In a circular conducting coil, when current increases from `2A` to `18A` in `0.05` sec., the induced e.m.f. is `20V`. The self-inductance of the coil is

Options

A. 62.5 mH
B. 6.25 mH
C. 50 mH
D. Zero

Answer

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

\[ \text{Induced e.m.f.} (E) = -L \frac{di}{dt} \]

Where:
- \( E \) is the induced e.m.f. (in volts),
- \( L \) is the self-inductance (in henries),
- \( di \) is the change in current (in amperes),
- \( dt \) is the change in time (in seconds).

### Step 1: Calculate the change in current (\( di \))
The current increases from \( 2A \) to \( 18A \).

\[ di = 18A - 2A = 16A \]

### Step 2: Calculate the change in time (\( dt \))
The change in time is given as \( 0.05 \) seconds.

\[ dt = 0.05 \, \text{s} \]

### Step 3: Calculate the rate of change of current (\( \frac{di}{dt} \))
Now, we can calculate \( \frac{di}{dt} \):

\[ \frac{di}{dt} = \frac{di}{dt} = \frac{16A}{0.05 \, \text{s}} = 320 \, A/s \]

### Step 4: Use the induced e.m.f. to find self-inductance (\( L \))
We know the induced e.m.f. \( E = 20V \). Plugging in the values into the formula:

\[ 20V = L \cdot 320 \, A/s \]

### Step 5: Solve for \( L \)
Rearranging the equation to find \( L \):

\[ L = \frac{20V}{320 \, A/s} = \frac{20}{320} = \frac{1}{16} \, H \]

### Step 6: Convert to milliHenries
To convert henries to milliHenries:

\[ L = \frac{1}{16} \, H = 0.0625 \, H = 62.5 \, mH \]

### Final Answer
The self-inductance of the coil is:

\[ L = 62.5 \, mH \]

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