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 horiaontal, Fig. The circuit is closed through a perfert 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.

Explanation

Here, the component of magnetic field perpendicular the plane `=Bcos theta` Now, the conductor moves with speed v perpendicular to `Bcostheta` component of magnetic field. This causes motional emf across two ends of rod, which is given by `=v(Bcostheta)d`
This makes flow of induced current `i=(v(Bcostheta)d)/(R)`where, R is the resistance of rod. Now, current carrying rod experience force which is given by `F=iBd` (horizontally in backward direction). Now, the component of magnetic force parallel to incline plane along upward direction `=Fcostheta=(v(Bcostheta)d)/(R)Bdcostheta` when, `v=(dx)/(dt)`
Also, the component weight (mg) parallel to incline plane along donward direction `=mgsintheta`.
Now, by Newton's second law of motion
`m(d^2)/(dt^2)=mg sintheta -(Bcosthetad)/(R)((dx)/(dt))xx(Bd)costheta`
`(dv)/(dt)=gsintheta -(B^2d^2)/(mR)(cos theta)^2v`
`(dv)/(dt) +(B^2d^2)/(mR)(cos theta)^2v=gsintheta`
But this is the linear differential eqation. On solving we get
`v=(gsintheta)/((B^2d^2cos^2theta)/(mR))+Aexp(-(B^2d^2)/(mR)(cos^2theta)t)`
A is a constant to be determine by intial conditions.
The required expression of velocity as a function of time is given by
`=(mgRsintheta)/(B^2d^2cos^2theta)(1-exp(-(B^2d^2)/(mR)(cos^2theta)t))`