InterviewSolution
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InterviewSolution
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Two metal strip , eahc of lenght l , are clamped parallel to each other on a horizontal floor with a seperation b between them. A wire of mass m ies on them perpendicular as shown i figure. A vertically upward magnetic field of strenght B exists in the space. The metal strips are smooth but the coefficient of frication between wire and the floor is mu . A current i is established when the switch S is closed at the instant t=0.Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach ? |
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Answer» Solution :When current starts flowing in the wire, it experiences a force, F =Bibof CONSTANT MAGNITUDE due to which it accelerates on the metal strips. There after when wire falls on the floor, it retards due to friction and finallystops. Thus `a = (F)/(m) = (Bib)/(m)` The velocity gained by the wire on the strips `v^2 = 0 + 2al = 2 (Bib l)/(m)` Let x be the distance MOVED by the wire on the floor. Its final velocity becomes zero, and so `= v^2 -2a.x or x = (v^2)/(2a.)` Here ` a. = (mu mg)/(m) = mu g` `thereforex = ([(2Bib l)/(m)])/(2mug) = (Bib l)/(mu mg)` Note : TRY usiing work - energy theorem |
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