Question

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

${\overrightarrow{B}}_{ave}=\frac{{\mu }_0}{4\pi }\frac{2\overrightarrow{m}}{R^3}$

where$\overrightarrow{m}$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:

${\overrightarrow{B}}_{ave}=\frac{1}{\frac{4}{3}\pi R^3}\int \overrightarrow{B}d\tau$

Write$\overrightarrow{B}$as$\overrightarrow{\nabla }\times \overrightarrow{A}$ ,and apply Prob. 1.61(b). Now put in Eq. 5.65, and do the

surface integral first, showing that

$\int \frac{1}{r}d\overrightarrow{a}=\frac{4}{3}\pi r\text{'}$

(b) Show that the average magnetic field due to steady currents outsidethe 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 steadycurrents inside the sphere, is ${\overrightarrow{B}}_{ave}=\frac{{\mu }_0}{4\pi }\frac{2\overrightarrow{m}}{R^3}$.

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

Step-by-step solution

 Step 1: Given data

There is a sphere of radius R of steady current density $\overrightarrow{J}$

Step 2:

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

$\overrightarrow{B}=\overrightarrow{\nabla }\times \overrightarrow{A}$…… (1)

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

$\overrightarrow{A}=\frac{{\mu }_0}{4\pi }\int \frac{\overrightarrow{J}}{r}d\tau \text{'}$ …… (2)

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

The volume integral of the curl of a vector function

$\int \overrightarrow{\nabla }\times \overrightarrow{A}d\tau =-\oint \overrightarrow{A}\times d\overrightarrow{a}$ …… (3)

The magnetic moment of a current distribution is

$\overrightarrow{m}=\frac{1}{2}\int \overrightarrow{r}\times \overrightarrow{J}d\tau$ …… (4)

Step 3:Determine the average magnetic field inside the sphere

(a)

The average magnetic field over a sphere of radius R is

${\overrightarrow{B}}_{\text{ave}}=\frac{1}{\frac{4}{3}\pi R^3}\int \overrightarrow{B}d\tau$

Apply equation (1)

${\overrightarrow{B}}_{\text{ave}}=\frac{1}{\frac{4}{3}\pi R^3}\int \overrightarrow{\nabla }\times \overrightarrow{A}d\tau$

Use equation (3) to get,

${\overrightarrow{B}}_{\text{ave}}=-\frac{1}{\frac{4}{3}\pi R^3}\oint \overrightarrow{A}\times d\overrightarrow{a}$

Use equation (2) to get,

$\begin{array}{rcl}{\overrightarrow{B}}_{\text{ave}}& =& -\frac{1}{\frac{4}{3}\pi R^3}\frac{{\mu }_0}{4\pi }\oint \int \frac{\overrightarrow{J}}{r}d\tau \text{'}\times d\overrightarrow{a}\\ & =& -\frac{3{\mu }_0}{16{\pi }^2{R}^3}\int \overrightarrow{J}\times \left\{\oint \frac{d\overrightarrow{a}}{r}\right\}d\tau \text{'}\\ & & \end{array}$ …… (5)

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

$$\begin{gathered} r=\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos \theta } \\ d\overrightarrow{a}=R^2\sin \theta d\theta d\varphi \hat{r} \end{gathered}$$

The X and Y are components of the surface, integration is thus zero. The Z component is

$\begin{array}{rcl}\oint \frac{d\overrightarrow{a}}{r}& =& \oint \frac{R^2\sin \theta d\theta d\varphi \hat{z}\cos \theta }{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos \theta }}\\ & =& 2\pi R^2\hat{z}\underset{0}{\overset{\pi }{\int }}\frac{\sin \theta \cos \theta d\theta }{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos \theta }}\\ & & \end{array}$

Convert

$\begin{array}{rcl}u& =& \cos \theta \\ du& =& -\sin \theta d\theta \\ & & \end{array}$

Solve further as,

$\begin{array}{rcl}\oint \frac{d\overrightarrow{a}}{r}& =& -2\pi R^2\hat{z}\underset{1}{\overset{-1}{\int }}\frac{udu}{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}u}}\\ & =& 2\pi R^2\hat{z}{\left[-\frac{2\left\{2\left(R^2+{z\text{'}}^2\right)+2Rz\text{'}u\right\}}{3{\left(2Rz\text{'}\right)}^2}\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}u}\right]}_{-1}^1\\ & =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left|R-z\text{'}\right|+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right].....\left(6\right)\\ & & \end{array}$

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

$\begin{array}{rcl}\oint \frac{d\overrightarrow{a}}{r}& =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left(R-z\text{'}\right)+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right]\\ & =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\left[-R^3-R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3+R{z\text{'}}^2+R^3+R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3-R{z\text{'}}^2\right]\\ & =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\times 2{z\text{'}}^3\\ & =& \frac{4\pi z\text{'}\hat{z}}{3}\\ & & \end{array}$

Thus, from equation (5),

${\overrightarrow{B}}_{\text{ave}}=-\frac{3{\mu }_0}{16{\pi }^2{R}^3}\frac{4\pi }{3}\int \overrightarrow{J}\times \overrightarrow{r}\text{'}d\tau \text{'}$

Use equation (4) to get,

${\overrightarrow{B}}_{\text{ave}}=\frac{2{\mu }_0\overrightarrow{m}}{4\pi R^3}$

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

Average magnetic field outside the sphere

(b)

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

$\begin{array}{rcl}\oint \frac{d\overrightarrow{a}}{r}& =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left(-R+z\text{'}\right)+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right]\\ & =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\left[R^3+R{z\text{'}}^2+R^{2}z\text{'}-R^{2}z\text{'}-{z\text{'}}^3-R{z\text{'}}^2+R^3+R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3-R{z\text{'}}^2\right]\\ & =& \frac{2\pi \hat{z}}{3{z\text{'}}^2}\times 2R^3\\ & =& \frac{4\pi R^3\hat{z}}{3{z\text{'}}^2}\\ & & \end{array}$

Thus, from equation (5),

$\begin{array}{rcl}\backslash {\overrightarrow{B}}_{\text{ave}}& =& -\frac{3{\mu }_0}{16{\pi }^2{R}^3}\frac{4\pi R^3}{3}\int \overrightarrow{J}\times \frac{\hat{r}\text{'}}{{r\text{'}}^2}d\tau \text{'}\\ & =& \frac{{\mu }_0}{4\pi }\int \overrightarrow{J}\times \frac{\hat{r}\text{'}}{{r\text{'}}^2}d\tau \text{'}\\ & & \end{array}$

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