The average e.m.f. induced in a coil in which the current changes from 2 ampere to 4 ampere in 0.05 second is 8 volt . What is the self inductance of the coil ?

Options

A. (a)`0.1` H
B. (b)`0.2`H
C. (c)`0.4`H
D. (d)`0.8`H

Answer

To find the self-inductance of the coil, we can use the formula for the average induced electromotive force (e.m.f.) in a coil, which is given by:

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

Where:
- \(E\) is the average 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 1: Identify the values from the problem
From the problem, we have:
- The average e.m.f. \(E = 8 \, \text{volts}\)
- The initial current \(I_1 = 2 \, \text{amperes}\)
- The final current \(I_2 = 4 \, \text{amperes}\)
- The time interval \(t = 0.05 \, \text{seconds}\)

### Step 2: Calculate the change in current (\(di\))
The change in current (\(di\)) can be calculated as follows:

\[ di = I_2 - I_1 = 4 \, \text{A} - 2 \, \text{A} = 2 \, \text{A} \]

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

\[ \frac{di}{dt} = \frac{di}{t} = \frac{2 \, \text{A}}{0.05 \, \text{s}} = 40 \, \text{A/s} \]

### Step 4: Substitute the values into the e.m.f. formula
Now we can substitute the values of \(E\) and \(\frac{di}{dt}\) into the formula:

\[ 8 = L \cdot 40 \]

### Step 5: Solve for self-inductance (\(L\))
Now, we can solve for \(L\):

\[ L = \frac{8}{40} = 0.2 \, \text{H} \]

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

\[ L = 0.2 \, \text{H} \]