Two long straight parallel current carrying conductors each of length `l` and current `i` are placed at a distance `r_(0)`. Show that the total work done by an external agent in slowly reducing their distance of separation to `(r_(0))/(2)` is `(mu_(0))/(2pi)i^(2)ln(2)`

Two long straight parallel conductors of length l ,separated by a distance d and each carrying a current I attract each other with a force equal to

Two long straight wires, each carrying a current I in opposite directions are separated by a distance R . The magnetic induction at a point mid way between the wire is

Two long straight parallel conductors are carrying currents i_(1)andi_(2) in the same direction. Work done per unit length, when the distance between them is doubled is

A conductor AB of length l carrying current i is placed perpendicular to a long straight conductor carrying a current I as shown. Force on AB will be

A conductor AB of length l carrying current i is placed perpendicular to a long straight conductor carrying a current I as shown. Force on AB will be

Two long straight parallel conductors separated by a distance of 0.5m carry currents of 5A and 8A in the same direction. The force per unit length experienced by each other is

Two long straight parallel conductors separated by a distance of 0.5m carry currents of 5A and 8A in the same direction. The force per unit length experienced by each other is

Figure below shows two infinitely long and thin current carrying conductors X and Y kept in vacuum, parallel to each other, at a distance .a.. (i) How much force per unit length acts on the conductor y due to the current flowing through X ? Write your answer in terms of ((mu0)/(4pi)),I_(1),I_(2) and a. (Derivation of formula is not required.), (ii) Define ampere, in terms of force between two current carrying conductors.