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I How is there not an infinite regress in the EMF of a motor?

  1. Jul 6, 2016 #1
    Say we have a coil connected to a battery in a uniform magnetic field, perpendicular to the magnetic moment of the coil. This is a simple motor. There is a torque on the coil that varies with the angle θ between the field and the moment. Clearly, the angular velocity ω is not constant.

    Here is my concern: since the flux through the coil is changing in time, there is a back emf in it. his will reduce the net emf, which will reduce the current. That will reduce the torque on the coil which in turn will change the angular velocity in time (I imagine it is the angular velocity that determines dΦ/dt in the first place) and that should change the back emf...one can see where this is going.

    What is happening here? And how does one calculate dΦ/dt?

    Thanks.
     
  2. jcsd
  3. Jul 6, 2016 #2
    There are two back emf's:

    1) The first one from self induction of the coil. This quickly drops to zero
    2) The second one from the rotation of the coil in the external magnetic field (this is a mutual induction in case the external magnetic field is created by another coil). This second emf grows larger as angular velocity grows larger (due to the torque which is decreasing but still it is in the positive direction to increase angular velocity. In other words ##\frac{d\omega}{dt}## is decreasing but remains positive). At some point this emf is so large that it almost cancels the emf of the source, so the total current is very small and the torque is very small just to counterbalance any resistances/frictions.
     
  4. Jul 6, 2016 #3
    @Delta² Thank you for your answer! This is just the kind of explanation I was looking for! At the point where the torque is very small, is the angular velocity (more or less) at a maximum? Also, is there any way to calculate the back-emf as a function of time? I imagine that at this point, it too is at a maximum, and the current at its minimum, most stable value. Have I got it right?

    Thanks so much!
     
  5. Jul 6, 2016 #4
    Yes seems to me you got it right.

    But to calculate exactly the back em as a function of time is not so easy. It depends mainly on the geometry of the coil (its exact shape, how many turns, area per turn) and the magnetic field of the stator.

    The equation of the back emf is coupled to equation of the torque. if ##T(t)## is the torque then we ll have

    ##T(t)=Ia(t)=A\frac{d\omega}{dt}=A\frac{d^2\theta(t)}{dt^2}## (1)
    where A is the moment of inertia of the coil and depends on the geometry of the coil as well.

    Also we will have if E(t) is the back emf,

    ##E(t)=BScos(\theta(t))\frac{d\theta(t)}{dt}## (2)
    where B is the magnetic field of stator and S is another constant that depends on the geometry of the coil(number of turns and area per turn) as well.

    There are two more equations one involving how the Torque is generated from Laplace force due to the current in coil and it will be like

    ##T(t)=UBI(t)sin(\theta(t))## (3)
    where U another constants that depends again on number of turns and radius of each turn in coil, and I(t) is the current in coil.

    and the last equation is the differential equation for the current I(t) in the coil

    ##E-E(t)-L\frac{dI(t)}{dt}-I(t)R=0## (4)
    where R is the ohmic resistance of coil

    So all in all we have a system of four differential equations with four unknown functions, Torque T(t), back emf E(t), current I(t) and the angle ##\theta(t)##. As you see the equations are coupled in various ways, solving this system we can get all four of the unknown functions.
     
  6. Jul 6, 2016 #5
    @Delta² I've got nothing to say, except that YOU ARE brilliant!
    This had been bothering me for days, and I can finally have a good night's sleep.
     
  7. Aug 9, 2016 #6

    David Lewis

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    Gold Member

    Does B represent magnetic flux or magnetic flux density?
     
  8. Aug 10, 2016 #7
    B is the magnetic field vector(not sure if we can call it magnetix flux density), the flux through a surface S will be BScos(theta), where theta the angle between the vector B and the normal n to the surface S.

    I think I have a mistake in (2) and/or (3), both equations should have ##sin(\theta(t))## or ##cos(\theta(t)##, maybe there are other mistakes too, my post was a kind of "raw" attempt to show how even a simple motor involves solving a system of 4 differential equations.
     
  9. Aug 10, 2016 #8

    David Lewis

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    Gold Member

    Many thanks for your help. The physical quantity represented by B in textbooks seems to be sometimes flux (Wb), and other times flux density (T).

    It may be analagous to force versus stress in mechanics. One is newtons, and the other is newtons per square metre.
     
  10. Aug 10, 2016 #9

    cnh1995

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    Homework Helper

    In electrical machines, B is the flux density and 'Φ' is used to denote flux.
     
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