Physics, asked by thomasivf, 26 days ago

A bar magnet with poles 25 cm apart and pole strength 14.4 A-m is resting on a frictionless pivot. It is held in equilibrium at 30° in a uniform magnetic field of
0.25 T by applying a force F always at right angles to its axis at a point 2.5 cm from one end, then Fis

Answers

Answered by babalraj1976
0

Answer:

m rests with its centre on a frictionless pivot. It is held in equilibrium at 60^@ to a uniform magnetic field of induction 0*25T by applying a force F at right angles to its axis, 10cm from the pivot. B=0⋅25T,r=10cm=0⋅1m, F=?

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