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Asked by Prashant DIGHE 6th December 2019, 10:29 PM
if the conductor has jumped to a height h, then its initial velocity v from its resting position is given by

v = (2gh)1/2 ....................(1)

impulse due to force acting on conductor = F dt = d(mv) = (mv-0)= mv  ...............(2)

force F acting on conductor in magnetic field is given by,  F = B i l ................(3)

where B is magnetic field flux density, i is current passing through the conductor and l is length of conductor

impulse = F dt = ( B i l ) dt = B l q ......................(4)

where q = ( i dt )  is the charge passed to the conductor

from eqn.(2) and (4), we get,  mv = B l q    or   q = (m v ) / ( B l ) .........................(5)

Using eqn.(1), we get, q = [ m (2gh)1/2 ] / ( B l )
Answered by Expert 7th December 2019, 12:19 PM
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