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 solve the problem, we need to find the magnetic flux (Φ) through the cross-sectional area of a coil given its self-inductance (L), the number of turns (N), and the current (I) flowing through it.

### Given Data:
- Number of turns (N) = 500
- Self-inductance (L) = 50 mH = 50 × 10^(-3) H
- Current (I) = 8 mA = 8 × 10^(-3) A

### Step-by-Step Solution:

1. **Write the formula for magnetic flux (Φ)**:
The relationship between self-inductance (L), magnetic flux (Φ), number of turns (N), and current (I) is given by:
\[ N \cdot \Phi = L \cdot I \]
Rearranging this formula to find Φ gives:
\[ \Phi = \frac{L \cdot I}{N} \]

2. **Substitute the known values into the formula**:
Now we can substitute the values of L, I, and N into the formula:
\[ \Phi = \frac{(50 \times 10^{-3} \, \text{H}) \cdot (8 \times 10^{-3} \, \text{A})}{500} \]

3. **Calculate the numerator**:
First, calculate the product of L and I:
\[ L \cdot I = 50 \times 10^{-3} \cdot 8 \times 10^{-3} = 400 \times 10^{-6} \, \text{H·A} = 0.4 \times 10^{-3} \, \text{H·A} \]

4. **Divide by the number of turns (N)**:
Now divide this result by the number of turns:
\[ \Phi = \frac{0.4 \times 10^{-3}}{500} = 0.8 \times 10^{-6} \, \text{Wb} \]

5. **Convert to micro Weber**:
Since \(1 \, \text{Wb} = 10^6 \, \mu\text{Wb}\), we can express the flux in micro Weber:
\[ \Phi = 0.8 \, \mu\text{Wb} \]

### Final Answer:
The magnetic flux through the cross-sectional area of the coil while the current through it is 8 mA is:
\[ \Phi = 0.8 \, \mu\text{Wb} \]