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?

A wire of 60 cm length and mass 16 gm is suspended by a pair of flexib

1.	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?
Answer» 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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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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