A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is `mu`. If the wire carries a current I, what minimum magnetic field should exist in the space in order to slide the wire on the rails.

Explanation

(a)
The maximum friction that can act on wire
`f_(max) =muN = mu mg`
The magnetic force on wire
`F = BiL sin theta`
To move the wire
`F ge f_(max)`
`BiL sin theta ge mu mg`
`B ge (mu mg)/(iL sin theta)`
`B` will be minimum if `theta = 90^(@)`
`B ge (mu mg)/(iL)`
`B_(min) = (mu mg)/(iL)`, inside the plane of paper
(b)
Magnetic force on wire
`F = BiL` in downward direction

`uarr : N_(1) = mg + Bil, m`: mass of wire
`f_(max) = mu N_(1) = mu(mg + Bil)`
To move the wire, `F_(1) ge f_(max)`
`F_(1) = mu(mg + Bil)` ...(i)

Magnetic force `F = Bil`, in upward direction

`uarr: N_(2) + Bil = mg`
`N_(2) = mg - Bil`
`f'_(max) = muN_(2) = mu(mg - Bil)`
`F_(2) = mu(mg - Bil)`
`(i) - (ii)`
`F_(1) = F_(2) = mu xx 2Bil`
`mu = (F_(1) - F_(2))/(2Bil)`