Current in a coil increases from 0 to 1 ampere in 0.1 second. If self inductance of the coil is 5 millihenry, magnitude of induced emf is

Options

A. 5V
B. `5 xx 10^(-3) V`
C. 0.5V
D. `5 xx 10^(-2) V`

Answer

To solve the problem, we need to find the magnitude of the induced electromotive force (emf) in a coil when the current changes. We can use the formula for induced emf in a coil, which is given by:

\[ e = -L \frac{di}{dt} \]

where:
- \( e \) is the induced emf,
- \( L \) is the self-inductance of the coil,
- \( \frac{di}{dt} \) is the rate of change of current.

### Step 1: Identify the given values
- The self-inductance \( L \) of the coil is given as 5 millihenries (mH). We need to convert this to henries (H):
\[ L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \]
- The current changes from 0 to 1 ampere in a time interval of 0.1 seconds. Therefore, we can calculate the rate of change of current \( \frac{di}{dt} \):
\[ \frac{di}{dt} = \frac{1 - 0}{0.1} = \frac{1}{0.1} = 10 \, \text{A/s} \]

### Step 2: Calculate the induced emf
Now we can substitute the values of \( L \) and \( \frac{di}{dt} \) into the formula for induced emf:
\[ e = L \frac{di}{dt} = (5 \times 10^{-3} \, \text{H}) \times (10 \, \text{A/s}) \]
Calculating this gives:
\[ e = 5 \times 10^{-3} \times 10 = 5 \times 10^{-2} \, \text{V} \]

### Step 3: Conclusion
The magnitude of the induced emf is:
\[ e = 5 \times 10^{-2} \, \text{V} = 0.05 \, \text{V} \]

Thus, the correct answer is \( 5 \times 10^{-2} \, \text{V} \).