Answer
(a) $6.9\times 10^{-3}m^2$
(b) $0.42V$
Work Step by Step
(a) We know that
$A=\frac{lL}{\mu_{\circ}N^2}$
We plug in the known values to obtain:
$A=\frac{(0.24)(7.3\times 10^{-3})}{(4\pi\times 10^{-7})(450)^2}$
$A=6.9\times 10^{-3}m^2$
(b) We can find the induced emf as
$\epsilon=-(L)(\frac{\Delta I}{\Delta t})$
We plug in the known values to obtain:
$\epsilon=-(7.3)(-\frac{3.2}{55\times 10^{-3}})$
$\epsilon=0.42V$
