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Question: The current in a coil of \[L = 40mH\] is to be increased uniformly from 1A to 11A in 4 milliseconds....

The current in a coil of L=40mHL = 40mH is to be increased uniformly from 1A to 11A in 4 milliseconds. The induced e.m.f will be
A. 100V
B. 0.4V
C. 440V
D. 40V

Explanation

Solution

Hint: In order to find the required value of induced e.m.f in this question we have to write down the given data in the question and use the formula: ε=−Ldidt\varepsilon = - L\dfrac{{di}}{{dt}} where didt\dfrac{{di}}{{dt}} is the rate of decay of current and the L is the coefficient of magnetic induction.

Complete step-by-step answer:
To calculate the induced e.m.f
The formula used:- ε=−Ldidt\varepsilon = - L\dfrac{{di}}{{dt}} ……………….(i)

di = change in current dt= time interval  L= coefficient of induced emf  di{\text{ = change in current}} \\\ dt = {\text{ time interval }} \\\ L = {\text{ coefficient of induced emf}} \\\

Given:-
L=40mHL = 40mH
=40×10−3H= 40 \times {10^{ - 3}}H

i1=1A , i2=11A and dt=4 milliseconds  =4×10−3 sec  {i_1} = 1{\text{A , }}{i_2} = 11{\text{A and }}dt = 4{\text{ milliseconds}} \\\ {\text{ }} = 4 \times {10^{ - 3}}{\text{ sec}} \\\

Similarly,

di=i1−i2 ⇒di=−10A  di = {i_1} - {i_2} \\\ \Rightarrow di = - 10{\text{A}} \\\

Substituting the given values in eqn (i) we get;
ε=− 40×10−3×(−10)4×10−3 V\varepsilon = - {\text{ }}\dfrac{{40 \times {{10}^{ - 3}} \times ( - 10)}}{{4 \times {{10}^{ - 3}}}}{\text{ }}V
⇒ε=100 V\Rightarrow \varepsilon = 100{\text{ }}V
Hence, the correct answer is option (A).

Note: In order to solve such kinds of formula-based questions one should have to practice numerical problems and carefully convert the given physical quantities into their respective SI unit. The minus sign used in the formula indicates that the induced emf opposes the rate of change in current in the coil.