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:

Options

A. `(mu_0 qf)/(2R)`
B. `(mu_0 q)/(2fR)`
C. `(mu_0 q)/(2pi fR)`
D. `(mu_0 qf)/(2pi R)`

Explanation

To find the magnitude of magnetic induction (magnetic field) at the center of a rotating ring with charge \( q \) uniformly spread over it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have a thin ring of radius \( R \) with a total charge \( Q \) uniformly distributed over it.
- The ring rotates about its axis with a uniform frequency \( f \) (in Hz).

2. **Relate Frequency to Revolutions**:
- The frequency \( f \) indicates that the ring completes \( f \) revolutions in one second.

3. **Determine the Current**:
- Current \( I \) is defined as the charge flowing per unit time.
- In one complete revolution, the entire charge \( Q \) passes a point on the ring.
- Since the ring makes \( f \) revolutions in one second, the total charge that flows in one second is:
\[ I = Q \times f \]

4. **Use the Formula for Magnetic Induction**:
- The formula for the magnetic induction \( B \) at the center of a ring carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2R} \]
- Here, \( \mu_0 \) is the permeability of free space.

5. **Substitute the Current into the Formula**:
- Now substitute \( I = Q \times f \) into the magnetic induction formula:
\[ B = \frac{\mu_0 (Q \times f)}{2R} \]

6. **Final Expression**:
- Thus, the magnitude of the magnetic induction at the center of the ring is:
\[ B = \frac{\mu_0 Q f}{2R} \]

### Conclusion:
The magnitude of magnetic induction at the center of the ring is given by:
\[ B = \frac{\mu_0 Q f}{2R} \]