Answer
$emf=24\mu V$
Work Step by Step
We know that;
$emf=\frac{d\phi}{dt}$........eq(1)
But, $\phi=BA$. This can be rewritten as:
$\phi=\frac{1}{4}\pi r^2B$
Thus eq(1) becomes:
$emf=\frac{1}{4}\pi r^2(\frac{dB}{dt})$
We plug in the known values to obtain:
$emf=\frac{1}{4}(3.1416)(0.10)^2(0.003)$
$emf=24\times^{-6}$
$emf=24\mu V$
