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How to calculate force of electric current on magnet

  1. Jul 11, 2014 #1
    Hello,

    It is my understanding that a moving magnetic can induce an electrical current, and that conversely, and electrical current can move a magnet. I believe that this is the basis by which speakers work (changing electric field causes vibration of the magnet that generates the sound).

    My question is how would one calculate the amount a permanent magnet would move when exposed to an electric field. Another way to put it, how would you calculate the force of an electric field on a permanent magnet. Would there only be a force given an AC current of would there also be a force with a DC current?

    What factors would you need to know to be able to calculate this? I would assume you would need to know:

    frequency (hz) and power (Amp) of the electrical current
    mass of magnet
    strength of magnet (what units? gauss?)

    Thanks for any help one can offer.


    Tord
     
  2. jcsd
  3. Jul 12, 2014 #2
    A changing electric field generates a magnetic field which exerts force on a magnet (so its not the changing electric field by itself that exerts the force but the magnetic field that is generated by the changing electric field).

    But also a current (AC or DC ) generates a magnetic field.

    In order to calculate the force we ll need the equation B=B(I) that relates the magnetic field to the current I (for all frequencies below Ghz we neglect the magnetic field due to changing electric field unless we are studying electromagnetic fields from antennas that radiate electromagnetic energy).

    Also we ll need to know the current (amplitude, frequency and phase if possible). From the equation B=B(I) we will be able then to calculate B. The force on the magnet from the magnetic field B will be F=B*q where q is the quantity of magnetism of the magnet measured in Newton/Tesla. Mass of the magnet will come into play to equate the Force of the magnetic field as m*a, so it would B*q=m*a where m the mass of the magnet and a its acceleration.
     
  4. Jul 14, 2014 #3
  5. Jul 22, 2014 #4
    Thanks for the help, that is exactly the information I was looking for.


    Take care,

    T
     
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