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Total energy in a wire using energy densities

  1. Apr 14, 2012 #1
    Given a cylindrical wire of radius r, length L, carrying a current I, find the total energy stored inside the wire.

    From griffiths,

    uem= εE2/2 +B2/2μ

    and the tot energy is

    ∫uem dV


    I have my E and B fields, but my B field is a function of x where x<r, (E is uniform)

    B=kx/r2 (k=all the constants)

    my question is,

    it says inside the wire, does this mean i cannot put x=r and integrate easily to get

    Energy=(εE2/2 +B2/2μ)[itex]\pi r^2 L[/itex] ?

    Where B = k/r = K

    or do i have to integrate B seperately to get something like

    ∫∫∫Kx x dx dz d[itex]\phi[/itex] where K = k/r2 (since dV = x dx dz dthi in cylindrical)

    = (2/3) K x3 L pi

    If so would this be it? No bounds on x, giving the energy at some radius inside the wire?

    i guess the question is over what volume is considered inside the cylinder, some x<a or just x=a?

    Im unsure about this because of the (2/3)x3 in the second approach since if you put x=r here it will be different to the first approach because of the (2/3)
     
  2. jcsd
  3. Apr 14, 2012 #2

    BruceW

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    Almost, but not quite. You need to integrate B2/2μ You've got the right method, but you've not integrated the correct thing.

    You need to go back to do the calculation of the energy again, but still, you shouldn't be surprised if the answer is less than your first approach. In the first approach, you are effectively saying that the magnetic field is the same throughout the conductor, but actually it gets less, the further in the conductor you go. Your second approach is the correct method, because the integration takes account of this.
     
  4. Apr 14, 2012 #3

    rude man

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    How do you know E? E = V/L but all you know is L, current i plus wire radius r.

    But you could have a wire of low resistance R and low voltage V across it, or a wire of high R and high V across it. Current i could be the same in both cases, yet E = V/L, greater in the high-R wire than the low-R one.
     
  5. Apr 14, 2012 #4
    Ohh right,
    Thanks, so it should be

    ∫∫∫[itex] \frac{1}{2 μ_{0}} K^{2} x^{2} x dx dz d \phi[/itex] where K = k/r2 (since dV = x dx dz dthi in cylindrical)

    = ∫∫∫[itex] \frac{K^{2}}{2 μ_{0}} x^{3} dx dz d \phi[/itex]

    = [itex]2 \pi L \frac{K^{2} x^{4}}{8 μ_{0}} [/itex]

    = [itex] \frac{K^{2} x^{4} \pi L}{4 μ_{0}} [/itex]

    Does that look good?

    Thanks again!

    Hey rude man,

    I left out that there is a potential across the wire and it has resistance R, I left it out cause I'm happy with the E part of the integral I was just having trouble with the B part, =D
     
  6. Apr 15, 2012 #5

    rude man

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    OK Jessa. Forge ahead!
     
  7. Apr 15, 2012 #6

    BruceW

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    It is completely correct. And no worries, glad to have been of help. You seem to be pretty good at this stuff. So you've got the energy contained in the magnetic field within the radial distance x from the axis. And the question asks for the energy contained within the wire, so there is one last step (choosing the value to be used for x). It's only a small last step, but I thought I'd remind you. There are so many times when I got close to the end of the question, but forgot to do the last step, so I lost marks!
     
  8. Apr 15, 2012 #7
    ahaha yea,

    I'm worried about what to do with x, because in the question it says inside the wire, with inside in italics. Do you think it's safest to write this case for and place inside the wire, then under write, when x=r the energy is bla? Would the total energy inside the wire be when x=r?

    Could I ask one last thing, how come there is a factor of 2 difference between the first method and the second,

    These are the K's ive been talking about,


    [itex]k=\frac{{{\mu }_{0}}I}{2\pi },\,\,K=\frac{k}{{{r}^{2}}},\,\,{{K}^{2}}=\frac{{{\mu }_{0}}^{2}{{I}^{2}}}{4{{\pi }^{2}}{{r}^{4}}}[/itex]

    and the first way, is this way incorrect because I am saying I'm integrating the B field only at x=r through the volume of the cylinder?
    [itex]First\,way:B=\frac{kx}{{{r}^{2}}},B(r)=\frac{k}{r},\iiint{\frac{1}{2\mu }\frac{{{k}^{2}}}{{{r}^{2}}}x}dxdzd\phi =\frac{1}{2\mu }\frac{{{\mu }^{2}}{{I}^{2}}}{4{{\pi }^{2}}{{r}^{2}}}\pi {{r}^{2}}L=\frac{\mu {{I}^{2}}}{8\pi }L[/itex]

    and then the
    [itex]Second:B=Kx,\iiint{\frac{1}{2\mu }{{K}^{2}}{{x}^{2}}xdxdzd\phi =\frac{1}{2\mu }\frac{{{\mu }^{2}}{{I}^{2}}}{4{{\pi }^{2}}{{r}^{4}}}\frac{{{x}^{4}}}{4}2\pi L}=\frac{\mu {{I}^{2}}}{16\pi {{r}^{4}}}{{x}^{4}}L\,=\frac{\mu {{I}^{2}}}{16\pi }L\,\,\,(when\,x=r)[/itex]
     
  9. Apr 15, 2012 #8

    BruceW

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    Yes, the total energy within the wire is when you integrate all the way up to x=r. That's what the limits of an integral mean - that you are counting the integral all the way up to the limit. You could also write for the case when x is any old value less than r, but because the question asks for total energy within the wire, they will be expecting you to integrate up to x=r.

    Yes, in the first way, you're saying B=B(r) at all points within the conductor. But it is actually not, because B is not constant throughout the conductor. Which is why the first way gets an incorrect answer. Your second way does get you the correct answer.
     
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