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 solve the problem of finding the magnetic field at the center of a circular path along which a charge \( q \) is moving, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine the Current \( I \)**:
- The charge \( q \) completes \( n \) revolutions in one second. Therefore, the total charge passing through a point in one second (which is the definition of current) can be calculated as:
\[ I = \frac{q}{T} \]
- Here, \( T \) (the time period for one revolution) is given by:
\[ T = \frac{1}{n} \text{ seconds} \]
- Substituting \( T \) into the current equation gives:
\[ I = q \cdot n \]

3. **Use the Formula for Magnetic Field at the Center of a Circular Loop**:
- The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2r} \]
- Here, \( \mu_0 \) is the permeability of free space.

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

5. **Final Expression for the Magnetic Field**:
- Therefore, the magnetic field at the center of the circle is:
\[ B = \frac{\mu_0 q n}{2r} \]

6. **Direction of the Magnetic Field**:
- According to the right-hand rule, the direction of the magnetic field at the center of the circle is out of the plane of the circle (towards the observer).

### Final Answer:
The magnetic field at the center of the circle is given by:
\[ B = \frac{\mu_0 q n}{2r} \]
with the direction being out of the screen.

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