Calculations for the Magnetic Field Strength of Neodymium Magnets

Consider a magnet with a length $L$ along the y-axis, a width $w$ along the z-axis, a depth $z$ along the x-axis. Let this magnet be defined by $x\in [-\frac{d}{2},\frac{d}{2}]$, $y\in [0,L]$, and $z\in [0,w]$. To calculate the Magnetic Field Strength $B$ of a vector $\text{r}$ from the magnet to a point, the formula is $B=-{\mu }_0\nabla {\psi }_m$, where:

$${\psi }_m(\text{r})=\frac{1}{4\pi }{\oint }_S\frac{\text{M}({\text{r}}^{\prime })\cdot \hat{\text{n}}}{|{\text{r}}^{\prime }-\text{r}|}d{\text{A}}^{\prime }$$

and where $\text{M}$ is the magnetization. Suppose the point in question is above the top face of the magnet. Then, we have (as derived in Guide to modelling permanent magnets):

$$\text{M}=(0,M_y,0)=(0,\frac{B_r}{{\mu }_0},0)$$

Thus ${\psi }_m$ reduces to:

$${\psi }_m(\text{r})=\frac{1}{4\pi }\oint \frac{\text{M}({\text{r}}^{\prime })\cdot \hat{\text{n}}}{|{\text{r}}^{\prime }-\text{r}|}d{\text{A}}^{\prime }=\frac{B_r}{4\pi {\mu }_0}\oint \frac{d{\text{A}}^{\prime }}{|{\text{r}}^{\prime }-\text{r}|}$$

To translate the vector using cartesian coordinates:

$${\psi }_m(\text{r})=\frac{B_r}{4\pi {\mu }_0}\oint \frac{d{\text{A}}^{\prime }}{|{\text{r}}^{\prime }-\text{r}|}=>{\psi }_m(x,y,z)=\frac{B_r}{4\pi {\mu }_0}\oint \frac{d{\text{A}}^{\prime }}{\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2+(z-z^{\prime }{)}^2}}$$

Let the magnet be thin such that $d\to 0$. Then:

$${\psi }_m(x,y,z)=\frac{B_r}{4\pi {\mu }_0}\oint \frac{d{\text{A}}^{\prime }}{\sqrt{x^2+(y-y^{\prime }{)}^2+(z-z^{\prime }{)}^2}}$$

Because the point is above a single face of the magnet, the integral will only consider the face the point is above, so:

$${\psi }_m(x,y,z)=\frac{B_r}{4\pi {\mu }_0}{\int }_0^L{\int }_0^w\frac{d{z}^{\prime }d{y}^{\prime }}{\sqrt{x^2+(y-y^{\prime }{)}^2+(z-z^{\prime }{)}^2}}$$

After many calculations, the equation reduces to:

$${\psi }_m(x)=-\frac{B_r}{4\pi {\mu }_0}\frac{wL}{x}$$

Therefore:

$$B=-{\mu }_0\nabla {\psi }_m={\mu }_0(-\frac{d}{dx}(-\frac{B_r}{4\pi {\mu }_0}\frac{wL}{x}))={\mu }_0\frac{B_r}{4\pi {\mu }_0}\frac{d}{dx}\frac{wL}{x}=\frac{B_r}{4\pi }\frac{d}{dx}\frac{wL}{x}=-\frac{B_{r}wL}{4\pi x^2}$$

Apply this formula to the neodymium magnets being used for the project, $w=10\cdot 10^{-3}\text{m}=10^{-2}\text{m}=.01\text{m}$ and $L=60\cdot 10^{-3}\text{m}=6\cdot 10^{-2}\text{m}=.06\text{m}$. Therefore:

$$B=-\frac{B_{r}wL}{4\pi x^2}=-\frac{B_r\cdot .06\cdot .01}{4\pi x^2}=-\frac{B_r\cdot .0006}{4\pi x^2}=-.6\cdot 10^{-3}\frac{B_r}{4\pi x^2}$$

The calculate resonance, the magnets are advertised to have a given pull force of 33lbs. The force of a magnet is calculated by the following formula:

$$F=\frac{B_r^{2}A}{2{\mu }_{0}g}$$

Therefore, by converting this, it becomes:

$$\sqrt{\frac{2{\mu }_{0}gF}{A}}=B_r$$

Substituting $F=mg$ and $A=wd$, where $d=5\cdot 10^{-3}=.005\text{m}$, so:

$$\sqrt{\frac{2{\mu }_{0}m{g}^2}{wd}}=B_r$$

Therefore:

$$B=\frac{-.6\cdot 10^{-3}}{4\pi x^2}\sqrt{\frac{2{\mu }_{0}m{g}^2}{wd}}$$

Substituting for the given properties of the magnet:

$$B=\frac{-.6\cdot 10^{-3}}{4\pi x^2}\sqrt{\frac{2\cdot 4\pi \cdot 10^{-7}\cdot 15(9.81{)}^2}{.01\cdot .005}}=\frac{-0.407\cdot 10^{-2}}{x^2}$$