An e.m.f. of `5 "volt"` is produced by a self-inductance, when the current changes at a steady rate from `3 A` to `2A` `1`millisecond. The value of self-inductance is

Options

A. zero
B. 5H
C. 5000H
D. 5 mH

Answer

To solve the problem, we need to find the value of self-inductance (L) given the change in current and the induced electromotive force (e.m.f.).

### Step-by-Step Solution:

1. **Identify the given values:**
- e.m.f. (ε) = 5 volts
- Initial current (I_initial) = 3 A
- Final current (I_final) = 2 A
- Time interval (Δt) = 1 millisecond = \(1 \times 10^{-3}\) seconds

2. **Calculate the change in current (ΔI):**
\[ ΔI = I_final - I_initial = 2 A - 3 A = -1 A \]

3. **Calculate the rate of change of current (di/dt):**
\[ \frac{di}{dt} = \frac{ΔI}{Δt} = \frac{-1 A}{1 \times 10^{-3} s} = -1000 A/s \]

4. **Use the formula for self-induced e.m.f.:**
The formula for self-induced e.m.f. is given by:
\[ ε = -L \frac{di}{dt} \]
Substituting the known values:
\[ 5 V = -L \times (-1000 A/s) \]

5. **Rearranging the equation to solve for L:**
\[ L = \frac{5 V}{1000 A/s} = \frac{5}{1000} H = 0.005 H \]

6. **Convert the value of L to milliHenries:**
\[ L = 0.005 H = 5 \text{ milliHenries (mH)} \]

### Final Answer:
The value of self-inductance (L) is **5 milliHenries (mH)**.

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