A coil of self inductance 2H carries a 2A current If direction of current is reversed in 1 sec., then induced emf in it :-

Options

A. `-8V`
B. 8V
C. `-4V`
D. Zero

Answer

To solve the problem, we need to calculate the induced electromotive force (emf) in a coil when the current flowing through it is reversed. The self-inductance of the coil is given, and we can use the formula for induced emf in an inductor.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Self-inductance \( L = 2 \, \text{H} \)
- Initial current \( I_{\text{initial}} = 2 \, \text{A} \)
- Final current \( I_{\text{final}} = -2 \, \text{A} \) (since the direction is reversed)
- Time interval \( \Delta t = 1 \, \text{s} \)

2. **Calculate Change in Current (\( \Delta I \)):**
\[ \Delta I = I_{\text{final}} - I_{\text{initial}} = -2 \, \text{A} - 2 \, \text{A} = -4 \, \text{A} \]

3. **Calculate Change in Time (\( \Delta t \)):**
\[ \Delta t = 1 \, \text{s} - 0 \, \text{s} = 1 \, \text{s} \]

4. **Use the Formula for Induced emf:**
The formula for induced emf (\( \mathcal{E} \)) in an inductor is given by:
\[ \mathcal{E} = -L \frac{\Delta I}{\Delta t} \]

5. **Substitute the Values:**
\[ \mathcal{E} = -2 \, \text{H} \cdot \frac{-4 \, \text{A}}{1 \, \text{s}} = -2 \cdot (-4) = 8 \, \text{V} \]

6. **Final Answer:**
The induced emf in the coil is \( 8 \, \text{V} \).

### Summary:
The induced emf when the current is reversed in a coil of self-inductance 2 H carrying a 2 A current is \( 8 \, \text{V} \).