Magnetic Field of a Bar Magnet

[Zeophlite]

So I'm making a simulation to help me understand electromagnetics.
Basically I have a stationary bar magnet, along with a moving electron.
Each frame the electron has a force applied to it via the Lorentz force law, F = q(E+v cross B).
Now E = 0, but how do I calculate B?

The Biot-Savart law relates to currents, but a bar magnet doesn't have any current.

Cheers

 

[Gerenuk]

A bar magnet is like many tiny current loops. However the treatment is actually easier if you use the magnetic scalar potential (which works as long as the electron doesn't go through the current loop).

For a bar magnet with a constant magnetization parallel to the magnet you can calculate the magnetic scalar potential by assuming that there is a constant magnetic density m spread over both the pole surfaces. Each magnetic density bit contributes
$$\mathrm{d}V_m=\frac{m\mathrm{d}A}{4\pi r}$$
(where dA is a surface element) to the scalar magnetic potential. The magnetic field from this potential - once you have summed over both pole surfaces - is
$$\vec{H}=\frac{\vec{B}}{\mu_0\mu_r}=-\nabla V_m$$

 

[astrokreng]

Hello, it is great to hear that you are using simulation to understand electromagnetics. The magnetic field of a bar magnet can be calculated using the Biot-Savart law, which states that the magnetic field at a point is directly proportional to the current flowing through a wire and inversely proportional to the distance from the wire. In the case of a bar magnet, the current is essentially a collection of tiny moving charges within the magnet, which generates a magnetic field around it. This field can be calculated by considering the magnetic moments of these charges and using the Biot-Savart law. Alternatively, you can also use the magnetic dipole moment of the bar magnet to calculate the magnetic field. I would recommend researching more about magnetic dipole moments and their relationship to the magnetic field to better understand how to calculate the magnetic field of a bar magnet. Good luck with your simulation!