Figure shows a rectangular current carrying loop placed `2 cm` away from a long, straight, current-carrying conductor. What is the direction of the net force acting on the loop?

Explanation

Force between two parallel current carrying wires :`AB` and `CD` are two infinitely long conductors placed parallel to each other and separated by distance `r`.They carry currents `I_(1)` and `I_(2)` in the same direction.Magnetic field produced by current `l_(1)` at any point of `CD` is
`B_(1)=(pi_(0)l_(1))/(2pir)`
This field acts perpendicular to `CD` and into the plane of paper.It exerts a force on wire `CD` carrying current `I_(2)` Force exerted on unit length of `CD` is
`F=B_(1)=I_(2)l=(pi_(0)I_(1))/(2pir)xxI_(2)xx1` or `F=(pi_(1)I_(1)I_(2))/(2pir)`
or `pi_(0)/(4pi).(2I_(1)I_(2))/r`
By Fleming,s left hand rule, this force acts on `CD` towards `AB`.Similarlarly consuctor `CD` also exerts an equal force `AB` towards itself.Hence the two wires get attracted towards each other. Numerical:
Force an `AB F_(1)=(pi_(0))/(4pi)(2I_(1)I_(2))/r_(1)xx"length of "AB`
`=(10^(-7)xx2xx15xx25)/(12xx10^(-2))xx25xx10^(-2)`
`=1.5625xx10^(-4)N`(repulsive away form `XY`)
Net force on the loop `F=F_(1)-F_(2)`
`=(9.375-1.5625)xx10^(-4)`
`=7.8125xx10^(-4)` (attactive towards `XY`).