Electromagnetic Induction

JEE Main level, numerical answer, medium

91 of 153

Question 91 of 153

A circular coil of 1000 turns each with area 1m21{m}^{2} is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T0.07T. The maximum voltage generation will be _____ VV.

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Answer: 440

Given, N=1000,A=1m2N=1000,A=1{m}^{2}and B=0.07TB=0.07T
The maximum emf generated is ϵmax⁡=NBAω{\epsilon }_{\max }=NBA\omega,
=NBA(2πf)=NBA\left(2\pi f\right)
=1000×0.07×1×2π(1)=1000\times 0.07\times 1\times 2\pi \left(1\right)=140π=140\pi
Thus, maximum emf is 439.6V≈440V439.6V\approx 440V.

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