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

Options

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

Explanation

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

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

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

### Step-by-Step Solution:

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

2. **Substitute the values into the formula:**
\[ U = \frac{1}{2} \times (100 \times 10^{-3}) \times (1)^2 \]

3. **Calculate \( I^2 \):**
\[ I^2 = 1^2 = 1 \]

4. **Substitute \( I^2 \) back into the equation:**
\[ U = \frac{1}{2} \times (100 \times 10^{-3}) \times 1 \]

5. **Calculate the product:**
\[ U = \frac{1}{2} \times 100 \times 10^{-3} = 50 \times 10^{-3} \]

6. **Convert to standard form:**
\[ U = 0.05 \, \text{J} \]

### Final Answer:
The energy stored in the magnetic field is \( 0.05 \, \text{J} \).