Question

(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.

Short answer

(a) It is proved that the average magnetic field, over a sphere of radius $R$, due to steady currents inside the sphere, is $B_{\text{ave}}=\frac{{\mu }_0}{4\pi }\frac{2\stackrel{k}{m}}{R^3}$

(b) The average magnetic field due to steady currents outside the sphere is $\frac{{\mu }_0}{4\pi }\int J\times \frac{{\hat{r}}^{\text{'}}}{r^{\text{'}2}}d{\tau }^{\text{'}}$and is same as the field produced at the center.

Step-by-step solution

Given data

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

step 2

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

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

Determine the average magnetic field inside the sphere

(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$\frac{2{\mu }_0\stackrel{m}{m}}{4\pi R^3}$

Average magnetic field outside the sphere

(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..