Electromagnetic Induction

JEE Main level, single correct, medium

21 of 153

Question 21 of 153

The magnetic flux through a coil perpendicular to its plane is varying according to the relation ϕ=(5t3+4t2+2t−5)\phi =\left(5{t}^{3}+4{t}^{2}+2t-5\right) Weber. If the resistance of the coil is 5ohm5ohm, then the induced current through the coil at t=2st=2s will be,

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Answer: (A) 15.6A15.6A

Given, ϕ=5t3+4t2+2t−5\phi =5{t}^{3}+4{t}^{2}+2t-5
Magnitude of induced emf ϵ=∣dϕdt∣=15t2+8t+2\epsilon =\left|\dfrac{d\phi }{dt}\right|=15{t}^{2}+8t+2
At, t=2st=2s, ϵ=60+16+2=78V\epsilon =60+16+2=78V
Current, i=ϵR=785=15.6Ai=\dfrac{\epsilon }{R}=\dfrac{78}{5}=15.6A

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