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Is this true about the emf in a closed circular wire?

  1. Apr 15, 2009 #1
    the emf is defined as the potential difference between two points [itex]\varphi(\vec{r_1})-\varphi(\vec{r_2})[/itex].

    ok so lets say we make a trip round a closed circular wire with a battery to keep the current flowing then r1=r2 and so no emf has been done - is this true and if so why?
     
  2. jcsd
  3. Apr 15, 2009 #2

    Matterwave

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    Re: Emf

    The battery in general produces an emf. r1 does not equal r2 because you hit the battery on your walk around the loop. If r1=r2 you can't fit a battery in there.

    I'm guessing you're talking about Kirchoff's rules? If you're talking about why the line integral of the Electric field around a closed path is not identically 0, it's because currents are not static.
     
  4. Apr 15, 2009 #3
    Re: Emf

    i dont follow. i haven't covered kirchoffs rules yet. why does r1 not equal r2, these are just vectors from the origin to the point we're at in the circuit so surely regardless of whether we do a lap or not its still the same position vector from the origin that descirbes the point???
     
  5. Apr 15, 2009 #4

    Matterwave

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    Re: Emf

    When you walk around the loop, you will inevitably hit the battery or emf source. That source is like a discontinuity in your loop...when you "walk" over the emf source you need to add the voltage given off by the emf to your potential difference.
     
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