Question

The current in a coil of \( L = 40 \, \text{mH} \) is to be increased uniformly from 1 A to 11 A in 4 milli sec. The induced e.m.f. will be

• A. 100 V
• B. 4 V
• C. 40 V
• D. 440 V

Hint:

The induced e.m.f. in a coil can be calculated using Faraday's law: \( \text{e.m.f.} = -L \frac{dI}{dt} \), where \( L \) is the inductance and \( \frac{dI}{dt} \) is the rate of change of current.

Answer 1

The Correct Option is C

Step 1: Use the formula for induced EMF.
The induced e.m.f. in a coil is given by Faraday’s law of induction: \[ \text{e.m.f.} = -L \frac{dI}{dt} \] where \( L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} \), and the current changes from 1 A to 11 A in \( 4 \, \text{ms} = 4 \times 10^{-3} \, \text{s} \).
Step 2: Calculate the induced e.m.f.
The rate of change of current is: \[ \frac{dI}{dt} = \frac{11 - 1}{4 \times 10^{-3}} = 2500 \, \text{A/s} \] Thus, the induced e.m.f. is: \[ \text{e.m.f.} = 40 \times 10^{-3} \times 2500 = 100 \, \text{V} \]
Final Answer: \[ \boxed{40 \, \text{V}} \]