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← JEE Advanced 2019 · Paper 2
PhysicsNumerical+3Question 10 of 54

JEE Advanced 2019 · Paper 2, Physics question 10

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=1L = 1 mH and a resistance R=1 ΩR = 1\ \Omega as shown in figure. The horizontal rails, LL and RR lie in the same plane with a uniform magnetic field B=1B = 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×10−3x\times10^{-3} A, where the value of xx is ______.

[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given: e−1=0.37e^{-1} = 0.37, where ee is base of the natural logarithm]

Two parallel horizontal rails joined at the left end through an inductor L, a resistor R and a key S in series. A conducting wire PQ slides to the right at 1 cm/s across the rails in a uniform magnetic field B directed into the page.
Answer and worked solution

Official answer: 0.62

Circuit. The moving wire supplies a constant emf:
ε=Bvl=1×0.01×0.1=10−3 V.\varepsilon=Bvl=1\times0.01\times0.1=10^{-3}\text{ V}.
With τ=LR=1\tau=\dfrac LR=1 ms, the current grows as in an LR circuit.

After 1 ms.
i=εR(1−e−t/τ)=10−3(1−0.37)=0.63×10−3 A.i=\frac\varepsilon R\left(1-e^{-t/\tau}\right)=10^{-3}(1-0.37)=0.63\times10^{-3}\text{ A}.
Answer: 0.63 (key: 0.62 to 0.64)

Topics:LR circuitsMotional EMF
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