The current in a coil of self-inductance 2.0 | vedclass

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(A) The energy stored in an inductor is given by $E = \int L I \, dI$.Given $L = 2.0 \, H$ and $I = 2 \sin(t^2) \, A$.First,find the time interval. Wh

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$4$

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(A) The energy stored in an inductor is given by $E = \int L I \, dI$.Given $L = 2.0 \, H$ and $I = 2 \sin(t^2) \, A$.First,find the time interval. When $I = 0$,$2 \sin(t^2) = 0 \Rightarrow t = 0$.When $I = 2 \, A$,$2 \sin(t^2) = 2 \Rightarrow \sin(t^2) = 1 \Rightarrow t^2 = \frac{\pi}{2} \Rightarrow t = \sqrt{\frac{\pi}{2}}$.The energy spent is $E = \int_{0}^{2} L I \, dI = \int_{0}^{\sqrt{\pi/2}} L I \left( \frac{dI}{dt} ight) dt$.Since $I = 2 \sin(t^2)$,$\frac{dI}{dt} = 2 \cos(t^2) \cdot 2t = 4t \cos(t^2)$.Substituting these into the integral:$E = \int_{0}^{\sqrt{\pi/2}} 2 \cdot (2 \sin(t^2)) \cdot (4t \cos(t^2)) \, dt$$E = 16 \int_{0}^{\sqrt{\pi/2}} t \sin(t^2) \cos(t^2) \, dt$Using the identity $2 \sin 	heta \cos 	heta = \sin(2 	heta)$:$E = 8 \int_{0}^{\sqrt{\pi/2}} t \sin(2t^2) \, dt$Let $u = 2t^2$,then $du = 4t \, dt$,so $t \, dt = \frac{du}{4}$.When $t = 0, u = 0$. When $t = \sqrt{\pi/2}, u = \pi$.$E = 8 \int_{0}^{\pi} \sin(u) \frac{du}{4} = 2 \int_{0}^{\pi} \sin(u) \, du$$E = 2 [-\cos(u)]_{0}^{\pi} = 2 [-\cos(\pi) - (-\cos(0))] = 2 [1 + 1] = 4 \, J$.

The current in a coil of self-inductance $2.0 \, H$ is increasing according to $I = 2 \sin(t^2) \, A$. The amount of energy spent during the period when current changes from $0$ to $2 \, A$ is ...... $J$.

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