If a rate of change of current of 4 `As^(-1)` induces an emf of 20 mV in a solenoid, the self inductance of the solenoid is

Options

A. 5 mH
B. 80 mH
C. 0.25 mH
D. zero

Explanation

To find the self-inductance of the solenoid, we can use the formula that relates the induced electromotive force (emf), the self-inductance (L), and the rate of change of current (di/dt):

\[ \text{emf} = -L \frac{di}{dt} \]

### Step-by-step solution:

1. **Identify the given values**:
- Rate of change of current, \(\frac{di}{dt} = 4 \, \text{A/s}\)
- Induced emf, \(\text{emf} = 20 \, \text{mV} = 20 \times 10^{-3} \, \text{V}\)

2. **Rearranging the formula**:
We need to find the self-inductance \(L\). Rearranging the formula gives us:
\[ L = -\frac{\text{emf}}{\frac{di}{dt}} \]
Since we are dealing with magnitudes, we can ignore the negative sign.

3. **Substituting the values**:
Plugging in the values we have:
\[ L = \frac{20 \times 10^{-3} \, \text{V}}{4 \, \text{A/s}} \]

4. **Calculating L**:
\[ L = \frac{20 \times 10^{-3}}{4} = 5 \times 10^{-3} \, \text{H} \]

5. **Converting to milliHenries**:
\[ L = 5 \, \text{mH} \]

### Final Answer:
The self-inductance of the solenoid is \(5 \, \text{mH}\).

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