Question

(a) A phonograph record of radius $\mathit{R}$, carrying a uniform surface charge $\mathit{\sigma }$, is rotating at constant angular velocity $\mathit{\omega }$. Find its magnetic dipole moment.

(b) Find the magnetic dipole moment of the spinning spherical shell in Ex. 5.11. Show that for points$\mathit{r}\mathbf{>}\mathit{R}$ the potential is that of a perfect dipole.

Short answer

(a) The magnetic dipole moment is $\sigma \omega \mathrm{\pi }\frac{{\mathrm{R}}^4}{4}$.

(b) The magnetic dipole moment of the spinning spherical shell is $\frac{4\mathrm{\pi }}{3}\sigma \omega R^4\stackrel{⏜}{z}$.

For points $r>R$, the potential is that of a perfect dipole.

Step-by-step solution

Identification of the given data

The given data is listed below as:

• The radius of the phonograph record is $R$,
• The surface charge of the sphere is,$\sigma$
• The angular velocity of the phonograph is,$\omega$

Significance of the magnetic dipole moment

Themagnetic dipole moment is described as the product of the pole strength and the magnet’s magnetic length. However, the magnetic dipole moment also experiences a torque when placed inside a magnetic field.

(a) Determination of the magnetic dipole moment

The equation of the magnetic dipole moment for a ring is expressed as:

$m=I{\mathrm{\pi r}}^2$ …(i)

Here,$I$is the current and$r$is the radius of the ring.

The equation of the current carried by the ring is expressed as:

$I=\sigma vdr$

Here,$v$is the velocity,$\sigma$is the surface charge and$dr$is the small increase in the radius.

Substitute$\omega r$for$v$in the above equation.

$I=\sigma \omega rdr$

Substitute the above value in the equation (i).

$m={\int }_0^R\sigma \omega r{\mathrm{\pi r}}^2\mathrm{dr}$

Here, in the above equation, the limit is given as$R$ is the radius of the phonograph record.

The above equation can be calculated as:

$m=\sigma \omega \mathrm{\pi }\frac{{\mathrm{R}}^4}{4}$

Thus, the magnetic dipole moment is $\sigma \omega \mathrm{\pi }\frac{{\mathrm{R}}^4}{4}$.

(b) Determination of the magnetic dipole moment of the spinning spherical shell

The equation of the charge of the shaded ring is expressed as:

$dq=\sigma \left(2\mathrm{\pi R}\mathrm{sin\theta }\right)Rd\theta$ …(ii)

Here,$dq$is the total charge on the ring and$\theta$is the angle subtended by the ring.

The equation for the time for one revolution is expressed as:

$dt=\frac{2\mathrm{\pi }}{\omega }$ …(iii)

Here,$\omega$is the angular velocity of the phonograph.

The equation of the current in the ring is expressed as:

$I=\frac{dq}{dt}$

Here,$\frac{dq}{dt}$is the rate of change of charge with time.

Substitute the value of the equation (ii) and (iii) in the above equation.

$$\begin{gathered} \frac{dq}{dt}=\frac{\sigma \left(2\mathrm{\pi Rsin\theta }\right)Rd\theta }{2\mathrm{\pi }} \\ I=\frac{\sigma \left(2\mathrm{\pi Rsin\theta }\right)Rd\theta }{{\displaystyle \frac{2\mathrm{\pi }}{\mathrm{\omega }}}} \\ =\sigma \omega R^2\sin \theta d\theta \end{gathered}$$

The equation of the magnetic moment is expressed as:

$dm=I{\mathrm{\pi R}}^2{\sin }^2\mathrm{\theta }$

Substitute$\sigma \omega R^2\sin \theta d\theta$ for$I$ in the above equation.

$dm=\left(\sigma \omega R^2\sin \theta d\theta \right){\mathrm{\pi R}}^2{\sin }^2\mathrm{\theta }$

From the above equation, the equation of the magnetic dipole moment can be obtained.

The equation of the total magnetic dipole moment can be expressed as:

$m=\sigma \omega {\mathrm{\pi R}}^4{\int }_0^{\mathrm{\pi }}{\sin }^3\mathrm{\theta d\theta }$

The above equation can be solved as:

$m=\sigma \omega {\mathrm{\pi R}}^4\left(\frac{4}{3}\right)$

role="math" localid="1657621378712" $=\frac{4\mathrm{\pi }}{3}\sigma \omega R^4\stackrel{⏜}{z}$ …(iv)

Thus, the magnetic dipole moment of the spinning spherical shell is $\frac{4\mathrm{\pi }}{3}\sigma \omega R^4\stackrel{⏜}{z}$.

(b) Determination of the perfect dipole

For$r>R$, the equation (iv) can be expressed as:

$$\begin{gathered} m=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{4\mathrm{\pi }}{3}\sigma \omega R^4\frac{\sin \theta }{r^2}\stackrel{⏜}{\varphi } \\ =\frac{{\mu }_0\sigma \omega R^4}{3}\frac{\sin \theta }{r^2}\stackrel{⏜}{\varphi } \end{gathered}$$

Hence, the equation represents the potential for a perfect dipole.

Thus, for points $r>R$, the potential is that of a perfect dipole.