(a) A phonograph record carries a uniform density of "static electricity" $\mathit{\sigma }$.If it rotates at angular velocity $\mathbf{\omega }$,what is the surface current density Kat a distance r from the center?

(b) A uniformly charged solid sphere, of radius Rand total charge Q,is centered

at the origin and spinning at a constant angular velocity $\mathbf{\omega }$about the zaxis. Find

the current density J at any point $\left(r,\theta ,\varphi \right)$within the sphere.

A phonograph record carries a uniform density of "static electricity" $\sigma$and rotates at angular velocity $\omega$.

A uniformly charged solid sphere, of radius Rand total charge Q,is centered

at the origin and spinning at a constant angular velocity $\omega$about the zaxis.

The surface current density of a surface charge density $\sigma$moving with a speed v is

$\mathbf{K}\mathbf{=}\mathbf{\sigma v}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

The volume current density of a volume charge density $\rho$moving with a speed v is

$\mathit{J}\mathbf{=}\mathit{\rho }\mathbf{v}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The speed of the charge density rotating with angular velocity$\omega$in a circle of radius r is

$v=\omega r$

Substitute this in equation (1) to get

$K=\sigma \omega r$

Thus, the surface current density is $\sigma \omega r$.

The volume charge density of a sphere of radius R uniformly charged with a charge Q is

$$\begin{gathered} \rho =\frac{Q}{{\displaystyle \frac{4}{3}\pi R^3}} \\ =\frac{3Q}{4\pi R^3} \end{gathered}$$

The speed at any point in the sphere rotating with angular velocity $\omega$about the z at a radial distance r and making angle $\theta$with the z axis is

$v=\omega r\sin \theta$

Substitute these in equation (2) to get

$$\begin{gathered} J=\frac{3Q}{4\pi R^2}\times \omega r\sin \theta \\ =\frac{3Q\omega r\sin \theta }{4\pi R^3} \end{gathered}$$

Thus, the volume current density islocalid="1657774147962" $\frac{3Q\omega r\sin \theta }{4\pi R^3}$.