Answer
a. 18 square meters.
b. 7.5 m/s.
Work Step by Step
a. Use Example 29.3, generalized to N loops.
$$\epsilon_{max}=N\omega BA$$
$$A=\frac{\epsilon_{max}}{N \omega B}$$
$$A=\frac{9.0V}{(2000)(3.14rad/s)(8.0\times10^{-5}T)}=18m^2$$
b. Assume that the point on the circular coil is a distance r from the axis of rotation.
$$v=r\omega=\sqrt{\frac{A}{\pi}}\omega$$
$$v= \sqrt{\frac{18m^2}{\pi}}(3.14rad/s)=7.5m/s$$
