Answer
See answers.
Work Step by Step
a. As the rod moves, a motional emf occurs, given by equation 21–3,
$$\epsilon = B\mathcal{l}v = (0.35T)(0.300m)(1.6m/s)=0.168V\approx 0.17V$$
b. The induced current is clockwise, to make more magnetic field lines into the page, fighting the increasing number of B field lines pointing out of the page as the rod slides through the field.
So, the current flowing in the rod is from top to bottom. The amount is the induced emf divided by the circuit's total resistance.
$$I=\frac{\epsilon}{R}=\frac{0.168V}{21.0\Omega+2.5\Omega}=7.15 mA$$
c. By the right hand rule, there is a net magnetic force on the rod, to the left, $I\mathcal{l}B$. The external force to the right is equal and opposite since the rod moves at constant speed.
$$F= I\mathcal{l}B =(0.007149A)(0.300m)(0.35T)=7.5\times10^{-4}N$$
