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Electromagnetic Induction question

2019 · Shift 2 · Q50
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Electromagnetic Induction question

2019 · Shift 2 · Q50

JEE AdvancedPhysicsElectromagnetic InductionNumerical+3 / −1
A 10 cm long perfectly conducting wire PQ is moving with a velocity I cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1 Ω\OmegaΩ as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x ×\times× 10-3 A, where the value of x is ........... [Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given e-1 = 0.37, where e is base of the natural logarithm] JEE Advanced 2019 Paper 2 Offline Physics - Electromagnetic Induction Question 15 English
Numerical answer
View written solutionFree

Correct answer: 0.63

  1. Motional emf produced by the moving rod

The rod of length

l=10 cm=0.1 ml = 10\text{ cm} = 0.1\text{ m}l=10 cm=0.1 m

moves with speed

v=1 cm/s=0.01 m/sv = 1\text{ cm/s} = 0.01\text{ m/s}v=1 cm/s=0.01 m/s

in a uniform magnetic field

B=1 T.B = 1\text{ T}.B=1 T.

Hence the motional emf is

E=Blv=(1)(0.1)(0.01)=10−3 V.\mathcal E = Blv = (1)(0.1)(0.01) = 10^{-3}\text{ V}.E=Blv=(1)(0.1)(0.01)=10−3 V.
  1. Circuit after key is closed

After closing the key, the circuit is a series RLRLRL circuit driven by constant emf E=10−3\mathcal E = 10^{-3}E=10−3 V.

Given:

R=1 Ω,L=1 mH=10−3 HR = 1\,\Omega, \qquad L = 1\text{ mH} = 10^{-3}\text{ H}R=1Ω,L=1 mH=10−3 H

The current growth in an RLRLRL circuit is

i(t)=ER(1−e−Rt/L).i(t) = \frac{\mathcal E}{R}\left(1-e^{-Rt/L}\right).i(t)=RE​(1−e−Rt/L).
  1. Substitute values

First,

ER=10−31=10−3 A.\frac{\mathcal E}{R} = \frac{10^{-3}}{1} = 10^{-3}\text{ A}.RE​=110−3​=10−3 A.

Also,

RL=110−3=103 s−1.\frac{R}{L} = \frac{1}{10^{-3}} = 10^3\text{ s}^{-1}.LR​=10−31​=103 s−1.

At

t=1 ms=10−3 s,t = 1\text{ ms} = 10^{-3}\text{ s},t=1 ms=10−3 s,

we get

RtL=103×10−3=1.\frac{Rt}{L} = 10^3 \times 10^{-3} = 1.LRt​=103×10−3=1.

So,

i(1 ms)=10−3(1−e−1).i(1\text{ ms}) = 10^{-3}(1-e^{-1}).i(1 ms)=10−3(1−e−1).

Using

e−1=0.37,e^{-1} = 0.37,e−1=0.37, i(1 ms)=10−3(1−0.37)=10−3(0.63)=0.63×10−3 A.i(1\text{ ms}) = 10^{-3}(1-0.37)=10^{-3}(0.63)=0.63\times 10^{-3}\text{ A}.i(1 ms)=10−3(1−0.37)=10−3(0.63)=0.63×10−3 A.
  1. Value of xxx

Since current is written as

i=x×10−3 A,i = x\times 10^{-3}\text{ A},i=x×10−3 A,

we get

x=0.63.x = 0.63.x=0.63.
  1. Comparison with stored correct answer

Stored correct answer: 0.630.630.63

Derived answer: 0.630.630.63

They match.

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