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If I have two disc-shaped magnets (radius r=0.05m, width w=0.03m, Remanance Br=1.06 T) separated by a distance d, how can I calculate the magnetic flux density somewhere between them?

I found this webpage but I get a tiny value for B (10^-4 T) and I also have no idea where they get the formula from and whether it is trust-worthy.

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The given formula (in that website) is correct.

Assume we have a cylinder with permanent (axial) magnetization $\mathbf{M}$. It will cause a surface current density $\mathbf{K}$: $$\cases{\bf{K}=\bf M \times \hat n \\ \mathbf M=M_0\hat z}\to \mathbf{K}=M_0\hat \phi$$ So it is like a finite length solenoid. To find the field on it's axis ($z$, point $P$ in the below picture), from Biot–Savart law we'll arrive at:

enter image description here $$dB=dB_z=\frac{\mu_0K\mathrm{d}z}{2}\frac{R^2}{\xi^2}$$ $$\cases{dz=\frac{Rd\theta}{\sin^2\theta}\\ \frac{R}{z}=\tan \theta \\ \xi=\frac{R}{\sin \theta}}\to dB=-\frac{\mu_0 K}{2}\sin \theta d\theta$$ $$B=\int_{\theta_1}^{\theta_2}dB=\frac{\mu_0 K}{2}(\cos \theta_1-\cos \theta_2)$$ so $$\boxed {\mathbf{B}=\frac{\mu_0 M_0}{2}(\cos \theta_1-\cos \theta_2)\hat z}$$

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  • $\begingroup$ So then I assume remnance is $$\frac{\mu_0 M_0}{2}$$ and expanding the cos gives that formula. $\endgroup$ – geniass Aug 24 '13 at 22:30

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