A varying current ina coils changes from 10 A to zero in `0.5 s` .If the average emf induced in the coils is 220 V , the self- inductance of the coils is

Options

A. `5 H`
B. `6 H`
C. `11 H`
D. `12 H`

Explanation

To solve the problem of finding the self-inductance of the coil given the change in current and the average induced emf, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial current, \( I_i = 10 \, \text{A} \)
- Final current, \( I_f = 0 \, \text{A} \)
- Time taken for the change, \( t = 0.5 \, \text{s} \)
- Average induced emf, \( E = 220 \, \text{V} \)

2. **Calculate the Change in Current (\( \Delta I \)):**
\[ \Delta I = I_f - I_i = 0 - 10 = -10 \, \text{A} \]

3. **Calculate the Rate of Change of Current (\( \frac{di}{dt} \)):**
\[ \frac{di}{dt} = \frac{\Delta I}{t} = \frac{-10 \, \text{A}}{0.5 \, \text{s}} = -20 \, \text{A/s} \]

4. **Use Faraday's Law of Induction:**
The induced emf \( E \) is related to the self-inductance \( L \) and the rate of change of current by the formula:
\[ E = -L \frac{di}{dt} \]
Since we are given the average induced emf as positive, we can ignore the negative sign for our calculation of \( L \).

5. **Rearranging the Formula to Solve for Self-Inductance \( L \):**
\[ L = \frac{E}{-\frac{di}{dt}} = \frac{E}{20} \]

6. **Substituting the Values:**
\[ L = \frac{220 \, \text{V}}{20 \, \text{A/s}} = 11 \, \text{H} \]

7. **Conclusion:**
The self-inductance of the coil is \( L = 11 \, \text{H} \).

### Final Answer:
The self-inductance of the coil is \( 11 \, \text{H} \).

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