The magnet has magnetization $\vec{M}=M_0 \hat{k}$, radius $r=b$, and length $L=2l$, we need to find the potential $A$ along its $z$ axis and then the magnetic field $\vec{B}$. The equation where one can start is

\begin{equation} \vec{A}(\vec{r})=\frac{\mu _0}{4 \pi} \left[ \int \frac{\vec{J}_b}{|\vec{r}-\vec{r}´|}dV + \int\frac{\vec{K}_b}{|\vec{r}-\vec{r}´|}dS \right] \end{equation} where $\vec{J}_b=\vec{\nabla}\times \vec{M}=0$ and $\vec{K}_b=\vec{M} \times \hat{n}=M_0 \hat{\phi}$.

Now we only have to solve the second term and it is quite easy with

$$ \vec{r}´=b\hat{r}'+z'\hat{k}'$$ $$ \vec{r}=z\hat{k}$$ $$ dS'=bd\phi' dz'$$

so the potential takes the form

\begin{equation} \vec{A}(\vec{r})=\frac{\mu_0 }{4 \pi} \int\frac{M_0bd\phi' dz'}{((z-z')^2+b^2)^{1/2}}dS\quad\hat{\phi} \end{equation} which gives

\begin{equation} \vec{A}(z)=\frac{\mu_0 M_0 }{2b} \left[ \frac{(z-l)}{((z-l)^2+b^2)^{1/2}}- \frac{(z+l)}{((z+l)^2+b^2)^{1/2}} \right] \quad\hat{\phi} \end{equation}

so the magnetic field we just apply the roational in cylindricals $\vec{B}=\vec{\nabla} \times \vec{A}$

$$\vec{\nabla} \times \vec{A}=-\frac{1}{r}\frac{\partial A_{\phi}}{\partial z}\hat{r}+\frac{A_{\phi}}{r}\hat{k}$$

Here is the problem, I've got a component in $\hat{r}$ where it should not be there, and if we only use the component in $\hat{k}$ there is an extra factor of $\frac{1}{rb}$ which shouldn't be there as well and I don't know why.

The magnetic field is suppose to be

\begin{equation} \vec{B}(z)=\frac{\mu_0 M_0 }{2} \left[ \frac{(z-l)}{((z-l)^2+b^2)^{1/2}}- \frac{(z+l)}{((z+l)^2+b^2)^{1/2}} \right] \quad\hat{k} \end{equation}


1 Answer 1


Since the observation point is on the z-axis ($\vec{r}=z\hat{k}$), the system is axisymmetric, and the excitation condition is oriented in the $\hat{\phi}$ direction, $\vec{A}=\vec{0}$ is obtained. In your calculation, \begin{equation} \int_0^{2\pi}\vec{K}_bd\phi=\int_0^{2\pi}M_0\hat{\phi}d\phi=\int_0^{2\pi}M_0(-\text{sin}\phi,\text{cos}\phi)d\phi=0. \end{equation} should be considered. Of course this does not imply $\vec{B}=\vec{0}$ on the z-axis.


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