Q.

A rod of mass 0.720 kg and radius 6 cm rests on two parallel rails that are d=12 cm apart and length of rails is  L=49 cm. The rod carries a current of i=48 A in the direction shown in figure. And rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails.

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a

1.12 m/s

b

2.12 m/s

c

3.12 m/s

d

4.12 m/s

answer is A.

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Detailed Solution

Using work energy theorem,  Wby Magnetic force +Wby friction=ΔKE ⇒FL=12mv2+12Iw2   [Static friction does no work during pure rolling] ⇒Bid L=12mv2+1212mr2(vr)2 ∴ V=43idBLm=4×48×0.12×0.240×0.493×0.720=1.12m/s
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A rod of mass 0.720 kg and radius 6 cm rests on two parallel rails that are d=12 cm apart and length of rails is  L=49 cm. The rod carries a current of i=48 A in the direction shown in figure. And rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails.