Question

Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f\,Hz$ . The magnitude of magnetic induction at the center of the ring is

• A. $\frac{\mu_0 qf}{2\pi R}$
• B. $\frac{\mu_0 qf}{2 R}$
• C. $\frac{\mu_0 q}{2 \, fR}$
• D. $\frac{\mu_0 q}{2\pi fR}$

Answer 1

The Correct Option is B

To find the magnitude of the magnetic induction at the center of a ring due to its rotation, we consider the following physical concepts:

1. When a charge $q$ is uniformly distributed on a thin ring of radius $R$ and the ring rotates about its axis with a uniform frequency $f$ Hz, a current is produced, as the movement of charge constitutes an electric current.
2. The current $I$ is given by the formula: $$ I = q \times \text{Frequency} = qf $$.
3. Using Ampere’s Law, the magnetic induction $B$ at the center of a current-carrying loop is given by: $$ B = \frac{\mu_0 I}{2R} $$, where $(\mu_0)$ is the permeability of free space.
4. Substitute the value of $I$ from step 2 into the expression for $B$: $$ B = \frac{\mu_0 qf}{2R} $$.
5. Hence, the magnitude of the magnetic induction at the center of the ring is: $$ \frac{\mu_0 qf}{2R} $$.

This corresponds to the correct answer, $\frac{\mu_0 qf}{2R}$, confirming the outcome predicted in the initial information.