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QJEEAdv19Ph6b78: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph6b78 Solution: A 10 cm long perfectly conducting wire PQ is moving with a…

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\Omega 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 13 English

Step-by-Step Verified Solution

Motional emf,

e=(v×B)dl=10−2×1×10−1e = (v \times B)dl = {10^{ - 2}} \times 1 \times {10^{ - 1}}

e = 10−3V{10^{ - 3}}V

τL=LR=(10−3)(1)=10−3s=1 ms{\tau _L} = LR = ({10^{ - 3}})(1) = {10^{ - 3}}s = 1\,ms

i=i0(1−e−t/τL)=10−31(1−e−1)i = {i_0}(1 - {e^{ - t/{\tau _L}}}) = {{{{10}^{ - 3}}} \over 1}(1 - {e^{ - 1}})

i=10−3(1−0.37)i = {10^{ - 3}}(1 - 0.37)

i = 0.63 mA

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