Electromagnetic Induction and Alternating Current 7 Question 29

####32. Two long parallel horizontal rails, a distance d apart and each having a resistance λ per unit length, are joined at one end by a resistance R. A perfectly conducting rodMN 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 rodMN such that, as the rod moves, a constant current i flows through R.

Find the velocity of the rod and the applied force F as functions of the distance x of the rod from R.v

(1988,6M)

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Answer:

Correct Answer: 32. v=(R+2λx)iBd,F=2λi2mB2d2(R+2λx)2+idB

Solution:

  1. Total resistance of the circuit as function of distance x from resistance R is Rnet =R+2λx

Let v be velocity of rod at this instant, then motional emf induced across the rod, e=Bvd

Current i=eRnet =BvdR+2λxv=(R+2λx)iBd

Net force on the rod, Fnet =mdvdt=2λimBd(R+2λx)dxdt

but

dxdt=v=(R+2λx)iBdFnet =2λi2mB2d2(R+2λx)2

This net force is equal to FFm where Fm=idB

F=Fnet +Fm=2λi2mB2d2(R+2λx)2+idB



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