Question

A charge $\mathbf{q}$ is distributed uniformly around a thin ring of radius$\mathbf{r}$ . The ring is rotating about an axis through its center and perpendicular to its plane, at an angular speed$\mathbf{\omega }$. (a) Show that the magnetic moment due to the rotating charge has magnitude $\mathbf{\mu }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{q\omega r}}^{\mathbf{2}}$. (b) What is the direction of this magnetic moment if the charge is positive?

Short answer

(a) The magnitude of the magnetic moment due to the rotating chargeis$\mathrm{\mu }=\frac{1}{2}{\mathrm{q\omega r}}^2$

(b) 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.

Step-by-step solution

 Step 1: Identification of the given data

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,$\mathrm{\omega }$

Representation of various formulae

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,

$\mathbf{\mu }\mathbf{=}\mathbf{iA}$

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,

$\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{t}}$

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

Step 3:(a) Determination of the magnitude of the magnetic moment due to the rotating charge

Substitute$\frac{\mathrm{q}}{\mathrm{t}}$for $i$, and${\mathrm{\pi r}}^2$ for$\mathrm{A}$ in$\mathrm{\mu }=\mathrm{iA}$ .

$\mathrm{\mu }=\frac{\mathrm{q}}{\mathrm{t}}\times {\mathrm{\pi r}}^2$

Rearrange$\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{t}}$ .

$\mathrm{t}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}$

Substitute the above value in $\mathrm{\mu }=\frac{\mathrm{q}}{\mathrm{t}}\times {\mathrm{\pi r}}^2$.

$\begin{array}{c}\mathrm{\mu }=\frac{\mathrm{q\omega }\times {\mathrm{\pi r}}^2}{\left(2\mathrm{\pi }\right)}\\ =\frac{1}{2}{\mathrm{q\omega r}}^2\end{array}$

Thus, the magnitude of the magnetic moment due to the rotating charge is$\mathrm{\mu }=\frac{1}{2}{\mathrm{q\omega r}}^2$

Step 4:(b) Determination of direction of magnetic moment if charge is positive

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.