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AC Circuitry/ Magnetic Flux

  1. Apr 7, 2005 #1

    cAm

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    After everyone in our class got less than 50% on our first physics test this term, (AC, magnetic flux, frequency/oscillations), our teacher decided to give us a retest. Though, he hasn't gone over anything else, so i thought i'd ask a couple questions here.

    1. I was taught that Inductive Reactance (XL) = wl, and that Inductive Capacitance (XC) = wc. I heard today though, that XL actually = iwl, (i = sqrt(-1)), and XC actually = 1/(iwc). Is this true? If it's true, then is it also true that these reactances work similar to resistances in DC? (ie EMF=I(XL + XC + R) for series L, C, and R.

    2. Is there a simple way to prove that E(field) = (V x B) for a spherical conductor? One of the 'simple' problems on the test had to have this to be able to solve it, but my teacher has been too busy to explain it.

    that's all i can think of for now...
     
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  3. Apr 7, 2005 #2

    cepheid

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    That doesn't look right. But before I can tell for sure... what is V?
     
  4. Apr 7, 2005 #3

    cAm

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    Sorry, V is velocity.

    I did see a proof for it, but it was waay too complicated to be something for an 'easy' problem. We weren't given this equation either, so...
     
  5. Apr 7, 2005 #4

    cepheid

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    Remember the Lorentz force law? It states that the magnetic force on charges in motion with velocity v is given by:

    [tex] F = q(\mathbf{v} \times \mathbf{B}) [/tex]

    That is the only formula in which I've ever seen v cross B. When we were first introduced to it, we were told it was an experimental result. I think it is possible to derive it theoretically, but I haven't the foggiest idea how.

    So where do we go from here? I think I need more info about the specific situation, as it's still pretty vague. Why is there a B field involved at all? Was it given in the problem that there was an externally applied B field on this conductor?

    Also, why are the charges in motion with velocity v?
     
  6. Apr 7, 2005 #5

    cAm

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    Ok, i'll just type out the problem. Maybe there is an easier way to sort it out.

    "A wire lies along the Z axis and carries current I = 20A in the positive Z direction. A small conducting sphere of radius R = 0.02m is initially at rest on y axis at a distance h = 45m above the wire. The sphere is dropped at the time t=0. A. Assuming that the only magnetic field produced by the wire, determine the electric field at the center of the sphere after t=3s. What is the voltage across the sphere at t=3s?"

    I do have the derivation of Efield = V x B (i think i can remember it correctly), but i wanna see if someone else can figure this out.
     
  7. Apr 8, 2005 #6

    cepheid

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    Okay, I will give it a shot, but I'm no expert. First of all, we need to know the magnetic field produced by the current in the wire, as a function of the radial distance, r, of the centre of the sphere from the wire. That way, if we know the position of the sphere, we know the magnetic field it experiences at that instant, and how it changes with time (as r changes). Use ampere's law to find the B field, with an Amperian loop around the wire:

    [tex] \oint{\mathbf{B}(r, \theta, \phi) \cdot d\mathbf{l} } = \mu_0 I [/tex]

    But from what we know about magnetic fields, this one's magnitude depends only on the radius so it is uniform all around the Amperian loop. ie B is a function of r only: B = B(r) The integral reduces to:

    [tex] B(r) * (2\pi r) = \mu_0 I [/tex]

    [tex] B(r) = \frac{\mu_0 I}{2\pi r} [/tex]

    So, the question asks: what is the electric field at the centre of the sphere? In general, we know that a changing magnetic field induces an electric field, and by Faraday's law of induction, the curl of the electric field is equal to the negative time derivative of B, so that's the general strategy. But I have a problem with this* that'll I'll go over later. First let's figure out how the B field, as experience by the sphere, changes with time. The B field is actually static, it is function of r only, but r changes with time because the sphere is falling towards the wire, so the magnetic field increases in magnitude with time. Finding r as a function of time is easy because the sphere is in free fall so:

    [tex] r(t) = h - \frac{1}{2}gt^2 [/tex]

    So we can easily calculate how B(r) changes with time, and so we can use Faraday's law:

    [tex] \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} [/tex]

    *But what bothers me about this is that in the rest frame of the wire the B field is static and its partial time derivative is zero. There is no induced E field and the force on the charges that creates the circular currents is entirely magnetic, as far as I know, and given by the Lorentz force law as stated above. In the rest frame of the sphere the B field is increasing with time and Faraday's law applies, so there is an induced EMF (and associated rotational E field), pushing the charges. I know I know! Special relativity says the two situations must be the same, and the measured fields are just manifestations of the same field, but the fact remains that depending on how you calculate it, there either is an electric field in the sphere or there isn't. So how do they expect you to answer the question? Somebody else had better take it from here!!!
     
  8. Apr 8, 2005 #7

    cAm

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    Yeah, i've been having some issues with this problem. Our teacher still considers it an 'easier' problem...

    Here's what i figured out, with the help of another teacher (but he's been out of the field for a while, and isn't entirely sure about it all). (actually i didn't figure out much of this at all)

    If you look at it from a completely different perspective, you can do this (i think, someone point out an error if you see one, please)

    EMF = Work/charge, right? Work is the same as F*d(istance). F, for a sphere in a changing B field, is q(F x V(elocity)). So, if you plug it in, you get q(F x V)*d/(q). q's cancel, and you get EMF = (F x V) * d. (here is the part im most uncertain about). E(field)*d = EMF, (right? =/). So, if you substitute, can't you get E(field) = (F x V)?


    I'm not entirely sure what you mean in your * paragraph either. There is a time derivative of B field, because the Sphere is falling further into it....
     
  9. Apr 8, 2005 #8

    cepheid

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    First problem right here. The force on moving charges in a magnetic field is given by F = q(v x B) as I wrote above. What you have written (F x V), is that just a typo or....???



    If you pick a reference frame in which the wire is stationary, then B is static. I agree with you that B(r) at the centre of the sphere changes with time, but that is ONLY because (as you said, the sphere is falling) ie because r changes with time. To emphasize this point: B is static. It is a function of r only. B at the centre of the sphere changes with time only because r changes with time. B does not change with time INDEPENDENTLY of the change due to r changing with time. At any given r, B is constant in time. Otherwise it would be B(r,t).

    Remember, the PARTIAL derivative of B with respect to t refers to the rate at which B changes with t alone, independently of its change due to any other variables, (such as r) changing. So the partial derivative is zero in this case, since at a fixed r, B does not change in time, and that's how you evaluate the partial derivative (by assuming all other independent variables constant).

    Of course, if you pick the rest frame of the sphere, then problem solved. Arrgh, I never understood relativistic electrodynamics, not even the basic points like this. It gives me a headache. I'm sure someone else will solve my little dilemma. In the mean time, can you fix up that equation and maybe present that alternative approach again? It looks promising, except for the E = Vd part, which is true only for uniform E fields.
     
    Last edited: Apr 8, 2005
  10. Apr 8, 2005 #9

    cAm

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    Sorry, the FxV part Was a typo. I'll rewrite it more cleanly here:

    EMF = W/q
    W = F*d
    F = q(B x V)
    EMF = {q(B x V)*d}/q
    EMF = (B x V)*d
    EMF = E*d (isn't there spherical symmetry, so doesn't it still work? Or might it even be uniform?)
    E = (B x V)


    And, for what you were saying about B being only a function of r and t. That IS true, but why does it matter? we're interested in the flux of the sphere. So, as the magnetic field on the SPHERE changes w/ respect to time, there is a flux on the sphere, and thus a E field.

    Though, i could be completely wrong here. This is the equivelant of physics 106 (physics C), so i really haven't been 'tought' much more than the very basics.

    (and, any chance you know the answer to my AC question? :smile: )
     
  11. Apr 8, 2005 #10

    cepheid

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    W/q = F*d / q is only true if the force is constant and always points in the direction of the movement of the charges, which is not really true here. In fact it seems to me the situation is very complicated because the velocity is changing with time, it is changing in direction, and therefore so is the force, which changes the direction of the velocity still more, which changes the direction of the force...and so on and so forth.

    Technically speaking W/q = EMF =

    [tex] \oint{(\mathbf{v \times B) } \cdot d\mathbf{l}} [/tex]

    in this case

    but I think that your equations might be okay "at an instant" i.e the proof takes a few shortcuts and is not very mathematically rigorous but still arrives at the answer you need. I'm not totally sure on that.

    As for my other points, I was simply going into what is needless detail at your stage. I agree that no matter which point of view you take, there is a change in flux and therefore an EMF is produced. Whether it is a motional emf (due to the magnetic field and whatever force keeps the ball in motion) or an induced emf (due to an induced electric field) depends on how you choose to look at it, but in the end, by relativity, both ways of looking at it must be equivalent so the two descriptions refer to exactly the same physical phenomenon, and differ only in the traditional words (magnetic vs electric field) used to describe what is happening. So don't worry about it, just find out some way of calculating what the question asks for.

    I'm supposed to know all of that ac circuit stuff but don't remember it too well. I'l try to refresh (I need to for finals anyway) and get back to you tomorrow, if no one responds before then.
     
    Last edited: Apr 8, 2005
  12. Apr 8, 2005 #11

    cAm

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    ok, thanks for the help :smile: This one has just been bothering me for a while, particularly considering we have never been given EMF = E*d, nor EMF = W/q, nor has he ever even mentioned laurentz's (sp?) law (F = q(V x B)). These are all things i learned when talking to someone who knew a bit about this sort of thing. I had been wondering if we were just missing something simple, that we di d have the info to do, or if it was just a crappy problem. I'm going with crappy problem. :wink:
     
  13. Apr 11, 2005 #12

    cepheid

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    Okay, your notation is a bit confusing to me, so I'll present it the way I've seen it. In general, in a circuit in the AC steady state, the effective impedance Z of the circuit is a complex number that expresses the equivalent impedance of the various circuit elements making up that circuit:

    Z = R + iX (or jX if you prefer that notation. Either way, i or j = sqrt(-1))

    The real part of this number, denoted Re Z = R, the "resistance".

    The imaginary part of any complex number, denoted Im Z = X, the "reactance".

    Note that "imaginary part" is poor terminology because Im Z is a real number!! What they mean is that X is the real number you need to multiply by i in order to get the imaginary term of Z.

    I'm sure that you can see from all this that:

    An inductive reactance is given by:

    [tex] X = \omega L [/tex]

    The corresponding impedance of this inductor is:

    [tex] Z_L = i \omega L [/tex]

    Note that the real part is zero => the impedance of an inductor is entirely reactive. Same with a capacitor:

    [tex] X = \frac{1}{\omega C} [/tex]

    [tex] Z_C = \frac{1}{i \omega C} = \frac{-i}{ \omega C} [/tex]

    To answer your question, if you have a series LRC circuit with a source voltage (emf) Vs, then you can indeed use Ohm's law to relate this source voltage to the current outputed by that source (Is) *if* you write the voltage and current as phasors (complex numbers that indicate the amplitude and phase of your time varying, sinusoidal current and voltage signals). You might wish to read up on phasors if you aren't already familiar with them. So, using an overbar for phasor notation, we have:

    [tex] \bar{V}_s = \bar{Z} \bar{I}_s [/tex]

    [tex] = (\bar{Z_L} + \bar{Z_R} + \bar{Z_C}) \bar{I}_s [/tex]

    [tex] = (i \omega L + R - \frac{i}{ \omega C}) \bar{I}_s [/tex]
     
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