A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is `L`. A conducting massless rod of resistance `R` can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass `m` tied to the other end of the string hangs vertically. A constant magnetic field `B` exists perpendicular to the table. If the system is released from rest, calculate

a. the terminal velocity achieved by the rod and
b. The acceleration of the mass of the instant when the velocity of the rod is half the terminal velocity.

Explanation

(a) If `v` is the instantaneous velocity of the rod at any time `t`, the induced emf will be `epsilon = BvL`.
`:.` Induced current in rod
`I = (epsilon)/( R) = (RvL)/( R)` …(i)
Due to this current, the rod in magnetic field `B` will experience a force
`F = BiL` opposite to motion. (ii)
If `v_(T)` is terminal velocity of rod, and `T` the tension in the string, ythen free-body diagrams of rod and mass `m` are shown in Fig. 3.99.
So equation of motion of mass `m` is `mg - T = ma` (iii)
For terminal velocity, `a = 0`
`mg - T = 0` `rarr` `T = mg` (iv)
Equation of motion of rod is
`T - BiL = 0 xx a = 0`(as rod is massless)
`T = BiL`. (v)

Using Eq. (iv), we get, `mg = BiL`
`mg = B((Bv_(T)L)/(R))L` `rarr` `v_(T) = (mgR)/(B^(2)L^(2))` (vi)
(b) When velocity of the rod is half the terminal velocity
`v = (v_(T))/(2) = (1)/(2)(mgR)/((B^(2)L^(2))` (vii)
From Eq. (iii), acceleration
`A = g - (T)/(m) = g - (BiL)/(m)` [using `v`]
`= g - (BL)/(m)(BL)/(R)((1)/(2)mgR)/(B^(2)L^(2))`
`= g - (g)/(2) = (g)/(2)`