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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Physics
Chapter
Electromagnetic Induction

JEE Advanced 2019 · Paper 2 · Physics · Electromagnetic Induction

A 10cm long perfectly conducting wire PQ is moving, with a velocity...

A 10cm10cm long perfectly conducting wire PQPQ is moving, with a velocity 1cm/s1cm / s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L=1mHL = 1mH and a resistance R=1ΩR = 1Ω as shown in figure. The horizontal rails, LandRL and R lie in the same plane with a uniform magnetic field B=1TB = 1T perpendicular to the plane. If the key SS is closed at certain instant, the current in the circuit after 11 milli second is x×10−3A,x \times {10}^{- 3}A, where the value of xx is_______.
[Assume the velocity of wire PQPQ remains constant (1cm/s)\left(1cm / s\right) after key SS is closed. Given: e−1=0.37,{e}^{- 1} = 0.37, where ee is base of the natural logarithm]
A 10cm long perfectly conducting wire PQ is moving, with a velocity 1cm / s on a pair of horizontal rails of zero

Correct answer

0

Step-by-step solution

⇒ ε = \left(\overset{\to}{v} \times \overset{\to}{B}\right)\overset{\to}{d}l = {10}^{- 2} \times 1 \times {10}^{- 1}\left[Motionalemfinducedintherod∑ind=(V→×B→)d→l=B.V.l\begin{array}{ll}Motionalemfinducedintherod \\ ∑ ind = \left(\overset{\to}{V} \times \overset{\to}{B}\right)\overset{\to}{d}l \\ = B . V . l\end{array}\right]
ε=10−3voltε = {10}^{- 3}volt
⇒i=10−31(1−e−1)→[EquationofgrowthofcurrentinL−Rcircuiti=εk(1−e−t.RL)]⇒ i = \dfrac{{10}^{- 3}}{1}\left(1 - {e}^{- 1}\right) \to \left[EquationofgrowthofcurrentinL - Rcircuiti = \dfrac{ε}{k}\left(1 - {e}^{- t . \dfrac{R}{L}}\right)\right]
i=10−3(1−0.37)i = {10}^{- 3}\left(1 - 0.37\right)
i=0.63mAi = 0.63mA
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