The magnetic potential energy stored in a certain inductor is `25mJ`, when the current in the inductor is `60 mA`. This inductor is of inductance

Options

A. 1.389 H
B. 138.88 H
C. 0.138 H
D. 13.89 H

Explanation

To find the inductance of the inductor given the magnetic potential energy and the current, we can use the formula for the energy stored in an inductor:

\[ U = \frac{1}{2} L I^2 \]

Where:
- \( U \) is the magnetic potential energy,
- \( L \) is the inductance,
- \( I \) is the current.

### Step-by-Step Solution:

1. **Identify the given values**:
- Magnetic potential energy, \( U = 25 \, \text{mJ} = 25 \times 10^{-3} \, \text{J} \)
- Current, \( I = 60 \, \text{mA} = 60 \times 10^{-3} \, \text{A} \)

2. **Substitute the values into the energy formula**:
\[ 25 \times 10^{-3} = \frac{1}{2} L (60 \times 10^{-3})^2 \]

3. **Calculate \( (60 \times 10^{-3})^2 \)**:
\[ (60 \times 10^{-3})^2 = 3600 \times 10^{-6} = 3.6 \times 10^{-3} \]

4. **Rearranging the equation to solve for \( L \)**:
\[ 25 \times 10^{-3} = \frac{1}{2} L (3.6 \times 10^{-3}) \]
\[ 25 \times 10^{-3} = 1.8 \times 10^{-3} L \]

5. **Isolate \( L \)**:
\[ L = \frac{25 \times 10^{-3}}{1.8 \times 10^{-3}} \]

6. **Calculate \( L \)**:
\[ L = \frac{25}{1.8} \approx 13.89 \, \text{H} \]

### Final Answer:
The inductance \( L \) of the inductor is approximately \( 13.89 \, \text{H} \).

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