Force of interaction between magnets

[Meir Achuz]

The flat end of a bar magnet acts like a uniformly charge sheet with surface charge $\sigma_m=M$,in the same way that the end of an electrically polarized rod has a surface charge $\sigma=P$. Then, just as
$E=2\pi\sigma$ just outside a charged surface, $H=2\pi\sigma_m=2\pi M$, and B=H outside the magnet.
If this doesn't help, you may have to read a book. Also, read my post #29 carefully

 

[akashverma]

The flat end of a bar magnet acts like a uniformly charge sheet with surface charge $\sigma_m=M$,in the same way that the end of an electrically polarized rod has a surface charge $\sigma=P$. Then, just as
$E=2\pi\sigma$ just outside a charged surface, $H=2\pi\sigma_m=2\pi M$, and B=H outside the magnet.
If this doesn't help, you may have to read a book. Also, read my post #29 carefully

Thanks for the feedback. But in the formation of your expression you have taken $B=2\pi(M+M')=4\pi M$ i.e $M=M'$ but if we go for derivation of the force expression by using $B=2\pi(M+M')$ for two different M and M' we get ##F=1/2{π(M^2+M'^2+2MM')A}.
Also please refer me a good literature for this.

 

[Meir Achuz]

The correct force is in my post #8. The force can be given in terms of B ONLY IF M'=M.
https://www.amazon.com/dp/0805387331/?tag=pfamazon01-20
discusses this on p. 237 in Section 8.9.