A 10 cm long perfectly conducting wire PQ is moving, with a velocity 1cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L=1mH and a resistance R=1Ω as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B=1T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x×10−3A, 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]
Answer: 0.63
Solution
The induced emf is constant: ε=Bℓv=1×0.1×0.01=10−3 V.
Current in LR circuit:
i=Rε(1−e−Rt/L)
At t=1 ms = 10−3 s:
i=110−3(1−e−1×10−3/(10−3))=10−3(1−e−1)
i=10−3(1−0.37)=10−3×0.63=0.63×10−3 A
Therefore, x=0.63.
JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer
