A `100mH` coil carries a current of `1` ampere. Energy stored in its magnetic field is

Options

A. `0.5J`
B. `1J`
C. `0.05J`
D. `0.1J`

Explanation

To find the energy stored in the magnetic field of a coil, we can use the formula for the energy stored in an inductor:

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

Where:
- \( U \) is the energy stored in joules,
- \( L \) is the inductance in henries,
- \( I \) is the current in amperes.

### Step-by-Step Solution:

1. **Identify the values given:**
- Inductance \( L = 100 \, \text{mH} = 100 \times 10^{-3} \, \text{H} = 0.1 \, \text{H} \)
- Current \( I = 1 \, \text{A} \)

2. **Substitute the values into the energy formula:**
\[ U = \frac{1}{2} L I^2 \]
\[ U = \frac{1}{2} \times 0.1 \, \text{H} \times (1 \, \text{A})^2 \]

3. **Calculate \( I^2 \):**
\[ I^2 = 1^2 = 1 \, \text{A}^2 \]

4. **Calculate the energy:**
\[ U = \frac{1}{2} \times 0.1 \times 1 \]
\[ U = \frac{0.1}{2} = 0.05 \, \text{J} \]

5. **Final Result:**
The energy stored in the magnetic field of the coil is \( 0.05 \, \text{J} \) or \( 50 \, \text{mJ} \).