Problem 22.13

22.13.
First off, let's find the magnetic moment of the loop.
$\vec{\mu} = NI\vec{A}$
$\vec{\mu} = (100)(1.2 A)(.12 m^{2})$ in a direction normal to the plane of the loops
$\vec{\mu} = 14.4 A\cdot m^{2}$ normal to the loops
Now, we know that the torque is given by
$\vec{\tau} = \vec{\mu} \times \vec{B}$
$\vec{\tau} = (14.4 A \cdot m^{2})(.800 T) \sin 60{}^{\circ}$ in the $- \hat{y}$ direction
$\vec{\tau} = - 9.98 N \cdot m ~\hat{y}$
The loop will tend to rotate to aline the magnetic moment, $\vec{\mu}$ up with the magnetic field, $\vec{B}$. Looking down from the positive y axis, the loop will rotate clockwise.