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The vector potential for a magnetic dipole is given by the following formula:

$$ \vec{A} = \frac{1}{4\pi\epsilon_0 c^2}\frac{\vec{m}\times\hat{e}}{R^2} $$

From this expression we can deduce that for a dipole $\vec{m} = m\hat{k}$ the vector potential along the z axis would be zero, due to the vector product. Then we can deduce that the magnetic field of the dipole would be zero along the z axis.

But if we start from the magnetic field formula:

$$ \vec{B} = \frac{\mu}{4\pi}\left(\frac{3\vec{r}(\vec{m}\cdot\vec{r})}{r^5}-\frac{\vec{m}}{r^3}\right) $$

It is clear that there is a non-zero component along the z-axis. What am I doing wrong with my reasoning. Is this somewhat related to the Gauge freedom of the vector potential?

Georgios Filios
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2 Answers2

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the vector potential along the z axis would be zero, due to the vector product. Then we can deduce that the magnetic field of the dipole would be zero along the z axis.

No, you cannot deduce that.

For calculating the magnetic field $\vec{B}(\vec{r})$ at position $\vec{r}$ you not only need the vector potential $\vec{A}(\vec{r})$ at that position. Because of the differential operator in $\vec{B}(\vec{r})=\vec{\nabla}\times\vec{A}(\vec{r})$ you also need the vector potential in the immediate neighborhood around this position $\vec{r}$. So it may well be $\vec{B}(\vec{r})\ne\vec{0}$ even when $\vec{A}(\vec{r})=\vec{0}$ at that position, just because $\vec{A}$ is non-zero in the neighborhood around it.

Thomas Fritsch
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"Then we can deduce that the magnetic field of the dipole would be zero along the z axis". That's not correct because $\vec{B}=\vec{\nabla}\times\vec{A}$

$\vec{\nabla}\times\vec{A}=\begin{bmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_{x} & A_{y} & 0 \end{bmatrix}$

Clearly, there is a $z$ component.

Alberto Navarro
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