Find the torque on `M_(1)` due to `M_(2)`

Explanation

Due to `M_(2)`, magnetic fields at `S` and `N` of `M_(1)` are `B_(1)` and `B_(2)` respectively. The forces on `-m` and `+m` are `F_(1)` and `F_(2)` as shown in the figure.The torque (about the centre of the dipole `m_(1)`) will be
`F_(1)a+F_(2)a=(F_(1)+F_(2))a`
`[(mu_(0)/(4pi))M_(2)/((r-a))m+(mu_(0)/(4pi))M_(2)/((r+a))m]a`
`(mu_(0)/(4pi))M_(2)m(1/r^(3)+1/r^(3))a :' alt lt r`
`(mu_(0)M_(2)m)/(4pi)2/r^(3)a=(mu_(0)M_(1)M_(2))/(4pi^(3))`