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`
`therefore 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, 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.