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Torque equation of an electric motor

  1. Jan 29, 2012 #1
    The example I am looking at in my text book starts by considering and area of the rotor surface of width w and length L.

    Then the axial current flowing in the width w is equal to I=wA which is exposed to a radial flux density B

    So from the Lorentz force F=IBxL the equation becomes

    F= wABxL

    so the force per unit area is F/wL which becomes:

    F=BA

    Then to obtain the torque the force per area is multiplied by the entire area of the rotor (2∏rL) then multplied by the radius of the rotor

    So the overall torque equation becomes:

    T=BA x 2∏rL x r

    What doesn't make sense is how can the current be equal to wA? by the Lorentz equation the force on a current carrying conductor is IBxL so the width and area of the conductor carrying the current I does not matter? so why does it apply here?

    Regards
    Dan
     
  2. jcsd
  3. Jan 29, 2012 #2
    I agree with you. Without seeing any diagram I would say the force on the side of the coil of length L will be BIL. The torque produced by the coil will be BILd where d is the separation of the sides of length L. Ld is the area of the coil.
    If the coil is in a uniform field then the torque will be BIASinθ where θ is the angle between the field and the normal to the plane of the coil.
    If the coil has N turns then the torqe is BNIASinθ
    I don't understand what seems to be in your book !!!!!
    hope this is some help
     
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