Question

Two short bar magnets of magnetic moments $0.125A{m}^2$ and $0.512A{m}^2$ are placed with their like poles facing each other. If the distance between the centres of the magnet is $0.26m$. The distance of neutral point from the weaker magnet is

Answer 1

$B_{axial}=\frac{{\mu }_{0}2M}{4\pi r^3}$
Equating the magnetic fields due to the two bar magnets, $\frac{M_1}{x^3}=\frac{M_2}{{(d-x)}^3}$
Solving the above for x (distance from the center of weaker magnet to the neutral point), gives the formula in the hint. Putting in the values, we get, $x=0.1m$