Electromagnetic Induction

JEE Main level, numerical answer, medium

25 of 153

Question 25 of 153

In the given figure the magnetic flux through the loop increases according to the relation ϕB(t)=10t2+20t{\phi }_{B}(t)=10{t}^{2}+20t, where ϕB{\phi }_{B} is in milliwebers and tt is in seconds. The magnitude of current through R=2ΩR=2\Omega resistor at t=5st=5s is _____ mA.mA.

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Answer: 60

∣ϵ∣=dϕdt=20t+20mV|ϵ|=\dfrac{d\phi }{dt}=20t+20mV
∣i∣=∣ϵ∣R=10t+10mA|i|=\dfrac{|ϵ|}{R}=10t+10mA
At
t=5t=5
∣i∣=60mA|i|=60mA

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