Derivation of $F = mH$

As someone else pointed out, i think you mean F=mH.

There is probably a typo in your expression: $F=mB$. It should be $F=mH$

Given $$F=frac{mu m_1m_2}{4pi r^2}tag1$$ $$mu=frac{B}{H}tag 2$$ $$H=frac{mu m}{4pi r^2} tag3$$ Now, diving (1) by (3) (eq(2) is superfluous i.e. not required), $$frac{F}{H}=frac{mu m_1m_2}{4pi r^2}cdot frac{4pi r^2}{mu m}$$

$$ frac FH=frac{m_1m_2}{m}$$ Assuming magnetic dipole moments $m_1=m_2=m$, we get $$frac FH=frac{mcdot m}{m}=m$$ $$therefore F=mH$$