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 of finding the magnetic flux through the cross-section of a coil with given self-inductance and current, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Self-inductance (L) = 10 mH = \(10 \times 10^{-3}\) H
- Number of turns (N) = 400 turns
- Current (I) = 2 mA = \(2 \times 10^{-3}\) A

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

3. **Substitute the Values**:
Substitute the known values into the formula:
\[ \Phi = \frac{10 \times 10^{-3} \, \text{H} \cdot 2 \times 10^{-3} \, \text{A}}{400} \]

4. **Calculate the Numerator**:
Calculate the product of L and I:
\[ 10 \times 10^{-3} \cdot 2 \times 10^{-3} = 20 \times 10^{-6} \, \text{H·A} \]

5. **Calculate the Magnetic Flux**:
Now divide the result by the number of turns:
\[ \Phi = \frac{20 \times 10^{-6}}{400} \]

6. **Perform the Division**:
\[ \Phi = 0.05 \times 10^{-6} \, \text{Wb} = 5 \times 10^{-8} \, \text{Wb} \]

### Final Answer:
The magnetic flux through the cross-section of the coil corresponding to a current of 2 mA is:
\[ \Phi = 5 \times 10^{-8} \, \text{Wb} \]