A charge q coulomb moves in a circle at n revolution per second and the radius of the circle is r metre. Then magnetic feild at the centre of the circle is

Options

A. `(2piq)/(nr)xx10^-7NA^-1m^-1`
B. `(2piq)/rxx10^-7NA^-1m^-1`
C. `(2pinq)/rxx10^-7NA^-1m^-1`
D. `(2piq)/rxx10^-7NA^-1m^-1`

Explanation

To find the magnetic field at the center of a circular path where a charge \( q \) is moving, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Charge \( q \) (in coulombs)
- Frequency of revolution \( n \) (in revolutions per second)
- Radius of the circle \( r \) (in meters)

2. **Relate Current to Charge and Time**:
- The current \( I \) due to the moving charge can be expressed as:
\[ I = \frac{Q}{T} \]
- Where \( T \) is the time period for one revolution. Since the frequency \( n \) is given, we can find \( T \) as:
\[ T = \frac{1}{n} \]

3. **Substitute for Current**:
- Substitute \( T \) into the current equation:
\[ I = Qn \]
- Therefore, the current \( I \) is equal to \( q \cdot n \).

4. **Use the Formula for Magnetic Field**:
- The magnetic field \( B \) at the center of a circular path due to a current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2r} \]
- Where \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

5. **Substitute the Current into the Magnetic Field Formula**:
- Substitute \( I = qn \) into the magnetic field equation:
\[ B = \frac{\mu_0 (qn)}{2r} \]

6. **Final Expression for Magnetic Field**:
- Now substituting the value of \( \mu_0 \):
\[ B = \frac{4\pi \times 10^{-7} \cdot qn}{2r} \]
- Simplifying this gives:
\[ B = \frac{2\pi qn}{r} \times 10^{-7} \, \text{T} \]

### Conclusion:
The magnetic field at the center of the circle is:
\[ B = \frac{2\pi qn}{r} \times 10^{-7} \, \text{T} \]