Coefficient of mutual inductance for two coils is 1 H. Current in one of the coils is increased from 4 to 5 A in 1 ms. What average emf will be induced in the other coil?

Options

A. 500 V
B. 250 V
C. 750 V
D. 1000 V

Explanation

To solve the problem, we need to calculate the average electromotive force (emf) induced in the second coil when the current in the first coil changes. We can use the formula for induced emf in terms of mutual inductance.

### Step-by-Step Solution:

1. **Identify the given values:**
- Coefficient of mutual inductance (M) = 1 H
- Initial current in the first coil (I1_initial) = 4 A
- Final current in the first coil (I1_final) = 5 A
- Time interval (Δt) = 1 ms = 0.001 s

2. **Calculate the change in current (ΔI):**
\[ \Delta I = I1_{final} - I1_{initial} = 5 \, \text{A} - 4 \, \text{A} = 1 \, \text{A} \]

3. **Calculate the rate of change of current (di/dt):**
\[ \frac{di}{dt} = \frac{\Delta I}{\Delta t} = \frac{1 \, \text{A}}{0.001 \, \text{s}} = 1000 \, \text{A/s} \]

4. **Use the formula for induced emf (E):**
The formula for the average induced emf in the second coil is given by:
\[ E = -M \frac{di}{dt} \]
Since we are asked for the average emf, we can ignore the negative sign (which indicates the direction of the induced emf) for the magnitude:
\[ E = M \frac{di}{dt} = 1 \, \text{H} \times 1000 \, \text{A/s} = 1000 \, \text{V} \]

5. **Conclusion:**
The average emf induced in the other coil is **1000 V**.