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Q.

A loop formed by three identical uniform conducting rods each of length a = 20 cm is suspended from one of its vertices (P) so that it can rotate about horizontal fixed smooth axis CD. Initially plane of loop is in vertical plane. A constant current i = 10A is flowing in the loop. Total mass of the loop is m = 60gm.  At t = 0, a uniform magnetic field of strength B directed vertically upwards is switched on. If the minimum value of B (in mT) is n, so that the plane of the loop becomes horizontal (even for an instant) during its subsequent motion. Then value of n/100 is  (Assume current in the loop is maintained constant).
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answer is 4.

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Detailed Solution

Applying energy conservation, initially
Kinetic energy = 0
Gravitational P.E. = 0 (say) & Magnetic P.E.=μB  
Where, μ = magnetic moment of the loop =i,(3a24)  
Finally when the loop becomes horizontal, Kinetic energy = 0
Gravitation P.E.=mg(a3) (because mg acts on the centre of mass) magnetic
P.E. = 0 

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