If a current of 4A produces a magnetic flux of `10^(-3)` Wb per turn in a coil of 1000 turns, then the self inductance of the coil is

Options

A. 0.1 H
B. 0.15 H
C. 0.25 H
D. 0.4 H

Explanation

To find the self-inductance \( L \) of the coil, we can use the relationship between the magnetic flux, the number of turns, current, and self-inductance. The formula we will use is:

\[ N \Phi = L I \]

Where:
- \( N \) = number of turns in the coil
- \( \Phi \) = magnetic flux per turn
- \( L \) = self-inductance of the coil
- \( I \) = current flowing through the coil

### Step-by-Step Solution:

1. **Identify the given values**:
- Current \( I = 4 \, \text{A} \)
- Magnetic flux per turn \( \Phi = 10^{-3} \, \text{Wb} \)
- Number of turns \( N = 1000 \)

2. **Calculate the total magnetic flux linked with the coil**:
The total magnetic flux \( \Phi_{\text{total}} \) linked with the coil can be calculated as:
\[ \Phi_{\text{total}} = N \Phi = 1000 \times 10^{-3} = 1 \, \text{Wb} \]

3. **Use the formula to find self-inductance \( L \)**:
Rearranging the formula \( N \Phi = L I \) to solve for \( L \):
\[ L = \frac{N \Phi}{I} \]
Substituting the known values:
\[ L = \frac{1 \, \text{Wb}}{4 \, \text{A}} = 0.25 \, \text{H} \]

4. **Final answer**:
The self-inductance of the coil is:
\[ L = 0.25 \, \text{H} \]