Answer
a. 99.0 A.
b. $1.4\times10^{-2}\;m^2$.
Work Step by Step
a. Find the peak current from the rms current.
$$I_{peak}=\sqrt{2}I_{rms}=\sqrt{2}(70.0A)=99.0A$$
b. The area is calculated from equation 21–5 and the peak induced emf.
$$V_{peak}=NB\omega A=\sqrt{2}V_{rms}$$
$$A=\frac{\sqrt{2}V_{rms}}{ NB\omega }=\frac{\sqrt{2}(150V)}{ (950)(0.030T)(2 \pi rad/rev)(85rev/s)}\approx 1.4\times10^{-2}\;m^2$$
