Electromagnetic Induction

JEE Main level, single correct, medium

111 of 153

Question 111 of 153

The self induced emf of a coil is 2525 volts. When the current in it is changed at uniform rate from 10A10A to 25A25A in 1s1s , the change in the energy of the inductance is:

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Answer: (C) 437.5J437.5J

Induced emf / back emf in an inductor coil is given by
∣ϵ∣=Ldidt\left|\epsilon \right|=\dfrac{Ldi}{dt}

⇒25=L×(25−10)1\Rightarrow 25=L\times \dfrac{\left(25-10\right)}{1}
⇒L=2515=53H\Rightarrow L=\dfrac{25}{15}=\dfrac{5}{3}H

Energy of inductor is given by
E=12Li2E=\dfrac{1}{2}{Li}^{2}
Energy change △E=12L(I22−I12)△E=\dfrac{1}{2}L\left({{I}_{2}}^{2}-{{I}_{1}}^{2}\right)

⇒∆E=12×53[625−100]J\Rightarrow ∆E=\dfrac{1}{2}\times \dfrac{5}{3}\left[625-100\right]J

∆E=437.5J∆E=437.5J

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