An electron moving in a circular orbit of radius makes n rotations per second. The magnetic field produced at the centre has magnitude :

Options

A. Zero
B. `(mu_(0)n^(2)e)/(r )`
C. `(mu_(0)n e)/(2r)`
D. `(mu_(0)n e)/(2pir)`

Answer

To solve the problem of finding the magnetic field produced at the center of a circular orbit by an electron making n rotations per second, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
We have an electron moving in a circular orbit of radius \( R \) and making \( n \) rotations per second.

2. **Identify the Current**:
The motion of the electron can be equated to a current in a circular loop. The current \( I \) can be defined as the charge passing through a point in the circuit per unit time. For an electron, the charge \( Q \) is equal to the elementary charge \( e \).

3. **Calculate the Time Period**:
The time period \( T \) for one complete rotation is the reciprocal of the frequency:
\[ T = \frac{1}{n} \]

4. **Calculate the Current**:
The current \( I \) can be expressed as:
\[ I = \frac{Q}{T} = \frac{e}{T} = \frac{e}{\frac{1}{n}} = e \cdot n \]

5. **Apply the Biot-Savart Law**:
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

6. **Substitute the Current**:
Now substitute the expression for \( I \) into the formula for \( B \):
\[ B = \frac{\mu_0 (e \cdot n)}{2R} \]

7. **Final Expression**:
Thus, the magnetic field produced at the center of the circular orbit is:
\[ B = \frac{\mu_0 e n}{2R} \]

### Final Answer:
The magnetic field produced at the center of the circular orbit is:
\[ B = \frac{\mu_0 e n}{2R} \]

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