91影视

There isa uniformly charged spherical shell, of radius Rand total charge Q,spinning at

constant angular velocity $蝇$.

The magnetic field at a distance $z$ on the axis of a circular coil of radius $a$ and carrying current$I$ is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}}\frac{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$ 鈥︹�� (1)

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

Consider a ring on the surface of the sphere at an angle $胃$ from the center.

From equation (1), the magnetic field at the center from that ring is

$\begin{array}{c}dB=\frac{{渭}_{0}dI}{2}\frac{{\left(R\sin 胃\right)}^2}{[{\left(R\sin 胃\right)}^2+{\left(R\cos 胃\right)}^2]}\\ =\frac{{渭}_0}{2R}{\sin }^2胃dI\text{鈥夆赌夆赌夆赌夆赌�}\mathrm{.....}\left(2\right)\end{array}$

Solve as:

$\begin{array}{c}dI=\frac{Q}{4蟺R^2}蝇R\sin 胃Rd胃\\ =\frac{Q蝇}{4蟺}\sin 胃d胃\end{array}$

To get the field from the full spherical surface, substitute this in equation (2) and integrate from $0$ to $蟺$,

$\begin{array}{c}B=\frac{{渭}_0}{2R}\underset{0}{\overset{蟺}{鈭�}}{\sin }^2胃\frac{Q蝇}{4蟺}\sin 胃d胃\\ =\frac{{渭}_{0}Q蝇}{8蟺R}\underset{0}{\overset{蟺}{鈭�}}{\sin }^3胃d胃\\ =\frac{{渭}_{0}Q蝇}{8蟺R}脳\frac{4}{3}\\ =\frac{{渭}_{0}Q蝇}{6蟺R}\end{array}$

Thus, the field is $\frac{{渭}_{0}Q蝇}{6蟺R}$.