- #1

- 1

- 0

~Robert

- Thread starter rwhitman
- Start date

- #1

- 1

- 0

~Robert

- #2

Hans de Vries

Science Advisor

Gold Member

- 1,089

- 23

Make sure you use the gradient of the inproduct of the magnetic moment and the

magnetic field rather than the gradient of the magnetic field itself.

[tex]\vec{F}_{magn}\ =\ \textsf{grad}\left( \vec{\mu}_e\cdot

\textsf{B} \right)[/tex]

What you could do is assuming that every little volume of magnetic material is a

point dipole field like this:

[tex]\textsf{B}~ = ~ \ \frac{\mu_o\mu_e}{4\pi r^3}\ \ \left(\ 3\ \frac{xz}{r^2}, \quad 3\ \frac{yz}{r^2}, \quad 3\ \frac{zz}{r^2}-1\ \right)[/tex]

Assuming that they all point in the same direction you can derive the total

magnetic field by integration (analytic or numerical)

Regards, Hans.

magnetic field rather than the gradient of the magnetic field itself.

[tex]\vec{F}_{magn}\ =\ \textsf{grad}\left( \vec{\mu}_e\cdot

\textsf{B} \right)[/tex]

What you could do is assuming that every little volume of magnetic material is a

point dipole field like this:

[tex]\textsf{B}~ = ~ \ \frac{\mu_o\mu_e}{4\pi r^3}\ \ \left(\ 3\ \frac{xz}{r^2}, \quad 3\ \frac{yz}{r^2}, \quad 3\ \frac{zz}{r^2}-1\ \right)[/tex]

Assuming that they all point in the same direction you can derive the total

magnetic field by integration (analytic or numerical)

Regards, Hans.

Last edited:

- Last Post

- Replies
- 6

- Views
- 6K

- Last Post

- Replies
- 10

- Views
- 386

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 4K

- Last Post

- Replies
- 4

- Views
- 688

- Last Post

- Replies
- 4

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 642

- Last Post

- Replies
- 7

- Views
- 4K

- Last Post

- Replies
- 5

- Views
- 819

- Last Post

- Replies
- 2

- Views
- 3K