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(D) The magnetic field $B$ inside the solenoid is $B = \mu_0 n I(t)$.The magnetic flux $\phi$ through the outer coil of radius $2R$ is due only to the

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(D) The magnetic field $B$ inside the solenoid is $B = \mu_0 n I(t)$.The magnetic flux $\phi$ through the outer coil of radius $2R$ is due only to the magnetic field within the solenoid (since the field outside an ideal solenoid is zero). Thus,$\phi = B \cdot A = (\mu_0 n I(t))(\pi R^2)$.The induced $EMF$ $\varepsilon$ is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt} = -\mu_0 n \pi R^2 \frac{dI}{dt}$.Given $I(t) = kt e^{-\alpha t}$,we find $\frac{dI}{dt} = k(e^{-\alpha t} + t(-\alpha)e^{-\alpha t}) = k e^{-\alpha t}(1 - \alpha t)$.Therefore,$\varepsilon = -\mu_0 n \pi R^2 k e^{-\alpha t}(1 - \alpha t)$.At $t = 0$,$\varepsilon = -\mu_0 n \pi R^2 k(1) = -	ext{constant}$. Since the induced current $i_{ind} = \frac{\varepsilon}{R_{coil}}$,the current at $t = 0$ is negative.As $t$ increases,the term $(1 - \alpha t)$ becomes zero at $t = 1/\alpha$,meaning the induced current crosses the zero axis.For $t > 1/\alpha$,the term $(1 - \alpha t)$ becomes negative,making the induced current positive.This behavior corresponds to the graph that starts at a negative value,crosses the $t$-axis at $t = 1/\alpha$,reaches a positive peak,and then decays to zero as $t 	o \infty$.

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