Answer
$emf=0.452V$
Work Step by Step
We know that
$emf=\frac{d\phi}{dt}$
This can be written as:
$emf=\frac {d(BA)}{dt}$
$emf=\frac{B \pi r^2}{dt}$
As $B$ is constant:
$emf=B\pi\frac{d(r^2)}{dt}$
$emf=B\pi\frac{2r(dr)}{dt}$
$emf=2\pi r B\frac{(dr)}{dt}$
We plug in the known values to obtain:
$emf=2(3.1416)(0.12)(0.800)(0.750)=0.452V$
