Electromagnetic Induction

JEE Main level, numerical answer, medium

130 of 153

Question 130 of 153

The current in a coil of self inductance L=2.0HL=2.0H is increasing according to the law i=2sin⁡(t2)i=2\sin \left({t}^{2}\right). Find the amount of energy spent (in JJ) during the period when the current changes from 00 to 2A2A.

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Answer: 4

Let the current be 2A2A at t=τt=\tau
Then 2=2sinτ2⇒τ=π/22=2sin{\tau }^{2}\Rightarrow \tau =\sqrt{\pi /2}
when the instantaneous current is ii, the
self-induced emf is Ldidt.L\dfrac{di}{dt}. If dqdq amount of
charge is displaced in time dtdt, then elementary work done
=(Ldidt)dq=Ldidtidt==\left(L\dfrac{di}{dt}\right)dq=L\dfrac{di}{dt}idt=LidiLidi
∴W=∫0tLidi∴W={\int }_{0}^{t}Lidi =∫0tL2sint2d(2sint2)={\int }_{0}^{t}L2sin{t}^{2}d\left(2sin{t}^{2}\right)
⇒W=∫0t8Lsint2cost2tdt=4L∫0tsin2t2dt\Rightarrow W={\int }_{0}^{t}8Lsin{t}^{2}cos{t}^{2}tdt=4L{\int }_{0}^{t}sin2{t}^{2}dt
Let θ=2t2⇒dθ=4tdt\theta =2{t}^{2}\Rightarrow d\theta =4tdt
∴∴ the integral
=4L∫sinθdt/4L(−cosθ)=−Lcos2t2=4L\int sin\theta dt/4L(-cos\theta )=-Lcos2{t}^{2}
∴W=−[cos2t2]0π/2=2L=4J∴W=-{\left[cos2{t}^{2}\right]}_{0}^{\sqrt{\pi /2}}=2L=4J

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