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 =1 Ω as shown in the 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 × 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]

Physics · Electromagnetic Induction

JEE Advanced 2019 — Paper 2 — Question 11

A 10 cm long perfectly conducting wire PQ is moving with a velocity $1 \mathrm{~cm} / \mathrm{s}$ on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor $\mathrm{L}=1 \mathrm{mH}$ and a resistance $\mathrm{R}=1 \Omega$ as shown in the figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field $B=1 \mathrm{~T}$ perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is $\mathrm{x} \times 10^{-3} \mathrm{~A}$ , where the value of x is $\_\_\_\_$ -. [Assume the velocity of wire PQ remains constant ( $1 \mathrm{~cm} / \mathrm{s}$ ) after key S is closed. Given: $\mathrm{e}^{-1}=0.37$ , where e is base of the natural logarithm]

Answer: 0.63

Numerical answer — enter this value.

Step-by-step solution

$I=\frac{\varepsilon}{R}\left(1-e^{-R t / L}\right)=\frac{B l v}{R}\left(1-e^{-R t / L}\right)$

$=\frac{1 \times 0.1 \times 10^{-2}}{1}\left(1-\mathrm{e}^{-1 \times 0^{-3} / 10^{-3}}\right)=\left(1-\mathrm{e}^{-1}\right)=0.63 \mathrm{~mA}$

Hence $\mathrm{x}=0.63$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF