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Home / Doubts and Solutions/JEE/Physics

Ques 7

It is multiple answer ques with more than one options correct

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Asked by rbhatt16 23rd August 2018, 7:55 AM
Answered by Expert
Answer:
Figure shows the various forces acting at the contact between vertical rails and the hanging rod. 
Force F due to the effect of magentic field on current carrying rod is F = B×i×L , where B is magnetic field,
i is the current passing through the rod and L is length of rod. This force acting in the direction so that
this force presses the rod against the rail. Hence we get equal Normal reaction force N and friction force is 
proportional to this normal reaction force N.
 
Let us calculate the friction force acting on it at t=0.
 
current = 100/5 = 20A ; Force F = B×i×L = 2×20×1 = 40 Newton
Normal reaction force = 40 Newton;  friction force = μN = (3/4)×40 = 30 N.
 
The rod will start fall, when friction force equals to the weight of the rod.
 
μN = mg or N = mg/μ = (2×10)/(3/4) = 80/3 Newton
 
current to get the normal reaction force (80/3):-  (80/3) = B×i×L = 2×i×1  hence current i = (40/3)A 
 
Hence resistance in the circuit = 100 / (40/3) = 7.5 Ω
 since initial resistance is 5Ω and rate of increasing the resistance is 0.5Ω/s , after 5 s we get 7.5Ω and the rod will start fall
Answered by Expert 23rd August 2018, 4:45 PM
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