A straight horizontal conducting rod of length of 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A 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 wire 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 wires) `g = 9.8 m s^(-2)`

Explanation

(a) Magnetic forece `F = B I l`. When this force is balanced by the downward wight, the net tension in the wires will be zero.
`B I l = mg`
`implies B = (m g)/(I l) = (0.06 xx 9.8)/(5.0 xx 0.45) = 0.26 T`
Thus, a horizontal magnetic field `vecB` of 0.26 T normal to the conductor be appled in such a direction that Fleming.s left hand rule gives a magentic force upwards.
(b) On reversing current the tension in the wires will be
`= B I l + m g = mg + mg = 2 mg = 2 xx 0.06 xx 9.8 N = 1.18 N`.