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

Options

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

Answer

To solve the problem of finding the magnetic field produced at the center of a circular orbit by an electron moving in that orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
An electron is moving in a circular path of radius \( r \) and completes \( n \) rotations per second. We need to find the magnetic field at the center of this circular path.

2. **Calculate the Current**:
The current \( I \) produced by the electron can be calculated using the formula for current:
\[ I = \frac{q}{T} \]
where \( q \) is the charge of the electron and \( T \) is the time period for one complete rotation. Since the electron makes \( n \) rotations per second, the time period \( T \) is given by:
\[ T = \frac{1}{n} \]
Thus, the current becomes:
\[ I = q \cdot n \]

3. **Substituting the Charge of the Electron**:
The charge of an electron \( q \) is approximately \( -1.6 \times 10^{-19} \) C. Therefore, we can express the current as:
\[ I = n \cdot (-1.6 \times 10^{-19}) \]

4. **Using Biot-Savart Law**:
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by the Biot-Savart law:
\[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

5. **Substituting the Current into the Magnetic Field Equation**:
Now substituting the expression for current \( I \) into the magnetic field equation:
\[ B = \frac{\mu_0 (n \cdot q)}{2r} \]

6. **Final Expression for Magnetic Field**:
Substituting \( q \) with the charge of the electron:
\[ B = \frac{\mu_0 (n \cdot (-1.6 \times 10^{-19}))}{2r} \]
This gives us the magnitude of the magnetic field produced at the center of the circular orbit.

### Final Answer:
The magnitude of the magnetic field at the center is:
\[ B = \frac{\mu_0 n (-1.6 \times 10^{-19})}{2r} \]