(a) Prove that the average magnetic field, over a sphere of radius R, due to steady currents inside the sphere, is

${\mathrm{B}}_{\text{ave}}=\frac{{\mu }_0}{4}\frac{2\dot{m}}{4}$

where $\stackrel{b}{m}$is the total dipole moment of the sphere. Contrast the electrostatic result, Eq. 3.105. [This is tough, so I'll give you a start:

$B_{\text{ave}}=\frac{1}{4}{}_3\pi R^{3}fBd$

Write $B^U$as $\nabla \times A$, and apply Prob. 1.61(b). Now put in Eq. $5.65$, and do the surface integral first, showing that

$\int \frac{1}{r}d\frac{4}{3}\text{,}$

(b) Show that the average magnetic field due to steady currents outside the sphere is the same as the field they produce at the center.

There is a sphere of radius $R$ of steady current density $J$.

The magnetic field as a function of the magnetic vector potential is

$\mathbf{B}=\nabla \times \hat{A}$

The magnetic vector potential corresponding to a current density $\overrightarrow{j}$is

Here, ${\mu }_0$is the permeability of free space.

The volume integral of the curl of a vector function

The magnetic moment of a current distribution is

$m=\frac{1}{2}\int x\times Jd$

(a)

The average magnetic field over a sphere of radius is

Apply equation (1)

$B_{\text{ave}}=\frac{1}{4}\int R^3\int \nabla \times Ad\tau$

Use equation (3) to get,

Use equation (2) to get,

$B_{\mathrm{ave}}=-\frac{1}{\frac{4}{3}\pi R^3}\frac{{\mu }_0}{4\pi }\oint \int \frac{\overrightarrow{J}}{r}d{\tau }^{\text{'}}\times da$

The point ${\overrightarrow{r}}^{\text{'}\text{'}}$is chosen to be on the $z$axis. Therefore,

$\backslash begin\left\{aligned\right\}r\&=\backslash sqrt\{R^\{2\}+z^\{\backslash prime2\}-2Rz^\{\backslash prime\}\backslash \cos \backslash theta\}\backslash \backslash d^\{\backslash lambda\}\&=R^\left\{2\right\}\backslash \sin \backslash thetad\backslash thetad\backslash phi\backslash hat\left\{r\right\}\backslash end\left\{aligned\right\}$

The $x$and $y$are components of the surface, integration is thus zero. The z component is

$\oint \frac{da^}{r}=\oint \frac{R^2\sin \theta d\theta d\varphi \hat{z}\cos \theta }{\sqrt{R^2+z^{\text{'}2}-2R{z}^{\text{'}}\cos \theta }}$

$=2\pi R^2\hat{z}{\int }_0^{\pi }\frac{\sin \theta \cos \theta d\theta }{\sqrt{R^2+z^{\text{'}2}-2R{z}^{\text{'}}\cos \theta }}$

Convert

$u=\cos \theta$

$du=-\sin \theta d\theta$

Solve further as,

$\oint \frac{da^}{r}=-2\pi R^2\hat{z}{\int }_1^{-1}\frac{udu}{\sqrt{R^2+z^{\text{'}2}-2R{z}^{\text{'}}u}}$

$=2\pi R^2\hat{z}{\left[-\frac{2\left\{2\left(R^2+z^{\text{'}2}\right)+2R{z}^{\text{'}}u\right\}}{3{\left(2R{z}^{\text{'}}\right)}^2}\sqrt{R^2+z^{\text{'}2}-2R{z}^{\text{'}}u}\right]}_{-1}^1$

$=\frac{2\pi \hat{z}}{3z^{\text{'}2}}\left[-\left(R^2+z^{\text{'}2}+R{z}^{\text{'}}\right)\left|R-z^{\text{'}}\right|+\left(R^2+z^{\text{'}2}-R{z}^{\text{'}}\right)\left(R+z^{\text{'}}\right)\right]$

Inside the sphere, $R>z^{\text{'}}$. Therefore,

$\backslash begin\left\{aligned\right\}\backslash prod\backslash frac\{dâ\}\left\{r\right\}\&=\backslash frac\{2\backslash pi\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash left[-\backslash left(R^\left\{2\right\}+z^\{\backslash prime2\}+Rz^\{\backslash prime\}\backslash right)\backslash left(R-z^\{\backslash prime\}\backslash right)+\backslash left(R^\left\{2\right\}+z^\{\backslash prime2\}-Rz^\{\backslash prime\}\backslash right)\backslash left(R+z^\{\backslash prime\}\backslash right)\backslash right]\backslash \backslash \&=\backslash frac\{2\backslash pi\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash left[-R^\{3\}-Rz^\{\backslash prime2\}-R^\{2\}z^\{\backslash prime\}+R^\{2\}z^\{\backslash prime\}+z^\{\backslash prime3\}+Rz^\{\backslash prime2\}+R^\{3\}+Rz^\{\backslash prime2\}-R^\{2\}z^\{\backslash prime\}+R^\{2\}z^\{\backslash prime\}+z^\{\backslash prime3\}-Rz^\{\backslash prime2\}\backslash right]\backslash \backslash \&=\backslash frac\{2\backslash pi\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash times2z^\{\backslash prime3\}\backslash \backslash \&=\backslash frac\{4\backslash piz^\{\backslash prime\}\backslash hat\{z\left\}\right\}\left\{3\right\}\backslash end\left\{aligned\right\}$

Thus, from equation (5),

$\backslash begin\left\{aligned\right\}B\_\{\backslash text\{ave\left\}\right\}^\left\{R\right\}\&=-\backslash frac\{3\backslash mu\_\{0\left\}\right\}\{16\backslash pi^\{2\}R^\{3\left\}\right\}\backslash frac\{4\backslash piR^\{3\left\}\right\}\left\{3\right\}\backslash intJ\backslash times\backslash frac\{\backslash hat\{r\}^\{\backslash prime\left\}\right\}\{r^\{\backslash prime2\left\}\right\}d\backslash tau^\{\backslash prime\}\backslash \backslash \&=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash int\backslash stackrel\left\{D\right\}\left\{J\right\}\backslash times\backslash frac\{\backslash hat\{r\}^\{\backslash prime\left\}\right\}\{r^\{\backslash prime2\left\}\right\}d\backslash tau^\{\backslash prime\}\backslash end\left\{aligned\right\}$

Use equation (4) to get,

$B_{\mathrm{ave}}=\frac{2{\mu }_0\tilde{m}}{4\pi R^3}$

Thus, the average field inside the sphere is$\frac{2{\mu }_0\stackrel{m}{m}}{4\pi R^3}$

(b)

Outside the sphere, $R<z^{\text{'}}$. Therefore from equation (6),

$\backslash begin\left\{aligned\right\}\backslash dot\left\{y\right\}\backslash frac\{da\}\left\{r\right\}\&=\backslash frac\{2\backslash pi\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash left[-\backslash left(R^\left\{2\right\}+z^\{\backslash prime2\}+Rz^\{\backslash prime\}\backslash right)\backslash left(-R+z^\{\backslash prime\}\backslash right)+\backslash left(R^\left\{2\right\}+z^\{\backslash prime2\}-Rz^\{\backslash prime\}\backslash right)\backslash left(R+z^\{\backslash prime\}\backslash right)\backslash right]\backslash \backslash \&=\backslash frac\{2\backslash pi\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash left[R^\{3\}+Rz^\{\backslash prime2\}+R^\{2\}z^\{\backslash prime\}-R^\{2\}z^\{\backslash prime\}-z^\{\backslash prime3\}-Rz^\{\backslash prime2\}+R^\{3\}+Rz^\{\backslash prime2\}-R^\{2\}z^\{\backslash prime\}+R^\{2\}z^\{\backslash prime\}+z^\{\backslash prime3\}-Rz^\{\backslash prime2\}\backslash right]\backslash \backslash \&=\backslash frac\{2\backslash pi\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash times2R^\left\{3\right\}\backslash \backslash \&=\backslash frac\{4\backslash piR^\{3\}\backslash hat\{z\left\}\right\}\{3z^\{\backslash prime2\left\}\right\}\backslash end\left\{aligned\right\}$

Thus, from equation (5),

$\backslash begin\left\{aligned\right\}B\_\{\backslash text\{ave\left\}\right\}^\left\{R\right\}\&=-\backslash frac\{3\backslash mu\_\{0\left\}\right\}\{16\backslash pi^\{2\}R^\{3\left\}\right\}\backslash frac\{4\backslash piR^\{3\left\}\right\}\left\{3\right\}\backslash intJ\backslash times\backslash frac\{\backslash hat\{r\}^\{\backslash prime\left\}\right\}\{r^\{\backslash prime2\left\}\right\}d\backslash tau^\{\backslash prime\}\backslash \backslash \&=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash int\backslash stackrel\left\{D\right\}\left\{J\right\}\backslash times\backslash frac\{\backslash hat\{r\}^\{\backslash prime\left\}\right\}\{r^\{\backslash prime2\left\}\right\}d\backslash tau^\{\backslash prime\}\backslash end\left\{aligned\right\}$

Thus, the average field outside the sphere is $\frac{{\mu }_0}{4\pi }\int \sqrt{J}\times \frac{{\hat{r}}^{\text{'}}}{r^{\text{'}2}}d{\tau }^{\text{'}}$which is also the field at the center..