I have two magnets A and B:

  • Magnet A is movable, Magnet B is fixed.
  • Magnet A is smaller and weaker (surface area of 1 Square cm) , where as magnet B is stronger and bigger (surface area of 4 Square cm).

If I have a magnet (A) moving at a high velocity perpendicular to the magnetic field of the other magnet (B):

  1. Would the Magnet A be attracted to Magnet B (both magnets have opposite poles facing each other - Magnet A has its south facing Magnet B north) ?
  2. Would the Magnet A just move over the fixed Magnet B without being attracted because of its high velocity ?

Assuming the distance between the two magnets is small (example 5 mm). The velocity of the movable Magnet A is 60 m/s.

  • \$\begingroup\$ There should be an attraction,but it may be a small one. \$\endgroup\$ – On the way to success Jul 20 '15 at 1:54

Yes, North and South poles will be attracted to each other. More when they're close together and aligned, than when they're far apart. As is normal for magnets.

The velocity of the magnet has no effect on the force. The force is the same at any velocity. *

The trajectory of the moving magnet will be affected less if it's moving faster, as the force acts for less time.

  • there could be eddy current effects if the magnets are electrically conductive. These would cause drag which increases with velocity.
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  • \$\begingroup\$ Thanks for the reply Tom. Would you know of a formula which can give me the change in trajectory (in millimeter) based on the magnetic field strength. I use the formula to calculate the force of the magnet (electromagnet) based on the distance: daycounter.com/Calculators/Magnets/… If my input parameters are: Current: 200 mA Turns: 1000 Area in sq cn: 1 Distance in mm: 5 The output is 0.101 N (or) 0.023 LBS Based on this how do i calculate the deflection of the trajectory? \$\endgroup\$ – codePriest Jul 21 '15 at 3:46
  • \$\begingroup\$ For a first approximation you could use ForcedeltaTime = MassdeltaVelocity. You know force and time and mass, so you can calculate the sideways velocity. Add this to the original velocity to find the trajectory. \$\endgroup\$ – tomnexus Jul 21 '15 at 10:43

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