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Step-by-Step Solution
1. Identify the Physical Situation
We have a long solenoid of radius $R$, carrying a time-dependent current
$I(t) = k t e^{-a t}$ (with $k > 0$). The solenoid’s current (viewed along its axis) flows counterclockwise for positive current.
A circular conducting coil of radius $2R$ is placed coaxially with the solenoid, in the same plane perpendicular to the solenoid’s axis. We want to find how the induced current in the outer coil behaves as a function of time, based on Lenz’s law and the solenoid’s changing current.
2. Write the Expression for the Magnetic Flux
Because the coil of radius $2R$ fully encloses the solenoid of radius $R$, the magnetic flux $\phi(t)$ through the coil depends on the uniform magnetic field inside the solenoid (assuming it to be sufficiently long).
The magnetic field inside the solenoid is
$$
B(t) = \mu_0 \, n \, I(t),
$$
where $n$ is the number of turns per unit length of the solenoid.
The flux through the coil (since it only depends on the solenoid’s cross-sectional area $\pi R^2$) is:
$$
\phi(t) = B(t)\,\times\,\text{(area of solenoid cross section)}
= \bigl(\mu_0 \, n \, I(t)\bigr)\,\pi R^2.
$$
3. Compute the Induced EMF
By Faraday’s law of electromagnetic induction, the induced emf
$ \varepsilon $ is
$$
\varepsilon(t) = -\,\frac{d\phi(t)}{dt} \;=\; -\,\frac{d}{dt}\Bigl(\mu_0 \, n \,\pi R^2 \, I(t)\Bigr).
$$
Factor out the constants:
$$
\varepsilon(t)
= -\,\mu_0 \, n \,\pi R^2 \,\frac{d}{dt}\bigl(I(t)\bigr).
$$
Given $I(t) = k t \, e^{-\,a t}$, let us compute its time derivative:
$$
\frac{d}{dt}\bigl(k \, t \, e^{-\,a t}\bigr) = k\,\bigl(e^{-\,a t} - a\,t\,e^{-\,a t}\bigr)
= k \, e^{-\,a t} \,\bigl(1 - a\,t\bigr).
$$
Hence,
$$
\varepsilon(t)
= -\,\mu_0 \, n \,\pi R^2
\bigl[
k \, e^{-\,a t} (1 - a\,t)
\bigr].
$$
4. Deduce the Sign and Time Dependence of the Induced Current
The negative sign indicates that the induced current will act (by Lenz’s law) to oppose the change in flux. At
$t = 0$, the derivative $ (1 - a \cdot 0) = 1 $, so
$ \varepsilon(0) \neq 0. $
As $t$ increases, the factor $(1 - a t)$ eventually becomes negative for sufficiently large $t$, and the exponential factor $e^{-\,a t}$ also continuously decreases with $t$. Therefore, the induced emf (and thus the induced current) starts from a nonzero value, changes sign around the time $t = 1/a$, and then tends to zero as $t \to \infty$ (due to the exponential decay).
By applying Lenz’s law (which ensures the coil’s current direction always opposes the solenoid flux change), the correct graphical representation for the outer coil’s induced current as a function of time matches the curve where the induced current starts at a positive value (or negative, depending on the reference for clockwise vs. counterclockwise) and then eventually changes sign, approaching zero.
5. Match with the Correct Option
Comparing with the given options, the correct choice that shows an initial nonzero value of current, a sign reversal around $t = \frac{1}{a}$, and then a decay to zero is depicted by:
Correct Answer:
