A coil of self inductance 3 H carries a steady current of 2 A. What is the energy stored in the magnetic field of the coil?

Answer

To find the energy stored in the magnetic field of the 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 the magnetic field (in Joules),
- \( L \) is the self-inductance of the coil (in Henries),
- \( I \) is the current flowing through the coil (in Amperes).

### Step-by-step Solution:

1. **Identify the values given in the problem:**
- Self-inductance \( L = 3 \, \text{H} \)
- Current \( I = 2 \, \text{A} \)

2. **Substitute the values into the energy formula:**
\[ U = \frac{1}{2} L I^2 \]
\[ U = \frac{1}{2} \times 3 \, \text{H} \times (2 \, \text{A})^2 \]

3. **Calculate \( I^2 \):**
\[ (2 \, \text{A})^2 = 4 \, \text{A}^2 \]

4. **Substitute \( I^2 \) back into the equation:**
\[ U = \frac{1}{2} \times 3 \, \text{H} \times 4 \, \text{A}^2 \]

5. **Perform the multiplication:**
\[ U = \frac{1}{2} \times 12 \, \text{H} \cdot \text{A}^2 \]

6. **Calculate the final energy:**
\[ U = 6 \, \text{J} \]

### Final Answer:
The energy stored in the magnetic field of the coil is \( 6 \, \text{J} \).

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