0
$\begingroup$

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}

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.