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Maxwell's 4th equation for a Toroid

  1. Jun 9, 2014 #1
    Thank you for reading this.

    the circuit is simply a SinWave voltage source (ideal) connected to a single coil with an closed iron core inside. It would like the picture below while ignore the labels:

    Only consider saturation and hysteresis of the core and assuming no resistance,no eddie current, no core loss or leakage, nothing else. the relation between the voltage of the source: e, the current in the coils I, and the flux ø, will look like this from ALL textbooks:


    SO many text books just say:" Farady said v = -dø/dt, so if v is sin(something) then ø is cos(something)". I totally disagree, that's the induced voltage from a driving flux. Flux has nothing to do with voltage. flux is the result of current: N*i. However I am also aware of the fact that i won't be sinusoidal, as the inductance of the coil will be none constant, but I can not explain why ø is not affected in shape at all..

    Now, when I look at the Maxwell's 4th equation
    the E inside Maxwell's 4th equation in this case means the E inside of the wires. and H it generated around the winding. that does imply a relationship of V_magnetizing = d( flux_result ) / dt... I thought 4th equation does not apply here, but apparently it does apply inside the winding wires.

    but... what about the current side of the story, the current flowing in the wires, current will generate flux as well.

    did I apply maxwell's 4th equation right?, if I did, what about the flux generated by the currrent?

    Thank you.
  2. jcsd
  3. Jun 10, 2014 #2

    Jano L.

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    If the current is not sinusoidal while the circuit is driven by sinusoidal source of voltage, the magnetic core behaves non-linearly. This can be described by time-varying inductance ##L(t)##. If the wire has zero resistance, the electric field inside the wire has to vanish. Therefore the electric field maintained by the source of voltage has to be counteracted by induced electric field due to coil, so the induced emf ##d/dt(Li)## has to have the same magnitude as ##u_0\cos \omega t##. The flux ##Li## is therefore sinusoidal, but ##i## and ##L## do not need to be.

    The electric field term in the equation
    \nabla \times \mathbf B = \mu_0 \mathbf j + \mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t}
    does not mean that "electric field generates magnetic field". In common electronic engineering situations where the period of oscillation is much greater than the time light needs to pass through the setup, magnetic field is accurately given by the Biot-Savart law, i.e. is a function of the current distribution. This is the case with ordinary circuits, also for your circuit with magnetic core. (This simplification fails for antennae and related stuff, where the frequency of oscillation is very high and the wavelength may be comparable to the antenna itself.)
  4. Jun 10, 2014 #3
    "Therefore the electric field maintained by the source of voltage has to be counteracted by induced electric field due to coil"

    Such a powerful explanation, Thank you, I will come back and post more.
  5. Jun 10, 2014 #4
    I am now reading your post carefully, I can follow the fact the L*i is closely related to Magnetic Flux. But, I do have a mathematical impasse right now: if v = L(t) di(t)/dt, but d(flux)/dt is going to be d( L(t)*i(t) )/dt. so V is not exactly d(flux)/dt. Judging by the 90 degree phase difference between V and Flux, v should be related to derivative of the flux.

    also, you are saying that maxwell 4th equation only applies for the total E inside an enclosed space right? and there is no net E in the coils? so the maxwell 4th equation is irrelevant?
  6. Jun 11, 2014 #5

    Jano L.

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    ##Li## is equal to magnetic flux ##\Phi## , because we define variable inductance ##L## by the formula
    \Phi(t) = L(t)i(t).

    Now whatever the inductance does, the emf in the circuit is given generally by the Faraday law:
    |emf| = \left|\frac{d\Phi}{dt}\right|.
    (the sign can be found but it is trouble to argue it clearly, so let's leave it unspecified for the time being).
    Since there is zero electric field throughout the wire, the induced electric field ##\mathbf E_i## due to coil is such that it counteracts the electric field due to voltage source ##\mathbf E_s##. Because of this, the integral of the induced field from one pole of the voltage source to the other (the emf) has minus value of the integral of the field ##\mathbf E_s##, which equals ##U_0\cos \omega t##. So we have

    emf = -U_0\cos \omega t.

    From the last two equations, we obtain
    \left| \frac{d}{dt}(Li) \right | = |U_0 \cos \omega t |
    |Li| = |\frac{U_0}{\omega} \sin \omega t.|
    If ##L## does not vary in time (core is made of non-magnetic material or vacuum), the current ##i## will vary sinusoidally. If ##L## varies in time, neither ##L## nor ##i## have to vary sinusoidally, only their product has to, because the source of voltage enforces it.
  7. Jun 11, 2014 #6
    Thank you very much for your very detailed explanation and step to step guide. Those equations were not easy to type.

    However, I just realized that we didn't solve my most fundamental impasse: Farady/maxwell3rd equation, v = dΦ/dt is ONLY for calculating voltage induced by flux. It has nothing to do with generating flux with a voltage. As far as I know, the only way of generating flux is get a collection of M dipoles or you create charge flows or get a changing E.

    the L*i is one way of looking at what is happening from a current point of view: since L(t) is inversely proportional to µ, and i(t) is proportional to µ, Φ might just as well be a sinusoids. Although I disagree upon your application of v = dΦ/dt for this scenario, but I somehow felt that Φ might indeed be sinusoid.

    I think current is flowing inside the winding. Without current, the only way to have flux generated is thru a changing E. But I agree with you the fact that counter EMF = V_input. So the current is always zero...? what about the value of d(i)/dt if i is always zero? I don’t think that’s how it works. if we have a voltage source and a lone resistor connected to it: voltage drop across the resistor will be the same as the voltage source, so the voltage of the resistor perfectly counters the voltage of the source. But current will still flows thru the resistor.

    But induction is more than that, V_source => currrent in winding generated=> Φ generated => counter EMF generated => current in winding reduced. Unfortunately the events seem to be all happening at the same time.

    OMFG, NVM. I just checked the real definition of L. I was totally misled by my stupid text book
    L = dΦ/di
    so V_across the windings = Ldi/dt = dΦ/dt, and we reached this conclusion NOT because of Farady’s law.
  8. Jun 12, 2014 #7

    Jano L.

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    Yes, except in the Faraday law, ##d\Phi/dt## does not give value of voltage, but of emf. The difference is that emf is due to solenoidal (circuital) induced electric field, which has no potential.

    Why do you think that?

    It has to be if induced emf is to counteract the imposed voltage (which happens for zero resistance wire).

    Yes, the source of voltage guaranties that.

    The emf counteracts voltage, and for zero resistance wire it does so completely. For wires that have some resistance, for the current to flow, there has to be some electric field inside the wire so the electric field of the source of voltage is not exactly cancelled by emf.

    I do not think that is a common way to define ##L##. I think the common definition is
    L = \frac{\Phi}{I}.
    If you use ##L=d\Phi/di##, then indeed ##Ldi/dt = d\Phi/dt##. But to reach the conclusion that this equals emf or -voltage due to source can be reached only because of the Faraday law.
  9. Jun 12, 2014 #8
    Sir Jano, Thank you so much for your extraordinary patience and kindness.

    If L=dΦ/di, then Ldi/dt=dΦ/dt ==> explains How that Flux was generated.
    counter EMF = dΦ/di ==> explains what the effect of that Flux should be on the windings.

    and from both views, both proved that MFlux was sinusoidal. Unified theory right there! Fully resolved.
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