Electromagnetic Induction

JEE Main level, numerical answer, medium

20 of 153

Question 20 of 153

Magnetic flux (in weber) in a closed circuit of resistance 20Ω20\Omega varies with time t(s)t\left(s\right) as ϕ=8t2−9t+5\phi =8{t}^{2}-9t+5. The magnitude of the induced current at t=0.25st=0.25s will be _____ mAmA.

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Answer: 250

Given that magnetic flux ϕ=8t2−9t+5\phi =8{t}^{2}-9t+5
According to Faraday's law of electromagnetic induction
emf=−dϕdt=−(16t−9)emf=-\dfrac{d\phi }{dt}=-\left(16t-9\right)
At t=0.25st=0.25s
emf=−[(16×0.25)−9]=5Vemf=-\left[\left(16\times 0.25\right)-9\right]=5V
Current =emfResistance=5V20Ω=\dfrac{emf}{Resistance}=\dfrac{5V}{20\Omega }
=14A=10004mA=250mA=\dfrac{1}{4}A=\dfrac{1000}{4}mA=250mA

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