An average induced e.m.f. of `1V` appears in a coil when the current in it is changed from `10` `A` in one direction to `10A` in opposite direction in `0.5` sec. self-inductance of the coil is

Options

A. (a)25 mH
B. (b)50 mH
C. (c)75 mH
D. (d)100 mH

Answer

To find the self-inductance of the coil, we can use the formula for induced electromotive force (e.m.f.) in terms of self-inductance and the rate of change of current:

\[ E = -L \frac{dI}{dt} \]

Where:
- \( E \) is the induced e.m.f. (1 V in this case)
- \( L \) is the self-inductance of the coil (which we need to find)
- \( \frac{dI}{dt} \) is the rate of change of current

### Step-by-Step Solution:

1. **Identify the change in current (\( dI \))**:
- The current changes from \( +10 \, A \) to \( -10 \, A \).
- Therefore, the total change in current is:
\[ dI = -10 \, A - (+10 \, A) = -20 \, A \]

2. **Determine the time interval (\( dt \))**:
- The time interval during which this change occurs is given as \( 0.5 \, s \).

3. **Calculate the rate of change of current (\( \frac{dI}{dt} \))**:
- The rate of change of current can be calculated as:
\[ \frac{dI}{dt} = \frac{dI}{dt} = \frac{-20 \, A}{0.5 \, s} = -40 \, A/s \]

4. **Substitute the values into the induced e.m.f. formula**:
- We know the induced e.m.f. \( E = 1 \, V \), so we can rearrange the formula to solve for \( L \):
\[ 1 = L \cdot (-40) \]
- Since we are interested in the magnitude of self-inductance, we can ignore the negative sign:
\[ L = \frac{1}{40} \, H \]

5. **Convert the self-inductance into millihenries**:
- To convert henries to millihenries, we multiply by 1000:
\[ L = \frac{1}{40} \times 1000 \, mH = 25 \, mH \]

### Final Answer:
The self-inductance of the coil is \( 25 \, mH \).

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