Electromagnetic Induction and Alternating Current 2 Question 8

9. 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 :

(1997, 5M)

(a) the terminal velocity achieved by the rod, and

(b) the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity.

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Answer:

Correct Answer: 9. (a) vT=mgRB2L2 (b) a=g2

Solution:

  1. (a) Let v be the velocity of the wire (as well as block) at any instant of time t.

Motional emf, e=BvL

Motional current, i=er=BvLR

and magnetic force on the wire

Fm=iLB=vB2L2R

Net force in the system at this moment will be

Fnet =mgFm=mgvB2L2R or ma=mgvB2L2Ra=gvB2L2mR

Velocity will acquire its terminal value i.e. v=vT when Fnet  or acceleration ( a ) of the particle becomes zero.

 Thus, 0=gvTB2L2mR or vT=mgRB2L2 When v=vT2=mgR2B2L2

(b) When

Then from Eq. (i), acceleration of the block,

a=gmgR2B2L2B2L2mR=gg2

or

a=g2



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