Answer
a.) $I(t) = 5\times10^{-4}\pi(1+t)$
b.)
$I(5) = 9.4\times10^{-3}$ Amps
$I(10) = 1.7\times10^{-2}$ Amps
Work Step by Step
Recall that Faraday's law is
$\displaystyle \varepsilon_{induced} = |\frac{d\Phi_B}{dt}| = |\vec{A}\frac{d\vec{B}}{dt} + \vec{B}\frac{d\vec{A}}{dt}|$
In this problem, the cross-sectional area of the loop isn't changing and we have an N-coil loop so our equation becomes
$\displaystyle \varepsilon_{induced} = N\cdot|\vec{A}\frac{d\vec{B}}{dt}|$
Since we know the total resistance, we can say
$I(t) = \displaystyle \frac{\varepsilon_{induced}}{R} = \frac{N\vec{A}}{R}\cdot\frac{d\vec{B}}{dt} = \frac{N\vec{A}}{R}\vec{B}$ $'(t)$
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a.)
Given $\vec{B}(t) = 0.020 + 0.010t^2 \qquad\rightarrow\qquad\vec{B}$ $'(t) = 0.020 + 0.020t$
$\quad \displaystyle \vec{A} = \pi (\frac{d}{2})^2 = \pi(0.025)^2$
$\therefore I(t) \displaystyle = \frac{20[\pi(0.025)^2]}{0.50}(0.020 + 0.020t) = 2.5\times10^{-2}\pi\cdot0.02(1+t)$
$I(t) = 5\times10^{-4}\pi(1+t)$
b.)
$I(5) = 5\times10^{-4}\pi(1+(5)) = 9.4\times10^{-3}$ Amps
$I(10) = 5\times10^{-4}\pi(1+(10)) = 1.7\times10^{-2}$ Amps
