# How is possible that the magnetic field around electron is dipole but it's one circle

1. Jul 14, 2007

### stmartin

How is possible that the magnetic field around electron is dipole but it's one circle?

Best regards.

p.s thanks for the help.

2. Jul 14, 2007

### Staff: Mentor

I'm sorry, I don't understand what you're asking. Perhaps more words would help, or maybe even a picture?

3. Jul 14, 2007

### olgranpappy

indeed, more words, please. "It's one circle" doesn't mean much to me. What is one circle? The electron? The electron is one circle? What does that even mean.

An electron is not a circle, it's a point-like particle. Also, the electron possesses an intrinsic spin of magnitude $$\hbar/2$$and this spin can be thought of in some sense as a magnetic dipole with magnitude $$g \mu_B /2$$. Where $$\hbar$$ is the reduced Planck Constant, $$\mu_B$$ is the Bohr magneton, and $$g$$ is a pure number approximately equal to 2 (it's actually a little bigger, closer to 2.002).

4. Aug 11, 2007

### Hans de Vries

The magnetic field around the electron associated with its spin is indeed
called a dipole moment. So your question is: How can a circular current
generate a dipole field.

The field of a rotating electric charge happens to be equal to the field of
a magnetic dipole. (two adjacent opposite magnetic charges: two monopoles)

This can be shown using Maxwell's laws:

$$\textsf{\textbf{E}}\ = -\textsf{grad}\ V - \frac{\partial \textbf{A}}{\partial t},\ \qquad \ \ \textsf{\textbf{B}}\ = \textsf{curl}\ \textbf{A}$$

Monopole field

The electric point charge (monopole) is a delta function.

$$Q\ =\ \delta(r)$$

It's potential field V is:

$$V\ =\ \frac{1}{r}$$

It's electric field is is found by differentiating in x,y and z (gradient):

$$\textsf{E}_x\ = \frac{x}{r^3}, \quad \textsf{E}_y\ = \frac{y}{r^3}, \quad \textsf{E}_z\ = \frac{z}{r^3}$$

Dipole field

If we differentiate a delta function we get two adjacent and opposite
delta functions (a dipole)

$$Q\ =\ \frac{\partial}{\partial z}\ \delta(r)$$

To obtain the dipole field we differentiate the monopole field in z as well:

$$V\ =\ \frac{\partial}{\partial z}\left(\frac{1}{r}\right)\ =\ \frac{z}{r^3}$$

Which gives the electric potentials after differentiation in x, y and z:

$$\textsf{E}_x\ = \frac{3z}{r^5}x, \qquad \textsf{E}_y\ = \ \frac{3z}{r^5}y, \qquad \textsf{E}_z\ = \ \frac{3z}{r^5}z - \frac{1}{r^3}$$

Rotating point charge field

This current gives rise to a magnetic vector potential A. The currents
in the x direction are given by a delta function differentiated in the y
direction, while the current in the y direction is give by a differentiation
in the x direction: Two orthogonal skews yield a rotation, like:

$$\partial_x - \partial_y \qquad = \qquad \uparrow\downarrow\ +\ \rightleftarrows \qquad = \qquad \circlearrowright$$

So we have for the electric currents Jx, Jy and Jz:

$$J_x \ \ =\ -\frac{\partial}{\partial y}\ \delta(r), \qquad J_y \ \ =\ \frac{\partial}{\partial x}\ \delta(r), \qquad J_z\ =\ 0$$

We obtain for the Vector Potential fields Ax, Ay and Az after
differentiating 1/r in the same way:

$$\qquad A_x\ =\ - \frac{y}{r^3}, \qquad A_y\ =\ \frac{x}{r^3}, \qquad A_z\ =\ 0$$

To obtain the magnetic fields B we take the curl of the vector
potential A and we get:

$$\textsf{B}_x\ = \frac{3z}{r^5}x, \qquad \textsf{B}_y\ = \ \frac{3z}{r^5}y, \qquad \textsf{B}_z\ = \ \frac{3z}{r^5}z - \frac{1}{r^3}$$

Which is mathematically equal to the electric dipole field.

Regards, Hans

Last edited: Aug 11, 2007