[III] The metal bar of length $L$, mass $m$, and resistance $R$ depicted in slides without friction on a rectangular circuit composed of resistanceless wire resting on an inclined plane. There is a vertical uniform magnetic field $\overrightarrow{\mathbf{B}}$. Find the terminal speed of the bar (that is, the constant speed it attains).
Gravity pulls the bar down the incline as shown in $\underline{\text { Fig. }} 32-7(b)$. Induced current flowing in the bar interacts with the field so as to retard this motion.
Because of the motion of the bar in the magnetic field, an emf is induced in the bar:
$$
\mathscr{E}=(B l v)_{\perp}=B L(v \cos \theta)
$$
This causes a current
$$
I=\frac{\mathrm{emf}}{R}=\left(\frac{B L v}{R}\right) \cos \theta
$$
in the loop. A wire carrying a current in a magnetic field experiences a force that is perpendicular to the plane defined by the wire and the magnetic field lines. The bar thus experiences a horizontal force $\overrightarrow{\mathbf{F}}_{h}$ (perpendicular to the plane of $\overrightarrow{\mathbf{B}}$ and the bar) given by
$$
F_{h}=B I L=\left(\frac{B^{2} L^{2} v}{R}\right) \cos \theta
$$
and shown in Fig However, we want the force component along the plane, which is
$$
F_{\text {up plane }}=F_{h} \cos \theta=\left(\frac{B^{2} L^{2} v}{R}\right) \cos ^{2} \theta
$$
When the bar reaches its terminal velocity, this force equals the gravitational force down the plane. Therefore,
$$
\left(\frac{B^{2} L^{2} v}{R}\right) \cos ^{2} \theta=m g \sin \theta
$$
from which the terminal speed is
$$
v=\left(\frac{R m g}{B^{2} L^{2}}\right)\left(\frac{\sin \theta}{\cos ^{2} \theta}\right)
$$
Can you show that this answer is reasonable in the limiting cases $\theta$ $=0, B=0$, and $\theta=90^{\circ}$, and for $R$ very large or very small?