A coil of self inductance 4H carries a 10 A current. If direction of current is reversed in 2 sec, then induced emf in it

Options

A. `-80`
B. `+80`
C. `-40`
D. `+40V`

Explanation

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Self-inductance (L) = 4 H (Henries)
- Initial current (I_initial) = 10 A (Amperes)
- Final current (I_final) = -10 A (since the direction is reversed)
- Time interval (Δt) = 2 s (seconds)

2. **Calculate the Change in Current (ΔI):**
\[ \Delta I = I_{final} - I_{initial} = -10 A - 10 A = -20 A \]

3. **Use the Formula for Induced EMF:**
The formula for induced emf (ε) due to self-inductance is given by:
\[ \epsilon = -L \frac{dI}{dt} \]
Since we are interested in the magnitude of the induced emf, we can ignore the negative sign for now:
\[ |\epsilon| = L \left|\frac{\Delta I}{\Delta t}\right| \]

4. **Substitute the Values into the Formula:**
\[ |\epsilon| = 4 H \times \left|\frac{-20 A}{2 s}\right| = 4 H \times 10 A/s = 40 V \]

5. **Final Result:**
The induced emf in the coil when the current is reversed is:
\[ \epsilon = 40 V \]