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Electric field inside a conducting wire

  1. Apr 29, 2003 #1
    Yo, d00dz, I can't remember how to find the electric field inside a conducting wire (actually a coaxial cable)

    Here's the exact text of the problem:

    A coaxial cable (inner radius a, outer radius b) is used as a transmission line between a battery E and a resistor R, as shown in Fig 19. (attached)

    (a) Calculate E, B, for a < r < b.

    There are some more parts, but I can probably handle them once I figure out how to calculate E, since B*c = E and I know the Poynting vector and how to integrate and whatnot

    (b) Calculate the Poynting vector S for a < r < b. (c) By suitably integrating the Poynting vector, show that the total power flowing across the annular cross section a<r<b is E^2/R. Is this reasonable? (D) show that the direction of S is always from the battery to the resistor, no matter which way the battery is connected.
     

    Attached Files:

  2. jcsd
  3. Apr 29, 2003 #2
  4. Apr 29, 2003 #3

    Tom Mattson

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    You know that the E field has no radial component, so the whole thing is parallel to the wire itself. You can get it via:

    E=-[nab]V,

    where V is the potential of the battery. In this case, you simply have E=V/d for the magnitude.

    For the B field, use the integral form of Ampere's law:

    &int;B.ds=&mu;0ienc

    Since the whole current of the inner cable is included in the Amperian loop, this is not a difficult task.

    Easy: S=(1/&mu;0)ExB.

    You need to get the directions of E and B right first.
     
    Last edited: Apr 29, 2003
  5. Apr 29, 2003 #4
    Sweet

    I figured it'd be that, but for some reason someone wrote some weird answer in the margins of my textbook, so I wasn't really sure

    The answer in the book was Efield = E/r*ln(a/b)
    and B = (mu 0)*E/(2*pi*R*r)
     
  6. Apr 29, 2003 #5

    Tom Mattson

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    That's the E field inside a cylindrical capacitor, if I'm not mistaken. We don't have that here, because there is no net charge on the cables.

    That's close, but instead of E/R you should have the enclosed current.
     
  7. Apr 29, 2003 #6
    Excellent, excellent, thanks again

    I'm glad to hear that the sensible answer was in fact the right one

    I could've sworn I'd seen that strange formula before, and now I know where

    I guess a previous owner of the book just ended up searching frantically for a formula that fit, and came up with the one for the E-field in a capacitor

    If you ever need help with something related to web programming, just drop me a line, d00d
     
  8. Apr 29, 2003 #7

    Tom Mattson

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    I might take you up on your offer, but only for making a simple webpage.

    Hope you stick around and use the rest of the forums!
     
  9. Apr 29, 2003 #8
    Hmm, I just got home and started on it again, and this is kind of throwing me off

    The E field is parallel to the conductor, right? And then the B field should be clockwise or counterclockwise around it, shouldn't it?

    Wouldn't that indicate that ExB is always radially outward from the center of the conductor?

    That seems ridiculous, since energy is flowing from the battery to the resistor and then out into the great unknown, but maybe the Poynting vector isn't really valid for this case, I dunno

    If this isn't too tricky, could you clear me up on it real quick-like?

    If it'd be a lot of trouble, don't worry about it
     
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