\left(\frac{R^2}{3}-\frac{r^2}{5}\right)\cos \theta \hat{r}-\left(\frac{R^2}{3}-\frac{2r^2}{5}\right)\sin \theta \hat{\theta }\right]$.