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Homework Help: Energy spent by motor in a period

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a single coil that is turning in a constant and uniform magnetic field {\bf B} thanks to a motor. The normal to the coil is given by:

    $${\bf u}(t)=\sin (\omega t){\bf u_x}+\cos(\omega t){\bf u_z}$$

    How can we obtain the energy that the motor has to spend in a period [itex]T=\frac{2\pi}{\omega}[/itex]?

    2. Relevant equations

    3. The attempt at a solution
    I haven't any good idea. I have thought that the energy given by the motor cold be trasformed in magnetic potential energy. Knowing that [itex]U_p=-{\bf m} \cdot {\bf B}[/itex] where [itex]{\bf m}=i \Sigma {\bf u_n}[/itex], [itex]U_{motor}=-U_p[/itex]. I could integrate it from 0 to T and I could obtain the energy spent by the motor during the time T. But I think that this procceding is wrong.
    Many thanks for your help.
  2. jcsd
  3. Sep 16, 2013 #2

    rude man

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    First you need the dimensions of the coil and the current in the coil.
  4. Sep 16, 2013 #3
    The coil is characterized by radius r, resistence R and inductance L.

    I have written I=emf/R where emf is electromotive force [itex]fem=\frac{-L di}{dt}[/itex].
  5. Sep 16, 2013 #4

    rude man

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    OK, it's a closed round coil with no initial current.

    So yes, compute the emf which is related to the flux thru the coil how?
  6. Sep 16, 2013 #5
    But, how can I "link" emf and energy spent by motor?
  7. Sep 16, 2013 #6

    rude man

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    Equate the energy dissipated by the coil with the energy supplied by the motor.
  8. Sep 16, 2013 #7
    And so the only energy given by the motor is the energy that is dissipated in the resistor? There aren't other "parts" of energy given by the motor?
  9. Sep 16, 2013 #8

    rude man

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    What other dissipation mechanisms can you think of? There is of course friction in both the motor and the generator but these have to be ignored if they're not given.
  10. Sep 17, 2013 #9
    My doubt is born when I have seen that the exercise said: "Find the expression of the energy spent by the motor during T, then find the expression of the energy dissipated in the resistor during T. Are the two expressions compatible?"

    So I have thought to potential magnetic energy and I have followed the proceeding written in the upper part of the topic. But, if I have correctly understood, you said to me that proceeding is wrong. And so I don't know what I have to think.. :( Are you sure that I don't have to consider the magnetic potential energy? Does the motor give energy to fight the emf? (a generator yes...)
  11. Sep 17, 2013 #10

    rude man

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    It's true that energy is stored and lost within a cycle, but averaged over a cycle there is no energy lost.

    It looks like you are meant to compute the energy dissipated per cycle in some manner other than resistance loss, then compare with resistance loss. This would not be difficult if the coil were rectangular. You could use the Bli law to determine the integral of torque times angle which would be the work done by the motor. But it would seem that your coil is round, which makes that calculation more difficult.

    Rather than proceed under that assumption I would prefer that you state the problem verbatim and in toto.
  12. Sep 17, 2013 #11
    mmh... Could you explain me (or give me a link) what is "Bli law"? How coul I find the work done by motor?

    A lot of thanks for your help!!
  13. Sep 17, 2013 #12

    rude man

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    The force on a wire of length l carrying a current i in a B field is F = Bli. This assumes B and l are orthogonal. F will be orthogonal to l and B. In vector notation, F = i l x B .

    So torque = r[/B] x F where r is the moment arm. In a rectangular coil that would be the distance from the axis of rotation to the sections of the coil that are perpendicular to B.

    Then energy = torque x angle and angle = 2pi. Since F is a function of the angle the coil makes with the B field you have to integrate torque over 1 cycle to determine the work done by the B field on the moving coil.

    There is a more advanced approach. It can be shown that input electrical power to the rotating coil is L i di/dt. So even for your round coil you can compute the energy applied to the rotating coil over 1 period of rotation. My refrence is a textbook by H H Skilling, 'Electromechanics', Part 1.
  14. Sep 17, 2013 #13
    i.e. energy=∫ torque dθ? with θ from 0 to 2pi?

    "the work done by the B field on the moving coil." that is the work done by the motor?
  15. Sep 17, 2013 #14

    rude man

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    I must admit I'm not sure how to apply the "advanced' method I gave you to this problem. If the coil were rectangular it would be straight-forward. I may re-post if I figure it out (before someone else does!).

    EDIT: the method assumes doubly-excited motors which is not the case here.

    EDIT EDIT: there is a way to apply the above 'advanced' method to your problem. The steps are somewhat convoluted (at least for me):

    1. Show that in a doubly excited coil system, if neither coil's self-inductance changes with time, then
    torque τ = i1*i2*dM/dθ where M(θ) is the mutual inductance between the two coils and θ is the angle between the two coils.

    2. Assume your B field is produced by a coil with current i1, but all you know is B. So by definition
    M = ψ21/i1 where ψ21 = flux thru coil 2 due to coil 1, and i1 = current thru coil 1.

    3. So τ = [i1*i2 dψ21/dθ]/i1 = i*B*dA/dθ. (Since there is only one current, i2 = i.)

    Now you should be able to figure out what dA/dθ is for your moving round coil. Then get τ and the work done by ∫τdθ from 0 to 2π.
    Last edited: Sep 17, 2013
  16. Sep 18, 2013 #15
    Many thanks for your help. I agree with you. These steps are convoluted! :(

    I was thinking... Considering a coil, we have that the torque is given by [itex]τ=i Ʃ {\bf u} \times {\bf B}[/itex]

    and it's potential magnetic energy is given by [itex]U_p = -i Ʃ {\bf u} \cdot {\bf B}[/itex]

    So we have [itex]τ=-\frac{d U_p}{dθ}[/itex]

    Considering that the mechanic torque of the coil is the opposite of the mechanic torque of the motor, I was thinking that maybe the energy given by the motor could be expressed using -U_p and (considering that the normal u depends on time) if we integrate from t=o to t=2pi/omega, we could obtain the total energy... what do you think about it?
  17. Sep 18, 2013 #16

    rude man

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    Looks interesting, but what is u? In your 1st post, u was a unit vector. I don't see how torque can be the product of current and a sum of unit vectors???

    Anyway, even if the coil is circular it is possible to compute the torque on the coil as a function of θ, then integrate from θ = 0 to 2π. It's just more involved than if the coil were rectangular since in the latter case the force is uniform across the two sections of the coil perpendicular to B.
  18. Sep 18, 2013 #17
    I'm sorry, I have forgotten to specify the symbolism...

    [itex]{\bf u}[/itex] is the vector normal to the surface of the coil

    [itex] \Sigma[/itex] is the area of the coil

    The two relations that I have written in my last post infer from the Ampere's Equivalence Principle between coil and magnetic dipole
    Last edited: Sep 18, 2013
  19. Sep 18, 2013 #18

    rude man

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    OK, I never heard of this principle. It must be similar to what I put in my post #14. I know it's a bit roundabout but essentially I equated your B field with that produced by a fictitious coil #1. But I did not derive τ = i1*i2 dM/dθ and maybe you're wonderig what M (mutual inductance) is anyway so I will look up the Ampere equivalence principle and try to get back to you.

    I do know that my method of post 14 works; I tried it with a rectangular coil for which the torques are directly and easily computed by the Bli law (τ = F r sinθ = Bli r sinθ, r = distance from coil pivot axis to windings, θ = angle between B field and coil normal).
  20. Sep 19, 2013 #19
    I'm going to translate for you what is written in my book about Ampere's Principle Equivalence:

    "A flat coil that has area [itex]d \Sigma[/itex], in which flows the current i, is equivalent (in relation to the magnetic effects) to an elemental magnetic dipole that has magnetic moment [itex]d{\bf m}=i (d \Sigma) {\bf u{_n}}[/itex] orthogonal to the plane of the coil and oriented (in relation to the current verse) according to the right-hand rule.

    Now we can define the potential energy [itex]U_p=-{\bf m} \cdot {\bf B}[/itex], that is "linked" to the angle respect with the magnetic field B".

    Thinking and thinking again, I remember that the work done for an infinitesimal rigid rotation of a circuit is given by

    [itex]dW=-dU_p=\tau_{\theta}d \theta[/itex]..

    Do you think that we are near to the conclusion of the topic? ;)

    Many thanks again!
    Last edited: Sep 19, 2013
  21. Sep 19, 2013 #20

    rude man

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    Can you fix up your latex for " [itex]d{\bf m}=i d{\Sigma {\bf u{_n}}[/itex] ". It didn't come out understandably.
    Have you tried this method to get an answer? No conclusion without answers!
  22. Sep 19, 2013 #21
    post edited

    Yes, but at this moment I don't find the sheet with proceeding. Tomorrow morning I'll re-do it and post it!
  23. Sep 19, 2013 #22

    rude man

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    Never mind, I think I figured it out. This is a good way to solve the problem and I picked up some learning I forgot long ago along the way!

    So your coil has a magnetic moment m = iA where i = current in coil and A = its area.

    And torque τ = m x B so now you can compute the torque about your coil as a function of the angle θ the normal A makes with the external B field, then integrate it over θ = 0 to 2π. The direction of m and A is the same.

    But you still have to first compute i, the current, again as a function of θ.
  24. Sep 22, 2013 #23
    Excuse me for the delay of my reply but I couldn't reply before.

    I'm starting from the beginning.

    The flux of B is given by [itex]B cos (\omega t) \pi r^2[/itex]

    So the emf is given by [itex]B \omega \pi r^2 \sin(\omega t)[/itex]

    and the inducted current is given by [itex]i=\frac{B \omega \pi r^2 \sin (\omega t)}{R}[/itex]

    Now, the current isn't constant, so we have auto-inducted fem [itex] emf_2=-\frac{d \phi}{dt}=- L \frac{di}{dt}= \frac{-LB \omega^2 \pi r^2 cos \omega t}{R}[/itex].

    So we have the autoinducted current given by [itex]i_2=-\frac{d \phi}{dt}=- L \frac{di}{dt}= \frac{-LB \omega^2 \pi r^2 cos \omega t}{R^2}[/itex]

    So, in conclusion, [itex]i(t)=
    \frac{B \omega \pi r^2 \sin (\omega t)}{R} -\frac{LB \omega^2 \pi r^2 cos \omega t}{R^2}[/itex]

    Do you agree?

    Now, the exercise says: suppose that we can say [itex]i(t)=i_0 sin (wt - \phi)[/itex], find i_0 and phi. I have thought to use the goniometric formulas, but with no results. How can I do?

    Then, for the energy of the motor: if I have i(t), I can calculate U_p ([itex]U_p=-{\bf m} \cdot {\bf B}=-i \pi r^2 B sin (\omega t)[/itex]), integrate it from t=0 to t=T and obtain the energy spent by motor. You suggest to integrate the torque from 0 to 2 pi. Ans this is the same that my proceeding. Is it correct?

    Many thanks
    Last edited: Sep 22, 2013
  25. Sep 22, 2013 #24

    rude man

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    I don't think you can compute the self-induced current the way you did, by taking the current due to the external B field and computing L di/dt from that.

    The total emf around the coil is emf = dψ/dt - L di/dt = iR,
    ψ = external flux = ABcos(wt).

    L di/dt + Ri = dψ/dt = d/dt [ABcos(wt)], A = πr2 = area of coil.
    So L di/dt + Ri = -wABsin(wt).

    You need to solve this differential equation for i(t), then you can use i in
    torque = m x B, m = iA.
  26. Sep 24, 2013 #25
    excuse me, why do you have dψ/dt - L di/dt = iR and not -dψ/dt - L di/dt = iR?

    When I write the equation of the circuit, do I have to keep the signs of emf inducted and self-inducted opposite?

    Many thanks again!!
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