Find the magnetic force on a short magnet of magnetic dipole moment `M_(2)` due to another short magnet of magnetic dipole moment `M_(1)`.

Explanation

To find the magnetic force we will use the formula of `B` due to a magnet.We will also assume `m` and `-m` as pole strengths of `N` and `S` of `M_(2)`.Also length of `M_(2)` as `2a.B_(1)` and `B_(2)` are the strengths of the magnetic field due to `M_(1)` at `+m` and `-m` respectively.They experience magnetic forces `F_(1)` and `F_(2)` as shown.
`F_(1)=2(mu_(0)/(4pi))M_(1)/(r-a)^(3)m`
and `F_(2)=2(mu_(0)/(4pi))M_(1)/(r-a)^(3)m`
`:. F_(res=F_(1)-F_(2)=2((mu_(0))/(4pi))M_(1)m[(1/(r-a)^(3))-(1/(r+a)^(3))]`
`=2((mu_(0))/(4pi))(M_(1)m)/r^(3)[(1-a/r)^(-3)-(1+a/r)^(-3)]`
By using acceleration, Binomial expansion, and neglecting terms of high power we get
`F_(res)=2((mu_(0))/(4pi))(M_(1)m)/r^(3)[1+(3a)/r-1+(3a)/r]`
`=2((mu_(0))/(4pi))(M_(1)m)/r^(3)(6a)/r`
`=2((mu_(0))/(4pi))(M_(1)3M_(2))/r^(4)`
`=6((mu_(0))/(4pi))(M_(1)M_(2))/r^(4)`
Direction of `F_(res)` is towards right.
Alternative Method
`=B((mu_(0))/(4pi))(2M_(1))/r^(3) rArr (dB)/(dr)=-(mu_(0))/(4pi)xx(6M_(1))/r^(4)`
`F=-M_(2)xx(dB)/(dr) rArr F=((mu_(0))/(4pi))(6M_(1)M_(2))/r^(4)`