Two long parallel horizontal rails a, a distance d aprt and each having a risistance `lambda` per unit length are joing at one end by a resistance R. A perfectly conduction rod MN of mass m is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that, as the rod moves a constant current flows through R.

(i) Find the velocity of the rod and the applied force F as function of the distance x of the rod from R.
(ii) What fraction of the work done per second by F is converted into heat?

Explanation

(a) If the rod has instantaneous velocity `v` at a distance `x` from `R`, the include emf is `Bvd`.
`:.` Induced `I = (Bvd)/(R + 2lambdax) = constant`
`:.` Velocity `v = ((R + 2lambdax) i)/(Bd)`
Megnetic force on the rod `= Bid`.
This force will be opposite to `F`. Hence, net force acting in rod,
`F - Bid = m(dv)/(dt)`
or `F - Bi d = m(dv)/(dx)(dx)/(dt) = mv(d)/(dx){((R + 2lambdax) i)/(bd)} = mv(2lambdai)/(Bd)`
`:.` `F = Bi d + 2mlambda((R + 2lambdax))/(B^(2)d^(2))i^(2)`
(b) Work done per second per second `= Fv`
Heat produced per second `= i^(2)(R + 2lambdax)`
`Required ratio = (i^(2)(R + 2lambdax)Bd)/(F( R + 2lambdax)i) = (iBd)/(F)`
`= (Bi d)/(Bi d + (2mlambda(R + 2lambdax)i^(2))/(B^(2)d^(2)))`
`= (1)/(1 + (2mlambda(R + 2lambdax)i^(2))/(B^(3)d^(3)))`