1.

A straight horizontal conducting rod of length `0*45m` and mass `60g` is suspended by two vertical wires at its end. A current of `5*0A` is set up in the rod through the wires. (a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? (b) What will be the total tension in the wires if the direction of current is reversed, keeping the magnetic field same as before. (Ignore the mass of the wire) `g=9*8ms^-2`.

Answer» Here, `l=0*45m`, `m=60g=60xx10^-3kg`, `I=5*0A`.
(a) Tension in the wire is zero if force on the wire carrying current due to magnetic field is equal and opposite to the weight of wire i.e. `BIl=mg` or `B=mg//Il=(60xx10^-3)xx9*8//(5*0xx0*45)=0*26T`
The force due to magnetic field will be upwards if the direction of field is horizontal and normal to the conductor.
(b) When direction of current is reversed, Bil and mg will act vertically downwards, the effective tension in the wires, `T=BIl+mg=0*26xx5*0xx0*45+(60xx10^-3)xx9*8=1*176N`


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