An e.m.f. of `12 "volts"` is induced in a given coil when the current in it changes at the rate of `48` amperes per minute. The self-inductance of the coil is

Options

A. (a)`0.5` henry
B. (b)15 henry
C. (c)`1.5` henry
D. (d)`9.6` henry

Answer

To find the self-inductance of the coil, we can use the formula that relates the induced electromotive force (e.m.f.) to the self-inductance and the rate of change of current. The formula is given by:

\[ E = L \frac{di}{dt} \]

Where:
- \(E\) is the induced e.m.f. (in volts),
- \(L\) is the self-inductance (in henries),
- \(\frac{di}{dt}\) is the rate of change of current (in amperes per second).

### Step-by-Step Solution:

1. **Identify the given values**:
- Induced e.m.f. \(E = 12 \, \text{volts}\)
- Rate of change of current \(\frac{di}{dt} = 48 \, \text{amperes per minute}\)

2. **Convert the rate of change of current to amperes per second**:
- Since there are 60 seconds in a minute, we convert:
\[ \frac{di}{dt} = \frac{48 \, \text{amperes}}{60 \, \text{seconds}} = 0.8 \, \text{amperes per second} \]

3. **Substitute the values into the formula**:
\[ 12 = L \cdot 0.8 \]

4. **Solve for self-inductance \(L\)**:
\[ L = \frac{12}{0.8} \]

5. **Calculate \(L\)**:
\[ L = 15 \, \text{henries} \]

### Final Answer:
The self-inductance of the coil is \(15 \, \text{henries}\).