A 50 mH coil carries a current of 2 ampere. The energy stored in joules is

Options

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

Explanation

To find the energy stored in a coil (inductor), we can use the formula:

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

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 given values:**
- Inductance \( L = 50 \) mH = \( 50 \times 10^{-3} \) H
- Current \( I = 2 \) A

2. **Substitute the values into the formula:**
\[ U = \frac{1}{2} \times (50 \times 10^{-3}) \times (2^2) \]

3. **Calculate \( I^2 \):**
\[ I^2 = 2^2 = 4 \]

4. **Now substitute \( I^2 \) back into the energy formula:**
\[ U = \frac{1}{2} \times (50 \times 10^{-3}) \times 4 \]

5. **Perform the multiplication:**
\[ U = \frac{1}{2} \times 200 \times 10^{-3} \]

6. **Calculate the final value:**
\[ U = 100 \times 10^{-3} = 0.1 \text{ joules} \]

### Final Answer:
The energy stored in the coil is \( 0.1 \) joules.

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