A wire of 60 cm length and mass 16 gm is suspended by a pair of flexible leads in a magnetic field of induction 0.40 T. What are the magnitude and direction of the current required to remove the tension in the supporting leads?
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We know that the magnetic force on a wire is given by `F=ilB sin theta` Here B is at right angle to I and hence `theta=90^@`. So, `F=ilB` Setting `F=mg`, the weight of the wire, the magnitude of the current required to remove the tension in the supporting lead is `i=(mg)/(lB)=((1.0xx10^-2kg)(10ms^-2))/((0.6m)(0.4Wbm^-2))=0.41A` Since the magnetic force should act in upward direction so the direction of current should be from left no right.
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