The data given is
B=0.4TR=2×10−2m
It is given that dtdR=10−3ms−1
Area of the loop is A=πR2
So, dtdA=2πRdtdR
The flux is,
ϕ=2πRdtdR×B
So, the magnitude of emf is
ϵ=2×722×2×10−2×10−3×0.4
=5.028×10−5V=50.28μV≈50μV
JEE Main 2023 — Physics Electromagnetism
A conducting circular loop is placed in a uniform magnetic field of 0.4T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1mms−1. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2cm will be _______ μV.
Held on 12 Apr 2023 · Verified 6 Jul 2026.
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