The self-inductane of a coil having 500 turns is 50 mH. The magnetic flux through the cross-sectional area of the coil while current through it is 8 mA is found to be

Options

A. `4xx10^(-4)Wb`
B. `0.04Wb`
C. `0.8muWb`
D. 40 m Wb

Explanation

To find the magnetic flux through the cross-sectional area of the coil when a current of 8 mA is flowing through it, we can use the formula relating self-inductance (L), current (I), and magnetic flux (Φ):

\[ \Phi = L \cdot I \]

Where:
- \( \Phi \) is the magnetic flux,
- \( L \) is the self-inductance of the coil,
- \( I \) is the current flowing through the coil.

### Step-by-Step Solution:

1. **Identify the given values:**
- Self-inductance \( L = 50 \, \text{mH} = 50 \times 10^{-3} \, \text{H} \)
- Current \( I = 8 \, \text{mA} = 8 \times 10^{-3} \, \text{A} \)

2. **Substitute the values into the formula:**
\[ \Phi = L \cdot I = (50 \times 10^{-3} \, \text{H}) \cdot (8 \times 10^{-3} \, \text{A}) \]

3. **Calculate the magnetic flux:**
\[ \Phi = 50 \times 8 \times 10^{-3} \times 10^{-3} = 400 \times 10^{-6} \, \text{Wb} = 0.4 \, \text{mWb} \]

4. **Final answer:**
The magnetic flux through the cross-sectional area of the coil when the current is 8 mA is \( 0.4 \, \text{mWb} \).