If the current rate of change is 7A, as induces an emf of 56mV in a solenoid. Then, calculate the self-inductance of the solenoid ?

Options

A. 8 mH
B. 27 mH
C. 18.7 mH
D. 6.7 mH

Explanation

To solve the problem of calculating the self-inductance of a solenoid given the rate of change of current and the induced emf, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Rate of change of current (di/dt) = 7 A/s (Ampere per second)
- Induced emf (ε) = 56 mV (millivolts)

2. **Convert Units**:
- Convert the induced emf from millivolts to volts for consistency in SI units:
\[ 56 \text{ mV} = 56 \times 10^{-3} \text{ V} = 0.056 \text{ V} \]

3. **Use the Formula for Self-Inductance**:
- The self-inductance (L) of a solenoid can be calculated using the formula:
\[ L = \frac{\epsilon}{\frac{di}{dt}} \]
- Here, ε is the induced emf and di/dt is the rate of change of current.

4. **Substitute the Values into the Formula**:
- Substitute ε and di/dt into the formula:
\[ L = \frac{0.056 \text{ V}}{7 \text{ A/s}} \]

5. **Calculate the Self-Inductance**:
- Perform the division:
\[ L = \frac{0.056}{7} = 0.008 \text{ H} \]

6. **Convert to Millihenries**:
- Since 1 H = 1000 mH, convert the result:
\[ L = 0.008 \text{ H} = 8 \text{ mH} \]

### Final Answer:
The self-inductance of the solenoid is **8 mH** (millihenries).

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