A wire of length `l`, mass `m` and resistance `R` slides without any friction down the parallel conducting rails of negligible resistance . The rails are connected to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and the rails form a closed rectangualr conducting loop. The plane of the rails makes an angle `theta` with the horizontal and a uniform vertical magnetic field of the inducetion `B` exists throughout the rregion. Find the steady state velocity of the wire.

Explanation

(a) First method: Let us assume that the rod has acquired a velocity v. By applying the concept of motional emf (Eq. 30-9), the emf of the rod is
`DeltaV=(vecvxxvecB).l=Blvcostheta`
This emf will be such that the current is flowing toward the observer as in Fig. 30-21. The resulting current at that instant is
`i=(DeltaV)/R=(Blvcostheta)/R`
Applying Eq. 28-29, we get
`abs(vecF)=abs(iveclxxvecB)=(B^(2)l^(2)vcostheta)/R`

figure 30-20 A conducting rod is released from rest on a smooth conducting frame inclined at an angle `theta` to the horizontal.
This force is directed horizontally. A component of this force will oppose the motion of the rod as can be seen in Fig. 30-21.
Hence the acceleration of the rod is given by
`mgsintheta-(B^(2)l^(2)vcos^(2)theta)/R=ma`
It can be easily seen that the acceleration of the rod decreases as its velocity increases. After a long time, the acceleration of the rod will tend to zero. The velocity of the rod at this instant is known as the terminal velocity:
`v_(T)=(mgRsintheta)/(B^(2)l^(2)cos^(2)theta)`
(b) Second method:
Let us find the rate of work done by gravity when the velocity of the rod is v.
We know that the power developed by gravity is
`p=vecF.vecv=mgvsintheta`
The rate at which power will be dissipated is
`P_(d)=((DeltaV)^(2))/R=((Blvcostheta)^(2))/R`

Figure 30-21 A figure showing the direction of current flow and the force on the rod.
In steady state,
`mgvsintheta=((Blvcostheta)^(2))/R`
`v_(T)=(mgRsintheta)/(B^(2)l^(2)cos^(2)theta)`
which is the same as before.
We can also find the velocity of the rod as a function of time. Put
`a=(dv)/(dt)`
We get
`(dv)/(dt)=gsintheta(1-v/(v_(t)))`
By integrating, we get
`v=v_(T)(1-e^(-gsinthetat//v_(r)))`
The equation can be easily checked. By substituting t = 0, the velocity turns out to be zero, as expected. At `t = oo`, the velocity turns out to be `v_(T)`.