Chapter 30, Problem 74CP

(a)

To determine

The magnetic field at the center of the sphere.

Answer to Problem 74CP

The magnetic field at the center of the sphere is $\frac{{\mu }_0\rho \omega R^2}{3}$.

Explanation of Solution

Consider the sphere as being built up by little spinning ring element of radius $r$, thickness $dr$ and height $dx$, centered on the rotation axis.

Write the equation for the charge on each ring.

    $\begin{array}{c}dQ=\rho dV\\ dQ=\rho \left(2\pi rdr\right)\left(dx\right)\end{array}$                                                                                                 (I)

Here, $Q$ is the charge and $\rho$ is the density.

Write the equation for the current through each ring.

    $dI=\frac{dQ}{T}$                                                                                                                  (II)

Here, $I$ is the current and $\omega$ is the angular speed.

Substitute $\rho \left(2\pi rdr\right)\left(dx\right)$ for $dQ$ and $\frac{2\pi }{\omega }$ for $T$ in equation (II) to find $dI$.

    $\begin{array}{c}dI=\frac{\left[\rho \left(2\pi rdr\right)\left(dx\right)\right]}{\left(\frac{2\pi }{\omega }\right)}\\ =\rho \omega rdrdx\end{array}$                                                                                                    (III)

Write the expression for the magnetic field at a distance $x$ from the center of the sphere on the rotation axis.

    $B={\displaystyle \underset{x=-R}{\overset{R}{\int }}{\displaystyle \underset{r=0}{\overset{\sqrt{R^2-r^2}}{\int }}\frac{{\mu }_0\omega \rho }{2}\frac{r^{3}drdx}{{\left(x^2+r^2\right)}^{3/2}}}}$                                                                             (IV)

Substitute $V$ for $x^2+r^2$ in equation (IV) to find $B$.

    $\begin{array}{c}B={\displaystyle \underset{x=-R}{\overset{R}{\int }}{\displaystyle \underset{x^2}{\overset{R^2}{\int }}\frac{{\mu }_0\omega \rho }{2}\frac{\left(V-x^2\right)dVdx}{2V^{\frac{3}{2}}}}}\\ B={\displaystyle \underset{x=-R}{\overset{R}{\int }}{\displaystyle \underset{x^2}{\overset{R^2}{\int }}\frac{{\mu }_0\omega \rho }{4}\left(V^{-\frac{1}{2}}-\frac{x^2}{V^{\frac{3}{2}}}\right)dVdx}}\end{array}$                                                                          (V)

Solve equation (V) further.

    $\begin{array}{c}B=\frac{{\mu }_0\rho \omega }{4}{\displaystyle \underset{-R}{\overset{R}{\int }}\left[{\displaystyle \underset{x^2}{\overset{R^2}{\int }}{V}^{-1/2}dV-x^2{\displaystyle \underset{x^2}{\overset{R^2}{\int }}{V}^{-3/2}dV}}\right]}dx\\ =\frac{{\mu }_0\rho \omega }{4}{{\displaystyle \underset{-R}{\overset{R}{\int }}\left[2V^{1/2}+\left(2x^2\right)V^{-1/2}\right]}}_{x^2}^{R^2}dx\\ =\frac{2{\mu }_0\rho \omega }{4}{\displaystyle \underset{0}{\overset{R}{\int }}\left[2\left(\frac{x^2}{R}\right)-4\left|x\right|+2R\right]}dx\end{array}$                                                           (VI)

Solve equation (VI) further to find $B$.

    $\begin{array}{c}B=\frac{2{\mu }_0\rho \omega }{4}{\left[\frac{2x^3}{3R}-\frac{4x^2}{2}+2xR\right]}_0^R\\ =\frac{2{\mu }_0\rho \omega }{4}\left[\frac{2R^3}{3R}-\frac{4R^2}{2}+2R^2\right]\\ =\frac{{\mu }_0\rho \omega R^2}{3}\end{array}$

Conclusion:

Therefore, the magnetic field at the center of the sphere is $\frac{{\mu }_0\rho \omega R^2}{3}$.

(b)

To determine

The magnetic moment of the sphere.

Answer to Problem 74CP

The magnetic moment of the sphere is $\frac{4\pi \rho \omega R^5}{15}$.

Explanation of Solution

Write the expression for the magnetic moment by the ring.

  $d\mu =AdI$                                                                                                                (V)

Here, $\mu$ is the magnetic moment.

Substitute $\pi r^2$ for $A$ and $\rho \omega rdrdx$ for $dI$ in equation (V) to find $d\mu$.

    $\begin{array}{c}d\mu =\left(\pi r^2\right)\left(\rho \omega rdrdx\right)\\ =\pi \rho \omega r^{3}drdx\end{array}$                                                                                          (VI)

Integrate equation (VI) to find $\mu$.

    $\begin{array}{c}\mu =\pi \omega \rho {\displaystyle \underset{x=-R}{\overset{R}{\int }}\left({\displaystyle \underset{r=0}{\overset{\sqrt{R^2-x^2}}{\int }}{r}^{3}dr}\right)}dx\\ =\pi \omega \rho {\displaystyle \underset{x=-R}{\overset{R}{\int }}\left(\frac{{\left(\sqrt{R^2-x^2}\right)}^4}{4}\right)}dx\\ =\pi \omega \rho {\displaystyle \underset{-R}{\overset{R}{\int }}\frac{{\left(R^2-x^2\right)}^2}{4}}dx\\ =\pi \omega \rho {\displaystyle \underset{-R}{\overset{R}{\int }}\left(R^4-2R^2{x}^2+x^4\right)}dx\end{array}$                                                                           (VII)

Solve equation (VII) further.

    $\begin{array}{c}\mu =\frac{\pi \omega \rho }{4}\left(R^4\left(2R\right)-2R^2\left(\frac{2R^3}{3}\right)+\frac{2R^5}{5}\right)\\ =\frac{\pi \omega \rho }{4}\left(2-\frac{4}{3}+\frac{2}{5}\right)\\ =\frac{\pi \omega \rho R^5}{4}\left(\frac{16}{15}\right)\\ =\frac{4\pi \rho \omega R^5}{15}\end{array}$

Conclusion:

Therefore, the magnetic moment of the sphere is $\frac{4\pi \rho \omega R^5}{15}$.