Electromagnetic Induction

JEE Main level, numerical answer, medium

52 of 153

Question 52 of 153

A square shaped coil of area 70cm270{cm}^{2} having 600600 turns rotates in a magnetic field of 0.4Wbm–20.4Wb{m}^{–2}, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at 60∘60^{\circ} with the field, will be ______ VV.
(Take π=227\pi =\dfrac{22}{7})

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Answer: 44

Given here, N=600N=600, A=70×10−4m2A=70\times {10}^{-4}{m}^{2}, and B=0.4TB=0.4T
Angular speed of coil is ω=2πn60=500×2π60=100π6rads−1\omega =\dfrac{2\pi n}{60}=\dfrac{500\times 2\pi }{60}=\dfrac{100\pi }{6}rad{s}^{-1}
Instantaneous emf in coil is given by ϵ=−Ndϕdt\epsilon =-N\dfrac{d\phi }{dt}, where, magnetic flux ϕ=BAcos⁡ωt\phi =BA\cos \omega t and ωt=30∘\omega t=30^{\circ} is angle between AA and BB.
So, ϵ=−Nd(BAcos⁡ωt)dt=−N(−BAωsin⁡ωt)=NBAωsin⁡ωt\epsilon =-N\dfrac{d\left(BA\cos \omega t\right)}{dt}=-N\left(-BA\omega \sin \omega t\right)=NBA\omega \sin \omega t
Putting the values, we get
ϵ=600×70×10−4×0.4×100π6×12\epsilon =600\times 70\times {10}^{-4}\times 0.4\times \dfrac{100\pi }{6}\times \dfrac{1}{2}
=44V=44V

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