91影视

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

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

$\stackrel{鈫�}{B}=\stackrel{鈫�}{鈭�}脳\stackrel{鈫�}{A}$鈥︹�� (1)

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

$\stackrel{鈫�}{A}=\frac{{渭}_0}{4蟺}鈭�\frac{\stackrel{鈫�}{J}}{r}d蟿\text{'}$ 鈥︹�� (2)

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

The volume integral of the curl of a vector function

$鈭�\stackrel{鈫�}{鈭�}脳\stackrel{鈫�}{A}d蟿=-鈭�\stackrel{鈫�}{A}脳d\stackrel{鈫�}{a}$ 鈥︹�� (3)

The magnetic moment of a current distribution is

$\stackrel{鈫�}{m}=\frac{1}{2}鈭�\stackrel{鈫�}{r}脳\stackrel{鈫�}{J}d蟿$ 鈥︹�� (4)

(a)

The average magnetic field over a sphere of radius R is

${\stackrel{鈫�}{B}}_{\text{ave}}=\frac{1}{\frac{4}{3}蟺R^3}鈭�\stackrel{鈫�}{B}d蟿$

Apply equation (1)

${\stackrel{鈫�}{B}}_{\text{ave}}=\frac{1}{\frac{4}{3}蟺R^3}鈭�\stackrel{鈫�}{鈭�}脳\stackrel{鈫�}{A}d蟿$

Use equation (3) to get,

${\stackrel{鈫�}{B}}_{\text{ave}}=-\frac{1}{\frac{4}{3}蟺R^3}鈭�\stackrel{鈫�}{A}脳d\stackrel{鈫�}{a}$

Use equation (2) to get,

$\begin{array}{rcl}{\stackrel{鈫�}{B}}_{\text{ave}}& =& -\frac{1}{\frac{4}{3}蟺R^3}\frac{{渭}_0}{4蟺}鈭�鈭�\frac{\stackrel{鈫�}{J}}{r}d蟿\text{'}脳d\stackrel{鈫�}{a}\\ & =& -\frac{3{渭}_0}{16{蟺}^2{R}^3}鈭�\stackrel{鈫�}{J}脳\left\{鈭�\frac{d\stackrel{鈫�}{a}}{r}\right\}d蟿\text{'}\\ & & \end{array}$ 鈥︹�� (5)

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

$$\begin{gathered} r=\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos 胃} \\ d\stackrel{鈫�}{a}=R^2\sin 胃d胃d蠒\hat{r} \end{gathered}$$

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

$\begin{array}{rcl}鈭�\frac{d\stackrel{鈫�}{a}}{r}& =& 鈭�\frac{R^2\sin 胃d胃d蠒\hat{z}\cos 胃}{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos 胃}}\\ & =& 2蟺R^2\hat{z}\underset{0}{\overset{蟺}{鈭�}}\frac{\sin 胃\cos 胃d胃}{\sqrt{R^2+{z\text{'}}^2-2Rz\text{'}\cos 胃}}\\ & & \end{array}$

Convert

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

Solve further as,

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

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

$\begin{array}{rcl}鈭�\frac{d\stackrel{鈫�}{a}}{r}& =& \frac{2蟺\hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left(R-z\text{'}\right)+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right]\\ & =& \frac{2蟺\hat{z}}{3{z\text{'}}^2}\left[-R^3-R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3+R{z\text{'}}^2+R^3+R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3-R{z\text{'}}^2\right]\\ & =& \frac{2蟺\hat{z}}{3{z\text{'}}^2}脳2{z\text{'}}^3\\ & =& \frac{4蟺z\text{'}\hat{z}}{3}\\ & & \end{array}$

Thus, from equation (5),

${\stackrel{鈫�}{B}}_{\text{ave}}=-\frac{3{渭}_0}{16{蟺}^2{R}^3}\frac{4蟺}{3}鈭�\stackrel{鈫�}{J}脳\stackrel{鈫�}{r}\text{'}d蟿\text{'}$

Use equation (4) to get,

${\stackrel{鈫�}{B}}_{\text{ave}}=\frac{2{渭}_0\stackrel{鈫�}{m}}{4蟺R^3}$

Thus, the average field inside the sphere is $\frac{2{渭}_0\stackrel{鈫�}{m}}{4蟺R^3}$

(b)

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

$\begin{array}{rcl}鈭�\frac{d\stackrel{鈫�}{a}}{r}& =& \frac{2蟺\hat{z}}{3{z\text{'}}^2}\left[-\left(R^2+{z\text{'}}^2+Rz\text{'}\right)\left(-R+z\text{'}\right)+\left(R^2+{z\text{'}}^2-Rz\text{'}\right)\left(R+z\text{'}\right)\right]\\ & =& \frac{2蟺\hat{z}}{3{z\text{'}}^2}\left[R^3+R{z\text{'}}^2+R^{2}z\text{'}-R^{2}z\text{'}-{z\text{'}}^3-R{z\text{'}}^2+R^3+R{z\text{'}}^2-R^{2}z\text{'}+R^{2}z\text{'}+{z\text{'}}^3-R{z\text{'}}^2\right]\\ & =& \frac{2蟺\hat{z}}{3{z\text{'}}^2}脳2R^3\\ & =& \frac{4蟺R^3\hat{z}}{3{z\text{'}}^2}\\ & & \end{array}$

Thus, from equation (5),

$\begin{array}{rcl}\backslash {\stackrel{鈫�}{B}}_{\text{ave}}& =& -\frac{3{渭}_0}{16{蟺}^2{R}^3}\frac{4蟺R^3}{3}鈭�\stackrel{鈫�}{J}脳\frac{\hat{r}\text{'}}{{r\text{'}}^2}d蟿\text{'}\\ & =& \frac{{渭}_0}{4蟺}鈭�\stackrel{鈫�}{J}脳\frac{\hat{r}\text{'}}{{r\text{'}}^2}d蟿\text{'}\\ & & \end{array}$

Thus, the average field outside the sphere is $\frac{{渭}_0}{4蟺}鈭�\stackrel{鈫�}{J}脳\frac{\hat{r}\text{'}}{{r\text{'}}^2}d蟿\text{'}$which is also the field at the center..