If the back e.m.f. induced in a coil, when current changes from 1A to zero in one millisecond, is 5 volts, the self-inductance of the coil is

Options

A. `5 H `
B. `1 H `
C. `5 xx 10 ^(-3) H`
D. ` 5 xx 10 ^(3) H `

Explanation

To solve the problem, we will use the formula for self-inductance and the relationship between induced electromotive force (e.m.f.), change in current, and time.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial current (I_initial) = 1 A
- Final current (I_final) = 0 A
- Change in current (ΔI) = I_final - I_initial = 0 A - 1 A = -1 A
- Time interval (Δt) = 1 millisecond = 1 × 10^-3 seconds
- Induced e.m.f. (ε) = 5 V

2. **Use the formula for induced e.m.f.:**
The induced e.m.f. (ε) in a coil is given by the formula:
\[ ε = -L \frac{dI}{dt} \]
where:
- L = self-inductance of the coil
- \(\frac{dI}{dt}\) = rate of change of current

3. **Calculate the rate of change of current (dI/dt):**
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{-1 \, \text{A}}{1 \times 10^{-3} \, \text{s}} = -1000 \, \text{A/s} \]

4. **Substitute the values into the e.m.f. formula:**
Rearranging the formula for self-inductance (L):
\[ L = -\frac{ε}{\frac{dI}{dt}} \]
Substituting the known values:
\[ L = -\frac{5 \, \text{V}}{-1000 \, \text{A/s}} = \frac{5}{1000} \, \text{H} = 0.005 \, \text{H} \]

5. **Convert the result to standard units:**
\[ L = 5 \times 10^{-3} \, \text{H} \]

### Final Answer:
The self-inductance of the coil is \(5 \times 10^{-3} \, \text{H}\) or 5 mH.

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