A long horizontal wire P carries a current of 50A. It is rigidly fixed. Another wire Q is placed directly above and parallel to P, as shown in figure below. The weight per unit length of the wire Q is 0.025 `Nm^(-1)` and it carries a current of 25A. Find the distance .r. of the wire Q from the wire P so that the wire Q remains at rest.

Explanation

The magnetic force per unit length on the wire Q due to current in the wire P is:
`(F)/(l)= (mu_(0))/(2pi) (i_(1)i_(2))/(r ) Nm^(-1)`

where, r is the distance of wire Q from the wire P. Since, the currents in P and Q are opposite, Q experiences an upward force and it is balanced by the downward weight of Q.
`therefore (F)/(l)= (mu_(0))/(2pi) (i_(1)i_(2))/(r )= W`
where, W is the weight per unit length of the wire Q.
`rArr r= (mu_(0))/(2pi) (i_(1)i_(2))/(W)`
`=((4pi xx 10^(-7))/(2pi)) xx (50 xx 25)/(0.025)`
`=1 xx 10^(-2)m= 1cm`