Problem 23.10

23.10.
We want to get an expression that will give us the current. We start by finding the average emf.
$\epsilon = -N \frac{d}{dt} B A \cos \theta$
converting the differential to a time average, we find
$\epsilon_{av} = -N B \cos \theta \frac{\Delta A}{\Delta t}$
$\epsilon_{av} = (-1)(.1 T)(\cos 0{}^{\circ})\frac{(3m \times 3m \times \sin 60{}^{\circ}) - (3m)^{2}}{.1 s}$
$\epsilon_{av}=1.21 V$
From Ohm's law,
$I = \frac{1.21 V}{10 \Omega}$
I = .121 A counterclockwise out of the page