Pull force of magnet calculated at an angle

[Mozez]

Hi PF,

A disclaimer, I am no good at Physics or math.

My question:

When calculating the pull force between two magnets it seems to always be calculated at an angle perpendicular to the surface of the magnet.

I was wondering what effect on the force required to "pull" one magnet away from the other would be when a force is applied at a different angle and can this be calculated easier when you already know the pull force at the perpendicular?

I guess I am looking for a way of recalculating the force needed to 'pull' a magnet away from another when the angle of the applied force changes.

Thank you for your time

 

[superg33k]

Force_at_angle = cos (angle_from_perpendicular) * Force_at_peroendicular

 

[Mozez]

Thank you for replying.

Wouldn't this imply that at 90 degrees to the perpendicular that there is no force required to pull the magnet seeing as cos(90) = 0?

Unless you mean angle_from_perpendicular in radians? But then the resulting force needed is only minutely different. Maybe this is the case? That the force difference between shearing a magnet away from another or pulling perpendicular are roughly the same? It doesn't seem to be so though.

I obviously still need a bit of guidance on this.

Thanks

 

[superg33k]

Good point. This is because at 90 degree's the magnetic force is from a different part of the magnet.

Magnetic forces are caused by moving charges (generally electrons) in the plane perpindicular to their travel. A moving charge does not create a magnetic force perpendicular to this, i.e. in its direction of travel.

Now a magnet is like a closed electric circuit, with electrons moving round the outside. If you put the north of the magnet facing upwards the electrons are moving round it in the clockwise direction.

So let's go for a cuboid magnet for my simplicity. I'm going to ignore the N and S end bits. But let's say its a cuboid with faces f1,f2,f3,f4 and areas a1,a2,a3,a4 and such that a1=a3 and a2=a4, i.e. these are opposite faces. Since its a cuboid all these faces will have the same height but different width.

If we know the force when perpindicular to one face (say f1, call this force F) and we want to know it when we move it around angle theta from the perpindicular. It would be:

F*cos(theta)+(a2/a1)*F*sin(theta)

Just to let you know this is just me thinking aloud, I'm always doing magnetic forces from currents or electrons, not bar magnets. A better resource might be:

http://en.wikipedia.org/wiki/Force_between_magnets#Calculating_the_magnetic_force

 

[sardar]

and expertise!

Hi there,

Thank you for your question. The pull force between two magnets is typically calculated using the equation F = (μ0/4π)(m1m2/r^2), where μ0 is the permeability of free space, m1 and m2 are the magnetic moments of the two magnets, and r is the distance between them. This equation assumes that the force is being applied perpendicular to the surface of the magnets, as you mentioned.

If the force is applied at an angle, the equation becomes a bit more complicated. The force can be broken down into components, one parallel to the surface and one perpendicular. The parallel component will have a smaller effect on the overall force, while the perpendicular component will still contribute to the overall force. The exact calculation will depend on the angle and the specific geometry of the magnets.

To make this calculation easier, you can use trigonometry to determine the components of the force at different angles. You can also use the law of cosines to find the total force at a specific angle. However, if you already know the pull force at the perpendicular angle, it may be easier to simply use that value as an estimate for the force at a different angle.

I hope this helps answer your question. If you have any further questions, feel free to ask. Best of luck with your research!

Sincerely,