Answer
$I(0) = 1.6$ A
$I(1) = 0$ A
$I(2) = -1.6$ A
Work Step by Step
Recall that Faraday's law is
$\displaystyle \varepsilon_{induced} = \left|\frac{d\Phi_B}{dt}\right| = \left|\vec{A}\frac{d\vec{B}}{dt} + \vec{B}\frac{d\vec{A}}{dt}\right|$
In this problem, the cross-sectional area of the loop isn't changing so our equation becomes
$\displaystyle \varepsilon_{induced} = \left|\vec{A}\frac{d\vec{B}}{dt}\right|$
Since we know the total resistance, we can say
$I(t) = \displaystyle \frac{\varepsilon_{induced}}{R} = \frac{\vec{A}}{R}\cdot\frac{d\vec{B}}{dt} = \frac{\vec{A}}{R}\vec{B}$ $'(t)$
---
Given $\vec{B}(t) = 4t-2t^2 \qquad\rightarrow\qquad\vec{B}$ $'(t) = 4-4t$
$\qquad$ $\displaystyle \vec{A} = s^2 = (0.20)^2 = 4\times10^{-2}$ m$^2$
$I(t) \displaystyle = \frac{4\times10^{-2}}{0.10}(4-4t) = 0.4\cdot4(1-t)$
$I(t) = 1.6(1-t)$
$\therefore $
$I(0) = 1.6$ A
$I(1) = 0$ A
$I(2) = -1.6$ A
