Question

A coil has a self-inductance of \(2H\); find the induced EMF if the current changes from \(5A\) to \(2A\) in \(0.1\) seconds.

• A. \(30V\)
• B. \(40V\)
• C. \(50V\)
• D. \(60V\)

Hint:

Induced EMF in inductors follows \[ e = L\frac{di}{dt} \] A faster change in current produces a larger induced EMF.

Answer 1

The Correct Option is D

Topic: Electromagnetic Induction
Step 1: Understanding the Question:
When the current through an inductor changes, an electromotive force (EMF) is induced. We are given the inductance, the change in current, and the time duration.
Step 2: Key Formula or Approach:
Faraday's law for self-induction states that the magnitude of the induced EMF (\(e\)) is proportional to the rate of change of current:
\[ e = L \left| \frac{di}{dt} \right| \]
Where \(L\) is self-inductance, \(di\) is change in current, and \(dt\) is the time interval.
Step 3: Detailed Explanation:
Given data:
Self-inductance, \(L = 2 \text{ H}\).
Initial current, \(i_1 = 5 \text{ A}\).
Final current, \(i_2 = 2 \text{ A}\).
Time interval, \(dt = 0.1 \text{ s}\).
Magnitude of change in current:
\[ |di| = |i_2 - i_1| = |2 - 5| = 3 \text{ A} \]
Applying the formula:
\[ e = 2 \times \frac{3}{0.1} \]
\[ e = 2 \times 30 = 60 \text{ V} \]
Step 4: Final Answer:
The magnitude of the induced EMF is 60 V.