The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is

Options

A. `4xx10^(-5)Wb`
B. `2xx10^(-3)Wb`
C. `3xx10^(-5)Wb`
D. `8xx10^(-3)Wb`

Explanation

To solve the problem, we need to find the magnetic flux through the cross-section of a coil when a current of 2 mA is flowing through it. We are given the self-inductance of the coil and the number of turns.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Number of turns (N) = 400 turns
- Self-inductance (L) = 10 mH = 10 × 10^(-3) H
- Current (I) = 2 mA = 2 × 10^(-3) A

2. **Use the Formula for Magnetic Flux**:
The magnetic flux (Φ) linked with the coil can be calculated using the formula:
\[ N \Phi = L \cdot I \]
where:
- \( N \Phi \) is the total magnetic flux through the coil,
- \( L \) is the self-inductance,
- \( I \) is the current.

3. **Substitute the Given Values into the Formula**:
Substitute the values of L and I into the equation:
\[ N \Phi = (10 \times 10^{-3} \, \text{H}) \cdot (2 \times 10^{-3} \, \text{A}) \]

4. **Calculate the Magnetic Flux**:
Now, perform the multiplication:
\[ N \Phi = 10 \times 10^{-3} \times 2 \times 10^{-3} = 20 \times 10^{-6} \, \text{Wb} \]
This can be simplified to:
\[ N \Phi = 2 \times 10^{-5} \, \text{Wb} \]

5. **Final Result**:
The magnetic flux through the cross-section of the coil corresponding to a current of 2 mA is:
\[ \Phi = 2 \times 10^{-5} \, \text{Wb} \]