A 10 c m long perfectly conducting wire P Q is moving, with a velocity 1 c m / s on a pair of horizontal…

A 10 cm long perfectly conducting wire PQ is moving, with a velocity 1 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=1Ω 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 milli second is x×10-3A, where the value of x is_______.
[Assume the velocity of wire PQ remains constant 1cm/s after key S is closed. Given: e-1=0.37, where e is base of the natural logarithm]

Solution

ε=v×Bdl=10-2×1×10-1 Motional emf induced in the rodind=V×Bdl=B.V.l
ε=10-3 volt
i=10-311-e-1Equation of growth of current in L-R circuit i=εk1-e-t.RL
i=10-31-0.37
i=0.63 mA .

Asked in: JEE Advanced 2019 (Paper 2)

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