Question

The current in a coil changes from 2A to 5A in 0.3s. The magnitude of emf induced in the coil is 1.0V. The value of self-inductance of the coil is

• A. 100 mH
• B. 0.1 mH
• C. 10 mH
• D. 1.0 mH

Answer 1

The Correct Option is A

Given Information: 
Initial current, \( I_i = 2\,A \)
Final current, \( I_f = 5\,A \)
Change in time, \( \Delta t = 0.3\,s \)
Induced emf, \( E = 1.0\,V \)

Step-by-Step Explanation:

Step 1: Using formula for induced emf due to self-induction:

Self-induced emf (\( E \)) in a coil is given by the formula:

\[ E = L \frac{\Delta I}{\Delta t} \]

Where \( L \) is the self-inductance, \( \Delta I \) is the change in current, and \( \Delta t \) is the time interval.

Step 2: Solve for self-inductance \( L \):

\[ L = \frac{E \times \Delta t}{\Delta I} \]

Calculate change in current, \( \Delta I \):

\[ \Delta I = I_f - I_i = 5\,A - 2\,A = 3\,A \]

Substitute values:

\[ L = \frac{1.0\,V \times 0.3\,s}{3\,A} = \frac{0.3}{3} = 0.1\,H \]

Step 3: Convert inductance into millihenry (mH):

\[ 0.1\,H = 0.1 \times 1000\,mH = 100\,mH \]

Final Conclusion:
The self-inductance of the coil is 100 mH.

Answer 2

The induced emf in a coil is given by the formula: \[ \text{emf} = L \cdot \frac{\Delta I}{\Delta t} \] where:
\( L \) is the self-inductance of the coil,
\( \Delta I \) is the change in current, and
\( \Delta t \) is the time interval. We are given:

\( \Delta I = 5 \, \text{A} - 2 \, \text{A} = 3 \, \text{A} \),
\( \Delta t = 0.3 \, \text{s} \),
emf = 1.0 V.

Using the formula, we solve for \( L \): \[ 1.0 = L \cdot \frac{3}{0.3} \] \[ 1.0 = L \cdot 10 \] \[ L = \frac{1.0}{10} = 0.1 \, \text{H} = 100 \, \text{mH} \] Thus, the self-inductance of the coil is 100 mH.