#### 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 (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

The component of magnetic field along the inclined plane will be $=B\sin \theta$ and other will be perpendicular i.e. $=B\cos \theta$. The conductor is moving perpendicular to $=B\cos \theta$. It is the vertical component of the magnetic field. The movement will cause motional emf across the two ends of the rod.

given by $=v(B \cos \theta )d$

This makes flow of induced current

$i=\frac{v(B \cos \theta )d}{R}$

where R is the resistance of rod. Now, current-carrying rod experience a magnetic force which is given by

$F_{m}=iBd$ (horizontally in backward direction).

Now, the component of magnetic force parallels to the inclined plane along the upward direction.

$F_{\parallel }=F_{m} \cos \theta =iBd \cos \theta =\left ( \frac{v(B \cos \theta )d}{R} \right )Bd \cos \theta$

Where,

$v=\frac{dx}{dt}$

Also, the component of weight (mg) parallel to the inclined plane along downward direction = $mg\; \sin \theta .$

Now, by Newton's second law of motion

$m\frac{d^{2}x}{dt^{2}}=mg \sin \theta -\frac{B \cos \theta d}{R}\left ( \frac{dx}{dt} \right )\times (BD)\; \cos \theta$

$\Rightarrow \frac{dv}{dt}=g\; \sin \theta -\frac{B^{2}d^{2}}{mR}(\cos \theta )^{2}v$

$\Rightarrow \frac{dv}{dt}+\frac{B^{2}d^{2}}{mR}(\cos \theta )^{2}v=g\; \sin \theta$

But, this is the linear differential equation.

On solving, we get

$v=\frac{g \sin \theta }{\frac{B^{2}d^{2}\cos^{2}\theta }{mR}}+ A\; exp\left ( -\frac{B^{2}d^{2}}{mR}(\cos ^{2}\theta )t \right )$

A is a constant to be determined by initial conditions.

The required expression of velocity as a function of time is given by

$\frac{mgR \sin \theta }{B^{2}d^{2}\cos^{2}\theta }\left ( 1-exp\left ( -\frac{B^{2}d^{2}}{mR}(\cos^{2}\theta )t \right ) \right )$