Electromagnetic Induction

JEE Main level, numerical answer, medium

93 of 153

Question 93 of 153

A circular coil of radius 8.0cm8.0cm and 2020 turns is rotated about its vertical diameter with an angular speed of 50rads−150rad{s}^{-1} in a uniform horizontal magnetic field of 3.0×10−2T3.0\times {10}^{-2}T. The maximum emf induced in the coil will be _______×10−2\times {10}^{-2} volt (rounded off to the nearest integer).

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Answer: 60

Emf=BωNAsinωtEmf=B\omega NAsin\omega t
Emfmax⁡=BωNA=BωN(πR2){Emf}_{\max }=B\omega NA=B\omega N\left({\pi R}^{2}\right)
=(3×10−2)×(50)×(20)×(π×64104)≈60×10−2V=\left(3\times {10}^{-2}\right)\times \left(50\right)\times \left(20\right)\times \left(\pi \times \dfrac{64}{{10}^{4}}\right)\approx 60\times {10}^{-2}V

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