91Ó°ÊÓ

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

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

$\stackrel{→}{B}=\stackrel{→}{∇}×\stackrel{→}{A}$…… (1)

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

$\stackrel{→}{A}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{\stackrel{→}{J}}{r}d\mathrm{τ}\text{'}$ …… (2)

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

The volume integral of the curl of a vector function

$∫\stackrel{→}{∇}×\stackrel{→}{A}d\mathrm{Ï„}=-âˆ\circledR \stackrel{→}{A}×d\stackrel{→}{a}$ …… (3)

The magnetic moment of a current distribution is

$\stackrel{→}{m}=\frac{1}{2}∫\stackrel{→}{r}×\stackrel{→}{J}d\mathrm{τ}$ …… (4)

(a)

The average magnetic field over a sphere of radius R is

${\stackrel{→}{B}}_{\text{ave}}=\frac{1}{\frac{4}{3}\mathrm{π}{R}^3}∫\stackrel{→}{B}d\mathrm{τ}$

Apply equation (1)

${\stackrel{→}{B}}_{\text{ave}}=\frac{1}{\frac{4}{3}\mathrm{π}{R}^3}∫\stackrel{→}{∇}×\stackrel{→}{A}d\mathrm{τ}$

Use equation (3) to get,

${\stackrel{→}{B}}_{\text{ave}}=-\frac{1}{\frac{4}{3}\mathrm{Ï€}{R}^3}âˆ\circledR \stackrel{→}{A}×d\stackrel{→}{a}$

Use equation (2) to get,

$\begin{array}{rcl}{\stackrel{→}{B}}_{\text{ave}}& =& -\frac{1}{\frac{4}{3}\mathrm{Ï€}{R}^3}\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}âˆ\circledR ∫\frac{\stackrel{→}{J}}{r}d\mathrm{Ï„}\text{'}×d\stackrel{→}{a}\\ & =& -\frac{3{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2{R}^3}∫\stackrel{→}{J}×\left\{âˆ\circledR \frac{d\stackrel{→}{a}}{r}\right\}d\mathrm{Ï„}\text{'}\\ & & \end{array}$ …… (5)

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

$$\begin{gathered} r=\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos \mathrm{θ}} \\ d\stackrel{→}{a}=R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}\hat{r} \end{gathered}$$

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

$\begin{array}{rcl}âˆ\circledR \frac{d\stackrel{→}{a}}{r}& =& âˆ\circledR \frac{R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\hat{z}\cos \mathrm{θ}}{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos \mathrm{θ}}}\\ & =& 2\mathrm{Ï€}{R}^2\hat{z}\underset{0}{\overset{\mathrm{Ï€}}{∫}}\frac{\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}}{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos \mathrm{θ}}}\\ & & \end{array}$

Convert

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

Solve further as,

$\begin{array}{rcl}âˆ\circledR \frac{d\stackrel{→}{a}}{r}& =& -2\mathrm{Ï€}{R}^2\hat{z}\underset{1}{\overset{-1}{∫}}\frac{udu}{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}u}}\\ & =& 2\mathrm{Ï€}{R}^2\hat{z}{\left[-\frac{2\left\{2\left(R^2+{z\text{'}}^2\right)+2Rz\text{'}u\right\}}{3{\left(2Rz\text{'}\right)}^2}\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}u}\right]}_{-1}^1\\ & =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left|R-z\text{'}\right|+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right].....\left(6\right)\\ & & \end{array}$

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

$\begin{array}{rcl}âˆ\circledR \frac{d\stackrel{→}{a}}{r}& =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left(R-z\text{'}\right)+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right]\\ & =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}\left[-R^3-R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3+R{z\text{'}}^2+R^3+R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3-R{z\text{'}}^2\right]\\ & =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}×2{z\text{'}}^3\\ & =& \frac{4\mathrm{Ï€}z\text{'}\hat{z}}{3}\\ & & \end{array}$

Thus, from equation (5),

${\stackrel{→}{B}}_{\text{ave}}=-\frac{3{\mathrm{μ}}_0}{16{\mathrm{π}}^2{R}^3}\frac{4\mathrm{π}}{3}∫\stackrel{→}{J}×\stackrel{→}{r}\text{'}d\mathrm{τ}\text{'}$

Use equation (4) to get,

${\stackrel{→}{B}}_{\text{ave}}=\frac{2{\mathrm{μ}}_0\stackrel{→}{m}}{4\mathrm{π}{R}^3}$

Thus, the average field inside the sphere is $\frac{2{\mathrm{μ}}_0\stackrel{→}{m}}{4\mathrm{π}{R}^3}$

(b)

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

$\begin{array}{rcl}âˆ\circledR \frac{d\stackrel{→}{a}}{r}& =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left(-R+z\text{'}\right)+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right]\\ & =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}\left[R^3+R{z\text{'}}^2+R^{2}z\text{'}-R^{2}z\text{'}-{z\text{'}}^3-R{z\text{'}}^2+R^3+R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3-R{z\text{'}}^2\right]\\ & =& \frac{2\mathrm{Ï€}\hat{z}}{3{z\text{'}}^2}×2R^3\\ & =& \frac{4\mathrm{Ï€}{R}^3\hat{z}}{3{z\text{'}}^2}\\ & & \end{array}$

Thus, from equation (5),

$\begin{array}{rcl}\backslash {\stackrel{→}{B}}_{\text{ave}}& =& -\frac{3{\mathrm{μ}}_0}{16{\mathrm{π}}^2{R}^3}\frac{4\mathrm{π}{R}^3}{3}∫\stackrel{→}{J}×\frac{\hat{r}\text{'}}{{r\text{'}}^2}d\mathrm{τ}\text{'}\\ & =& \frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\stackrel{→}{J}×\frac{\hat{r}\text{'}}{{r\text{'}}^2}d\mathrm{τ}\text{'}\\ & & \end{array}$

Thus, the average field outside the sphere is $\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\stackrel{→}{J}×\frac{\hat{r}\text{'}}{{r\text{'}}^2}d\mathrm{τ}\text{'}$which is also the field at the center..