91Ó°ÊÓ

The given data is listed below as:

• The radius of the phonograph record is $R$,
• The surface charge of the sphere is,$\mathrm{σ}$
• The angular velocity of the phonograph is,$\mathrm{Ó\neg }$

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.

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

$m=I{\mathrm{Ï€°ù}}^2$ …(¾±)

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:

$I=\mathrm{σ}vdr$

Here,$v$is the velocity,$\mathrm{σ}$is the surface charge and$dr$is the small increase in the radius.

Substitute$\mathrm{Ó\neg }r$for$v$in the above equation.

$I=\mathrm{σ}\mathrm{Ó\neg }rdr$

Substitute the above value in the equation (i).

$m={∫}_0^R\mathrm{σ}\mathrm{Ó\neg }r{\mathrm{Ï€°ù}}^2\mathrm{dr}$

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:

$m=\mathrm{σ}\mathrm{Ó\neg }\mathrm{Ï€}\frac{{\mathrm{R}}^4}{4}$

Thus, the magnetic dipole moment is $\mathrm{σ}\mathrm{Ó\neg }\mathrm{Ï€}\frac{{\mathrm{R}}^4}{4}$.

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

$dq=\mathrm{σ}\left(2\mathrm{Ï€¸é}\mathrm{²õ¾\pm ²Ôθ}\right)Rd\mathrm{θ}$ …(¾±¾±)

Here,$dq$is the total charge on the ring and$\mathrm{θ}$is the angle subtended by the ring.

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

$dt=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}$ …(¾±¾±¾±)

Here,$\mathrm{Ó\neg }$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.

$$\begin{gathered} \frac{dq}{dt}=\frac{\mathrm{σ}\left(2\mathrm{Ï€¸é²õ¾\pm ²Ôθ}\right)Rd\mathrm{θ}}{2\mathrm{Ï€}} \\ I=\frac{\mathrm{σ}\left(2\mathrm{Ï€¸é²õ¾\pm ²Ôθ}\right)Rd\mathrm{θ}}{{\displaystyle \frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}}} \\ =\mathrm{σ}\mathrm{Ó\neg }{R}^2\sin \mathrm{θ}d\mathrm{θ} \end{gathered}$$

The equation of the magnetic moment is expressed as:

$dm=I{\mathrm{Ï€¸é}}^2{\sin }^2\mathrm{θ}$

Substitute$\mathrm{σ}\mathrm{Ó\neg }{R}^2\sin \mathrm{θ}d\mathrm{θ}$ for$I$ in the above equation.

$dm=\left(\mathrm{σ}\mathrm{Ó\neg }{R}^2\sin \mathrm{θ}d\mathrm{θ}\right){\mathrm{Ï€¸é}}^2{\sin }^2\mathrm{θ}$

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:

$m=\mathrm{σ}\mathrm{Ó\neg }{\mathrm{Ï€¸é}}^4{∫}_0^{\mathrm{Ï€}}{\sin }^3\mathrm{θ»åθ}$

The above equation can be solved as:

$m=\mathrm{σ}\mathrm{Ó\neg }{\mathrm{Ï€¸é}}^4\left(\frac{4}{3}\right)$

role="math" localid="1657621378712" $=\frac{4\mathrm{Ï€}}{3}\mathrm{σ}\mathrm{Ó\neg }{R}^4\stackrel{Áåœ}{z}$ …(¾±±¹)

Thus, the magnetic dipole moment of the spinning spherical shell is $\frac{4\mathrm{Ï€}}{3}\mathrm{σ}\mathrm{Ó\neg }{R}^4\stackrel{Áåœ}{z}$.

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

$$\begin{gathered} m=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{4\mathrm{Ï€}}{3}\mathrm{σ}\mathrm{Ó\neg }{R}^4\frac{\sin \mathrm{θ}}{r^2}\stackrel{Áåœ}{\mathrm{Ï•}} \\ =\frac{{\mathrm{μ}}_0\mathrm{σ}\mathrm{Ó\neg }{R}^4}{3}\frac{\sin \mathrm{θ}}{r^2}\stackrel{Áåœ}{\mathrm{Ï•}} \end{gathered}$$

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

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