Answer
$\mathscr{E} = 80~\mu V$
Work Step by Step
$B = 4.0~t^2~y$
Note that the magnetic field changes over time, and the magnitude of the magnetic field varies linearly at different values of $y$
Since the y values of the loop vary from $y = 0$ to $y = 0.02~m$, the average magnitude of the magnetic field through the loop is as follows:
$B_{ave} = 4.0~t^2~(0.01) = 0.04~t^2$
$\frac{dB}{dt} = 0.08~t$
When $t = 2.5~s$, then $\frac{dB}{dt} = (0.08)~(2.5~s) = 0.20~T/s$
The magnetic flux through the loop also changes over time.
We can find the magnitude of the induced emf in the loop:
$\mathscr{E} = \frac{d\Phi}{dt}$
$\mathscr{E} = A~\frac{dB}{dt}$
$\mathscr{E} = (0.020~m)(0.020~m)~(0.20~T/s)$
$\mathscr{E} = 80~\mu V$
