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 hanges vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from rest, calculate.
(i) the terminal velocity achieved by the rod, and
the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity.

Answer

B

Explanation

If v is the velocity of the rod at any time t, induced emf is BvL and so induced current in the rod
I=(Induced e.m.f)/(R)=(BvL)/(R)
Due to this current, the rod in the field B will experience a force
`F=BIL=(B^(2)L^(2)v)/(R) (opposite to its motion) ...(i)`
So, equation of motion of the rod will be,
`T-F=0xxa, i.e., T=F (as rod is massless]`
`mg-T=ma implies a=g-(T)/(m)=g- (B^(2)L^(2)v)/(R)` ...(ii)
so rod will acquire terminal velocity when its accelaration is zero i.e.
`g-(B^(2)L^(2)v_(T))/(R)=0 i.e. (v_T)=(mgR)/(b^(2)L^(2)),`
For the case when velocity is `(v_T)/2`

`v=(v_T)/2 =(mgR)/(2B^(2)L^(2))`
Substituting this value of velocity in eq. (ii) we get
`a=g-(b^(2)L^(2))/(mR)xx1/2 (mgR)/(B^(2)L^(2))=g-1/2 g=g/2`.