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Homework Help: Ampere's Law: Conceptual

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A long straight cylindrical shell has an inner radius Ri and an outer radius Ro. It carries a current i, uniformly distributed over its cross section. A wire is parallel to the cylinder axis, in the hollow region (r < Ri). The magnetic field is zero everywhere outside the shell (r > Ro). We conclude that the wire:
    1. may be anywhere in the hollow region but must be carrying current i in the same direction as the current in the shell
    2. may be anywhere in the hollow region but must be carrying current i in the direction opposite to that of the current in the shell
    3. does not carry any current
    4. is on the cylinder axis and carries current i in the same direction as the current in the shell
    5. is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell

    (each option is a radio button)

    2. Relevant equations

    [tex]\oint \vec{B} \cdot d\vec{s} = \mu_0 i_{enc}[/tex]

    3. The attempt at a solution

    Total enclosed current is the algebraic sum of each portion of enclosed current. Therefore, to get [tex]\oint B \cdot ds = 0[/tex] any current within the amperian loop (in this case, loop is the outer surface of the shell) must be compensated for by current flowing in the opposite direction. I don't see why the relative positions of the current-carrying bodies are of any relevence. So I said that the answer was 3. But the book says the answer is 5: the wire must be on the cylinder's axis. Can I have a pointer as to why the position of the wire matters, as long as the wire is within the amperian loop?

    edit: stupid mistake. I said that the answer was 2 not 3.
    Last edited: May 8, 2007
  2. jcsd
  3. May 8, 2007 #2
    Just guessing again, but I also picked the answer 5.

    - It cannot be 3, because the the magnetic field outside the wire would not be 0 due to the current carried by the conductor
    -And, the ll wire cannot carry current in the same direction because, that would amplify the magnetic field created by the outer cylinder (since B is dir. proportional to the current)
    -and now, that leaves only 5 and 2.

    If, it is 2, then that means one side of the cylinder would be more closer to the ll wire than the other, so there would be different B fields on the two opposite sides of the cylinder.

    So, that means the answer is 5.

    I would also add that the current the ll wire carries is more than the current carried by the cylinder. Because the magnitude of the B field created by the ll wire must be equal, and opposite in the direction to the B field created by the cylinder.

    Use right hand rule, to find the directions of the B fields created by both things, and so perhaps that would surely help
  4. May 8, 2007 #3
    Lets look at the shell. Outside the shell the magnetic field is as if all the current is concentrated at the center. Now firstly inorder to cancel the magnetic field of the present shell we'll need a wire carying current in the opposite direction, using the right hand rule you will see that the fields of the shell will cancel with the fields of the wire everywhere outside the shell if the wire was placed in the center of the shell.

    This is similar to electric fields, if you have a uniformly distributed shell of charge q and measure its E outside the shell then it will apear as if all of q is concentrated in the center of the sphere. Similarly can be drawn for gravatational bodies.

    Magnetic fields are vectors and you can add them using super position. If you try placing the wire in other places the magnetic fields will not cancel everywhere outside the shell, you can use the right hand rule to see this.

    Hope this helps!
  5. May 8, 2007 #4
    erm... if I told you that I actually answered 2, and the 3 is a typo, would you believe me? Because it's true.:redface:

    My thoughts exactly.

    That makes sense, but how does the equation reflect that?

    I don't think so. The currents must be equal in order to cacel each other; it is the current density of the wire that must be greater than the current density of the cylinder.
  6. May 8, 2007 #5
    So far, so good.

    But why won't it cancel everywhere? Assume that the cylinder's magentic field acts like a long wire along the cylinder's axis. If I put a wire with equal and opposite current anywhere in the same amperian loop as this hypothetical wire, the integral will equal 0. The wires don't need to be on top of each other!
    (Clearly, I'm wrong. I just don't see why.)

    edit: And thanks, rootX and Four, for your help!
  7. May 9, 2007 #6
    I think they wudn't cancel cuz
    B is inv. proportional to r,

    so, if the ll wire is towards the left of the central axis, then it's B field wud be out of phase with the B field of the cylinder
    (I really cannot describe my visualization, but try to visualize it may help.)
  8. May 9, 2007 #7
    My visualization is of two sets of circular ripples... Oh, forget it. See attachment for rough sketch. Scenario A (wire and cylinder not co-axial) won't cancel out. Scenario B (wire on cylinder's axis) does cancel. (B is purple because how else can I draw red and blue circles on top of each other?)
    So when I visualize it, it makes sense.
    But why why why does the equation seem to neglect the positioning of the enclosed current-carrying objects?

    Attached Files:

    Last edited: May 9, 2007
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