2
$\begingroup$

I was wondering what happens when i put a wire in which current flows parallel to a magnet, something like this: image where the straight line is the wire, the circle is a cylindrical magnet(with outgoing magnetic field) and the red line is a line force for the magnetic field of the wire. In particular i was wondering if the force on the magnet is directly proportional to the magnetic field produced by the wire. Thanks.

$\endgroup$

2 Answers 2

1
$\begingroup$

In particular i was wondering if the force on the magnet is directly proportional to the magnetic field produced by the wire.

If the magnet is far away from the wire, the force on the magnet can be expressed as

$$ \mathbf F = \mathbf m \cdot \nabla \mathbf B_w, $$

where $\mathbf m$ is magnetic moment of the magnet and $\nabla\mathbf B_w$ is a gradient of magnetic field of the wire at the position of the magnet (a tensor).

However, if the magnet gets close enough to the wire, the above formula breaks down - there is no single relevant value of the tensor that could be easily found and put into the above formula to obtain the right value of the force. Instead, similar formula can be integrated over the region containing the magnet:

$$ \mathbf F = \int_{magnet} \mathbf M(\mathbf x) \cdot \nabla \mathbf B_w(\mathbf x) d^3\mathbf x $$

where $\mathbf M$ gives magnetic moment of the magnet per unit volume.

Still, the magnetic field $\mathbf B_w$ at all positions is proportional to the current in the wire. From this and the last equation, it follows that the force $\mathbf F$ on the magnet is proportional to the current in the wire.

$\endgroup$
2
  • $\begingroup$ Well, but then, if I assume $\mathbf B_w = \frac{\mathbf k}{\sqrt{\mathbf x^2 + \mathbf y^2}}$ and $\mathbf B \parallel m$ then I get $\mathbf F = - \mathbf m \mathbf k \left(\frac{\mathbf x}{(\mathbf x^2 + \mathbf y^2)^{3/2}; \frac{\mathbf y}{(\mathbf x^2 + \mathbf y^2)^{3/2}; 0\right)$ and if I assume that $\mathbf y \to 0$ then $\mathbf r \to \mathbf x$ and $\mathbf F \sim \frac{1}{\mathbf r^2}$ so at the end $\mathbf F \sim \mathbf B^2$? $\endgroup$
    – user417679
    Commented Jun 22, 2014 at 8:16
  • $\begingroup$ Can't edit the previous comment, but the missing part is that $\mathbf F = - \mathbf m \mathbf k \left(\frac{\mathbf x}{(\mathbf x^2 + \mathbf y^2)^{3/2}}; \frac{\mathbf y}{(\mathbf x^2 + \mathbf y^2)^{3/2}}; 0\right)$ $\endgroup$
    – user417679
    Commented Jun 22, 2014 at 8:26
1
$\begingroup$

Yes the force on the magnet or the wire is directly proportional to the the strength of the magnet and amount of current in the wire. F=BLI force on a wire= magnetic field strength x length of wire x current in wire

$\endgroup$
3
  • $\begingroup$ Could you recommend a source? Or is there any simple way to deduce it? $\endgroup$
    – user417679
    Commented Jun 21, 2014 at 14:53
  • $\begingroup$ f=BLI @user51851 force on a wire = magnetic field strength x length x magnitude of current $\endgroup$
    – Drew
    Commented Jun 21, 2014 at 15:18
  • $\begingroup$ Maybe I have not explained it well but I want to know the force on the magnet not on the wire. $\endgroup$
    – user417679
    Commented Jun 21, 2014 at 15:44

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.