Question

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 1

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.