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Two questions about introductory EM confusions.

  1. Jan 11, 2015 #1
    QUESTION 1. I have asked this question here a while before but it seems that I have still not gotten it. My apologies to the previous answerers of this one, I guess it's just me being slow.

    We defined the electromotive force ##\epsilon## for a charge that moves along some path while some force ##\vec{F}## of electromagnetic origin acts upon it C as: ##\int_C \frac{\vec{F}}{q}\vec{dl}##

    The electrostatic definition of a potential difference between point A and point B for any path C going from A to B is given by ##\Delta{V}=-\int_C \vec{E} \vec{dl}##

    It is easy to see that in purely the electrostatic case for a path C from point A to point B ##\Delta{V}=-\epsilon## Is there a way to do an exactly the same formal reasoning to find this exact same result for electrodynamics cases where the magnetic force is involved? Because right now I have the feeling that in electrodynamics the result is going to be ##\Delta{V}=\epsilon## without the minus sign.

    QUESTION 2. This question is about a step in a derivation I'm confused about. The point of the derivation is to show that the energy density of a created magnetic field goes with ##B^{2}##. Consider a single loop with a current ##I## running through it. The potential energy stored in the total flux through the surface is equal to ##U=\frac{I \Phi}{2}##

    And so we can associate a small potential energy with a small piece of flux through the loop as ##dU=\frac{I d\Phi}{2}## Now let's take the special geometric set of field lines that contain the same amount of field lines through space. I think they are called stream surfaces? Anyway the point is that with each ##d\Phi## through the surface we can associate such a stream surface through which the flux is constant.

    This gives ##dU=\frac{I \vec{B} \vec{dS}}{2}##. Where ##\vec{B}## and ##\vec{dS}## can be taken anywhere on the stream surface as long as they are both corresponding to eachother. This is because the amount of field lines and thus the flux through a stream surface is constant.

    Now we use Maxwell's magnetic circulation law in integral form along one field line in this stream surface. This means that ##\oint_C B dl = \mu_{0} I## where I already use that B and dl are parallel.

    So including this into our expression we find: ##dU= \frac{1}{2\mu_{0}} \oint_C B^{2} dl dS(x,y,z)## What basically happens here is that one B field is being integrated over. The other expression has the B field for the flux which can be any B value along the stream surface as long as it corresponds to a correct ##dS## of the stream surface. If I take the value for the latter B field equal to the value of the B field I'm taking a line integral over dS would become a function along the line integral. I get everything until here.

    The next step is the proof saying that if we integrate over all those ##dU## we find ##dU= \frac{1}{2\mu_{0}} \iiint_V B^{2} dV## and thus the value in the integral represents the energy density.

    I don't get the math or formal reasoning behind the last step, can someone explain or elaborate.

    Thanks a lot.

     
    Last edited: Jan 11, 2015
  2. jcsd
  3. Jan 11, 2015 #2

    BvU

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    For Q1 check out this thread from last Thursday. Pretty hefty stuff. Sorry, no time at the moment for Q2...
     
  4. Jan 11, 2015 #3
    I meant to imply something else with my first question than what is being discussed in that thread. I already broke my head a lot on what they talked about but luckily I already get that more or less. My question is about the fact that to me it seems that the relationship between ##\Delta{V}=-\epsilon## seems to have a different sign in electrostatics and electrodynamics cases. I can show ''formally'' what the sign has to be for electrostatic cases which I did but I wanted to see a similar reasoning for when the magnetic field is involved but seem to not succeed in that.
     
  5. Jan 11, 2015 #4

    BvU

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    Then probably this thread is also beside the point.

    Could you elaborate be a bit more on this expectation that ##\Delta{V}=+\epsilon## in electrodynamics cases ? V still the electrostatic potential ? What cases ? Examples ?
     
  6. Jan 12, 2015 #5
    I remember the following from class. This is not a correct reasoning below but this is just an example of which I remember a result along the lines of a positive relationship. Again, it's something from class a while ago and I don't remember how the reasoning went exactly:

    Take an open circuit and move it through a B field. The charge accumulation will happen in such a way that the E-field at each point counteracts ''v x B'', so that ##\vec{E}=-\vec{v}\times\vec{B}##. So taking the integral of the E field would be taking MINUS the integral of the resulting magnetic force per charge. So this has the form of ''positive integral of E, gives MINUS the work per charge by the magnetic force''.

    A positive integral of E also gives MINUS the potential difference. And from such a reasoning my suspicion arsises. I can't really put a finger on it hence I'm asking if someone can provide a thorough reasoning along these lines.
     
  7. Jan 12, 2015 #6

    BvU

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    The thread in post #4 IS beside the point -- either that or way too complicated. You're much better off studying the Lorentz force. It's very fundamental.

    (The following is not very authorative, just my way of looking at things) :

    I see how you work ##\vec v \times \vec B## to something with an ##\vec E## character, and then the countertaction into a minus sign. But I would advise against it:
    • ##F_{Lorentz} \perp \vec B## so the B field itself does NOT do any work. The work is done by whatever brings about ##\vec v##.
    • ##F_{Lorentz}## is the cause of charge accumulation, which in turn causes an E-field, which you can work around to a potential outside the conductor. Inside the conductor only if it has non-zero resistance! So if you close the loop outside the B field you can get a current. (Also if you rotate a loop within the B field, because then you change the direction of ##\vec v##). From these Lorentz force consequences the Faraday law can be derived. (I like the counterexamples :) -- there the Lorentz force pops up again, and "wins" -- but not to worry: the Maxwell-Faraday equation is always correct)
     
  8. Jan 12, 2015 #7
    I see, thanks for your time and help!
     
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