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Ampere's Law and two conducting parallel plates

  1. Apr 23, 2009 #1
    hey sorry for the prolific posting but exam's are coming up and i'm just working through past papers.

    two conducting parallel plates have uniform uniform current densities flowing through them parallel to their surface. Use Ampere's Law to find the magnetic field inside and outside the plates.

    im quite confused as to where to take my loops here in order to find the field at all the various places im reuqired to. any suggestions?

    also why does the following argument not work here:
    integrate round an amperian loop that doesn't intersect either plate and get
    B(r)-B(r+dr)=0 as no enclosed current
    gives B(r)=B(r+dr) implies B(r)=0 but B goes to 0 at infinity so B is zero outised the plates???
     
  2. jcsd
  3. Apr 23, 2009 #2

    dx

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    What direction do you think the field is in? Use symmetry.
     
  4. Apr 23, 2009 #3
    surely it depends on the direction the two currents are flowing in - if they're opposite then they will cancel. im not sure abou where to put the magnetic field lines when its a plane and not like a wire or something.

    also why does my arhument about the field being 0 outside break down here - isn't that legitimiate for showing the field is 0 outside of a solenoid?
     
  5. Apr 23, 2009 #4

    Matterwave

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    Use superposition. Find the field due to one plate, then the other (which will be either symmetric or anti-symmetric) and then add them up.
     
  6. Apr 23, 2009 #5
    ok but where do i place my amperian loops

    and why is the field not 0 outside the plates according to me argument at the end of post 1?
     
  7. Apr 23, 2009 #6

    Matterwave

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    The ampere loops should be rectangles that cross the plate, and have 2 sides parallel to the plate.

    I don't think I get what you mean by your previous argument.
     
  8. Apr 23, 2009 #7
    ok so if i integrat round said loop for the top plate i get

    int(B.dr) = mu_0 int(K.dA)
    int(B.dr)= BL -0.L where L is the length of the loop

    so B =mu_0/L int(K.dA)
    hows that?

    surely my previous argument proves there is no field outside a solenoid? why cant i use it similarly here?
     
  9. Apr 24, 2009 #8

    Matterwave

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    This isn't a solenoid, it's a plate...

    Uh. Int(K.dl) not dA since K is a surface current, you integrate over a line.

    Draw the thing and see that in the loop, B moves to the right above the plate, and to the left below the plate. Thus B is parallel to dl through the top and bottom lines and perpendicular to dl in the left and right lines.

    What do you get for that?
     
  10. Apr 24, 2009 #9
    BL-B(-L)=mu_0 KL implies B=mu_0 K/2 from each plate presumably

    two questions:
    (i) are the field lines from these plates similar to those for a current loop magnetic dipole? i.e. they run parallel to the surface and then curve back round onto the other end of the surface?
    (ii)we aren't told the directions of the two currents so how do we know whether they superimpose or cancel each other out?
     
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