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(A) Let the initial current be $I_0$. Since the current decreases uniformly to zero in time $\Delta t$,the current at any time $t$ is given by $I(t) =

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$\frac{4}{3}\frac{q^2 R}{\Delta t}$

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(A) Let the initial current be $I_0$. Since the current decreases uniformly to zero in time $\Delta t$,the current at any time $t$ is given by $I(t) = I_0(1 - \frac{t}{\Delta t})$.The total charge $q$ passed is the integral of current over time: $q = \int_0^{\Delta t} I(t) dt = \int_0^{\Delta t} I_0(1 - \frac{t}{\Delta t}) dt = I_0 [t - \frac{t^2}{2\Delta t}]_0^{\Delta t} = I_0(\Delta t - \frac{\Delta t}{2}) = \frac{I_0 \Delta t}{2}$.Thus,$I_0 = \frac{2q}{\Delta t}$.The heat generated $H$ is given by $H = \int_0^{\Delta t} I^2 R dt = R \int_0^{\Delta t} I_0^2 (1 - \frac{t}{\Delta t})^2 dt$.Let $u = 1 - \frac{t}{\Delta t}$,then $du = -\frac{dt}{\Delta t}$,so $dt = -\Delta t du$.When $t=0, u=1$; when $t=\Delta t, u=0$.$H = R I_0^2 \int_1^0 u^2 (-\Delta t du) = R I_0^2 \Delta t \int_0^1 u^2 du = R I_0^2 \Delta t [\frac{u^3}{3}]_0^1 = \frac{R I_0^2 \Delta t}{3}$.Substituting $I_0 = \frac{2q}{\Delta t}$,we get $H = \frac{R}{3} (\frac{2q}{\Delta t})^2 \Delta t = \frac{R}{3} \frac{4q^2}{\Delta t^2} \Delta t = \frac{4q^2 R}{3 \Delta t}$.

What amount of heat will be generated in a coil of resistance $R$ due to a charge $q$ passing through it if the current in the coil decreases to zero uniformly during a time interval $\Delta t$?

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