91Ó°ÊÓ

The uniformly distributed charge over the ring is,$\mathrm{q}$

The radius of the ring is, $\mathrm{r}$

The angular speed of the ring is,$\mathrm{Ó\neg }$

The magnetic moment of the rotating charge is equal to the product of the current passing through the ring and the area of the ring. It is expressed as follows,

$\mathbf{μ}\mathbf{=}\mathbf{iA}$

Here,$\mathbf{i}$ is thecurrent passing through the ring, and$\mathbf{A}$is the area of the ring.

The expression for thecurrent passing through the ring is expressed as follows,

$\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}$

Here,$\mathrm{q}$is the charge on the object, and$\mathrm{t}$is the time.

The expression for the angular speed of the ring is given as follows,

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{\mathrm{t}}$

Here, $\mathrm{t}$is the time taken.

Substitute$\frac{\mathrm{q}}{\mathrm{t}}$for $i$, and${\mathrm{Ï€°ù}}^2$ for$\mathrm{A}$ in$\mathrm{μ}=\mathrm{iA}$ .

$\mathrm{μ}=\frac{\mathrm{q}}{\mathrm{t}}×{\mathrm{Ï€°ù}}^2$

Rearrange$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{\mathrm{t}}$ .

$\mathrm{t}=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}$

Substitute the above value in $\mathrm{μ}=\frac{\mathrm{q}}{\mathrm{t}}×{\mathrm{Ï€°ù}}^2$.

$\begin{array}{c}\mathrm{μ}=\frac{\mathrm{qÓ\neg }×{\mathrm{Ï€°ù}}^2}{\left(2\mathrm{Ï€}\right)}\\ =\frac{1}{2}{\mathrm{qÓ\neg r}}^2\end{array}$

Thus, the magnitude of the magnetic moment due to the rotating charge is$\mathrm{μ}=\frac{1}{2}{\mathrm{qÓ\neg r}}^2$

According to the right-hand thumb rule, if the charge is positive, on curling the fingers of right hand in the direction of rotation, the thumb points in the direction ofthedipole moment.

Thus, the direction of the thumb of the right-hand points in the direction of dipole moment if the charge is positive and the fingers are rotating in the direction of rotation.