A thin ring of radius `R` metre has charge `q` coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of `f` revolution/s. The value of magnetic induction in `Wb m^(-2)` at the centre of the ring is

Options

A. a `(mu_(0)qf)/(2piR)`
B. b `(mu_(0)qf)/(2R)`
C. c `(mu_(0)q)/(2fR)`
D. d `(mu_(0)q)/(2pifR)`

Explanation

To find the magnetic induction (magnetic field) at the center of a rotating charged ring, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
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 frequency \( f \) revolutions per second. We need to find the magnetic induction \( B \) at the center of the ring.

2. **Magnetic Field Formula**:
The magnetic field \( B \) at the center of a current-carrying loop is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space and \( I \) is the current flowing through the loop.

3. **Relate Charge and Current**:
The current \( I \) can be expressed in terms of charge \( q \) and the time period \( T \) of one complete revolution:
\[ I = \frac{q}{T} \]
The time period \( T \) is related to the frequency \( f \) by:
\[ T = \frac{1}{f} \]
Therefore, we can substitute \( T \) into the current equation:
\[ I = qf \]

4. **Substituting Current into Magnetic Field Formula**:
Now, substitute \( I \) into the magnetic field formula:
\[ B = \frac{\mu_0 (qf)}{2R} \]

5. **Final Expression**:
Thus, the magnetic induction \( B \) at the center of the ring is:
\[ B = \frac{\mu_0 q f}{2R} \]

### Final Answer:
The value of magnetic induction at the center of the ring is:
\[ B = \frac{\mu_0 q f}{2R} \, \text{Wb/m}^2 \]