Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

2

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Answer 2

solution :

According to Faraday's law,

induced emf ,

$\boxed{ε = \frac{ - N dΦ}{dt}}$

$\implies \: - N × (B.A cosωt)/dt$

$\implies \: NBAω \: \: sinωt$

now

power produced ,

$\boxed{P = \frac{ ε ^{2}}{ R}}$

$\implies \: \frac{ (NBAω \: sinωt) ^{2} }{R}$

$\implies \: \frac{N ^{2} B ^{2} A ^{2} ω ^{2} sin ^{2} ωt}{R}$

area of cross sectional,

A = πab

$\implies \: \frac{N ^{2} B ^{2} A ^{2} ω ^{2} sin ^{2} ωt}{R}$

now average power ,

< P > = N²B²π²a²b²ω²/R <sin²ωt>

we know, <sin²ωt>

$\frac{1}{\pi - 0} \int _0 ^{\pi} {sin}^{2} \: wtdt = \frac{1}{2}$

Therefore,

$Average \: \: power = \frac{N ^{2} B ^{2} π ^{2} a ^{2} b ^{2} ω ^{2} }{2R}$