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Homework Help: Nature of charge at the junction of a composite wire

  1. Nov 10, 2018 #1
    1. The problem statement, all variables and given/known data
    Please help me answer thew question in the image.

    KVPY doubt.PNG

    2. Relevant equations
    Current Density J=conductivity X Electric field


    3. The attempt at a solution
    As the current density depends on the conductivity of the material through which the electrons constitutuing it are flowing hence the current density in copper should be more than iron due to greater conductivity of copper. Hence negative charge should accumulate at the junction of copper and iron.
     
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  3. Nov 10, 2018 #2

    haruspex

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    As the choices indicate, you need to distinguish the two sides of the junction. And don't forget that the conventional current flow is opposite to the electron flow.
     
  4. Nov 10, 2018 #3

    gneill

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    Could it perhaps be that any charges there are a result of an electrochemical reaction between dissimilar metals?
     
  5. Nov 10, 2018 #4

    Wrichik Basu

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    That depends on what the word "joined" connotes. If they have been just kept in contact, should any electrochemical reaction occur?
     
  6. Nov 10, 2018 #5

    gneill

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    Yes, I'd think so. The metals have different electronegativity, different "standard electrode potentials". Galvanic corrosion comes to mind. I am the first one to admit that I am not an expert in this area, but I remember the terms from high school and college chemistry :smile:
     
  7. Nov 10, 2018 #6

    haruspex

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    The doubt I have is that the question implies it is related to the applied current. I think I have come across this effect before on this forum.
     
  8. Nov 10, 2018 #7

    Borek

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    Sorry guys, as long as the metals just touch each other all this chemical mumbo-jumbo doesn't apply.
     
  9. Nov 10, 2018 #8

    haruspex

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    It looks like the same principle as this one:
    https://www.physicsforums.com/threads/resistor-made-from-two-materials.900569/#post-5670963
    Essentially, the voltage gradients in each wire imply different field strengths. Applying Gauss' law to the boundary yields a local static charge.
     
  10. Nov 10, 2018 #9

    Charles Link

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  11. Nov 10, 2018 #10

    TSny

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    Welcome to PF, Aishikdesto.

    This would imply that the current ##I## in the copper is greater than the current ##I## in the iron. This isn't correct under steady-state conditions. Can you see why the currents must be the same? What can you say about the current density ##j##?

    ##j = \sigma E##
    Which material has a greater value of ##\sigma##? A greater value of ##E##?
    Apply Gauss' law to an appropriate closed surface.
     
  12. Nov 10, 2018 #11

    haruspex

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    I'm reasonably sure my interpretation in post #8 is the right one. TSny seems to agree.
     
  13. Nov 11, 2018 #12

    Delta2

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    For the steady state I totally agree with @TSny post.

    But can anyone tell us what happens in the transient state during which the charge at the interface is building up? I think the OP's reasoning at post #1 maybe applies for the transient state.
     
  14. Nov 11, 2018 #13

    TSny

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    Yes, I agree. Somehow I overlooked post #8. My post was fairly redundant.
     
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