First slide
Magnetic flux and induced emf
Question

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 the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity (in m/s2).

Difficult
Solution

If v is the velocity of the rod at any instant, the emf induced in it will be Bvl. And as the resistance of the rod is R, the induced current will be

I=eR=BvlR

Due to this induced current a force will act on the rod which will oppose its motion.

FM=iBl=B2l2vR   ....(i)

So if T is the tension in the string, equation of motion of rod and mass m will be

TFM=0×a and mgT=ma

So eliminating T between these

mgFM=ma i.e., a=gFMm

and on substituting the value of FM from Eq. (i)

a=gB2l2mRv i.e.,  a=gvt with τ=mR(Bl)2    .......(ii)

So the rod will achieve terminal velocity vT when a= 0,

 i.e.,  gvTτ=0

 i.e.,  vT=τg=mgR(Bl)2   .......(iii)

 and so for v=12vT=12τg, from Eq. (ii), 

a=g1τ12τg=12g=5m/s2

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