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Question: Answered & Verified by Expert
A rectangular loop $\mathrm{PQMN}$ with movable arm $\mathrm{PQ}$ of length $12 \mathrm{~cm}$ and resistance $2 \Omega$ is placed in a uniform magnetic field of $0.1 \mathrm{~T}$ acting perpendicular to the plane of the loop as shown in figure. The resistance of the arms MN, NP and MQ are negligible. The current induced in the loop when arm $\mathrm{PQ}$ is moved with velocity $20 \mathrm{~ms}^{-1}$ is

PhysicsElectromagnetic InductionMHT CETMHT CET 2021 (24 Sep Shift 2)
Options:
  • A $0.12 \mathrm{~A}$
  • B $0.06 \mathrm{~A}$
  • C $0.24 \mathrm{~A}$
  • D $0.18 \mathrm{~A}$
Solution:
1227 Upvotes Verified Answer
The correct answer is: $0.12 \mathrm{~A}$
$\mathrm{B}=0.1 \mathrm{~T}, \ell=0.12 \mathrm{~m}, \mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{R}=2 \Omega$
emf induced $\mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.1 \times 0.12 \times 20=0.24 \mathrm{~V}$
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{0.24}{2}=0.12 \mathrm{~A}$

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