Answer
a. $2.25\times10^{-5} m\cdot N $
b. The north edge
Work Step by Step
a. See Figure 20-34c. The angle between a downward line perpendicular to the loop face, and the magnetic field, is 34 degrees. Use equation 20–10 to find the torque.
$$\tau=NIABsin\theta$$
$$\tau= (9)(7.20A)\pi(0.06m)^2(5.5\times10^{-5}T)sin34^{\circ}$$
$$2.25\times10^{-5} m\cdot N $$
b. On page 576 it is noted that if the current-carrying coil is free to turn, it rotates in such a way that that the angle $\theta$ is 0 degrees. In order for the angle between a downward line perpendicular to the loop face, and the magnetic field, to become 0 degrees, the north edge of the coil rises.
