Electromagnetic Induction

JEE Main level, single correct, medium

90 of 153

Question 90 of 153

A metallic conductor of length 1m1m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5rads−15rad{s}^{-1}. If the horizontal component of earth's magnetic field is 0.2×10−4T0.2\times {10}^{-4}T, then emf induced between the two ends of the conductor is

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Answer: (B) 50μV50\mu V

Given here, BH=0.2×10−4T{B}_{H}=0.2\times {10}^{-4}T, ω=5rads−1\omega =5rad{s}^{-1} and l=1ml=1m.
Emf induced in a rotating conducting rod is given by,
ϵ=12Bωl2\epsilon =\dfrac{1}{2}B\omega {l}^{2}
Substituting the given values, we obtain
ϵ=12×0.2×10−4×5×(1)2=0.5×10−4V=50μV\epsilon =\dfrac{1}{2}\times 0.2\times {10}^{-4}\times 5\times {\left(1\right)}^{2}=0.5\times {10}^{-4}V=50\mu V

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