$A$ rod of mass $m$ and resistance $R$ slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $\theta$ with respect to the horizontal as shown in the figure. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field $\vec{B}$ along the vertical direction. If the rod is initially at rest,find the velocity of the rod as a function of time.

(N/A) The component of velocity perpendicular to the magnetic field is $v \cos \theta$.
The induced emf in the rod is $\varepsilon = B(v \cos \theta) l$,where $l = d$.
Thus,$\varepsilon = B v d \cos \theta$.
The induced current is $I = \frac{\varepsilon}{R} = \frac{B v d \cos \theta}{R}$.
The magnetic force on the rod is $F = B I d$,acting horizontally. The component of this force parallel to the slope is $F_{\parallel} = F \cos \theta = B I d \cos \theta = \frac{B^2 d^2 v \cos^2 \theta}{R}$.
The net force along the slope is $F_{\text{net}} = mg \sin \theta - F_{\parallel} = mg \sin \theta - \frac{B^2 d^2 v \cos^2 \theta}{R}$.
Using Newton's second law,$m \frac{dv}{dt} = mg \sin \theta - \frac{B^2 d^2 v \cos^2 \theta}{R}$.
Rearranging gives the linear differential equation: $\frac{dv}{dt} + \left( \frac{B^2 d^2 \cos^2 \theta}{mR} \right) v = g \sin \theta$.
Solving this with the initial condition $v(0) = 0$,we get:
$v(t) = \frac{mgR \sin \theta}{B^2 d^2 \cos^2 \theta} \left( 1 - e^{-\frac{B^2 d^2 \cos^2 \theta}{mR} t} \right)$.