Find the force of interaction between two current carrying coils, the distance between the centres of which is much greater than their linear dimensions (Fig. 28.16).

Answer

`F=-(6mu_(0)p_(m)^2)/(4pir^4)`

Explanation

It is easily seen that the moving coils hanging freely will arrange themselves so that their planes will be perpendicular to their common axis. The force of interaction is `F=p_m (d B)/(dx)` Noting that
`(dB)/(dx)=(6mu_(0) p_(m)x)/(4pi (a^2+x^2)^(5//2))`
(see Problem 27.11) and that according to the condition of the problem `x approx r ge ge a,` we have `(dB)/(dx)=(6mu_(0)p_(m))/(4pir^4)`
The force of interaction is `F=p_(m) (d B)/(dx)=-(6mu_(0)p_(m)^2)/(4pir^4)`
It is quite similar to the force of interaction between electric dipoles.