A coil of self inductance 80 mH carries a current of 2A. What is the energy stored in the coil?

Options

A. 0.1 J
B. 1.6 J
C. 0.16 J
D. 0.4 J

Explanation

To find the energy stored in a coil with a given self-inductance and current, we can use the formula for energy stored in an inductor:

### Step-by-Step Solution:

1. **Identify the given values:**
- Self-inductance (L) = 80 mH = 80 × 10^(-3) H
- Current (I) = 2 A

2. **Use the formula for energy stored in an inductor:**
\[ E = \frac{1}{2} L I^2 \]
where \(E\) is the energy stored in joules, \(L\) is the self-inductance in henries, and \(I\) is the current in amperes.

3. **Substitute the values into the formula:**
\[ E = \frac{1}{2} \times (80 \times 10^{-3}) \times (2^2) \]

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

5. **Now substitute \(I^2\) back into the equation:**
\[ E = \frac{1}{2} \times (80 \times 10^{-3}) \times 4 \]

6. **Calculate \(E\):**
- First, calculate \( \frac{1}{2} \times 80 = 40 \)
- Then, multiply by 4:
\[ E = 40 \times 4 \times 10^{-3} = 160 \times 10^{-3} \text{ joules} \]

7. **Convert to standard form:**
\[ E = 0.16 \text{ joules} \]

### Final Answer:
The energy stored in the coil is **0.16 joules**.

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