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Magnetic field of a cylinder

  1. Dec 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider an infinitely long solid metallic cylinder having axis along kˆ.Consider a plane passing through axis of cylinder cutting it in two equal parts. In one part is a uniformly distributed current I1kˆ and in another part is a uniformly distributed current −I2kˆ. As always, task is simple, find the magnitude of magnetic field on the axis of cylinder in μT.


    2. Relevant equations
    For this problem, since we have an infinite cylinder, I have decided to try to use Ampere's law:

    [itex]\oint \vec{B} \vec{dl} [/itex] = μ0 i



    3. The attempt at a solution

    I treat this cylinder as 2 separate ones, each having differet current. Applying Ampere's law to the "first cylinder" I get:

    B1 (Rπ/2) = μ0 I1 ==> B1 = 2μ0 I1 /Rπ

    B2 is then:

    B2 = 2μ0 I2 /Rπ

    The resulting magnetic field will be [itex]\vec{B}[/itex] = [itex]\vec{B1}[/itex] + [itex]\vec{B 2}[/itex]

    where B2 should be negative, since I2 flows in an opposite direction.

    I would like to know if this method would work, and if I had written the equations properly. Thank you very much!
     
  2. jcsd
  3. Dec 24, 2013 #2

    Simon Bridge

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    I'm not clear if the plane cutting the cylinder is along the axis or perpendicular to it.
    If perpendicular, there is a problem with where the current goes.
    If along, then the currents occupy a "D" crossection of a cylinder, not the whole cylinder.

    The temptation, to me, is to use a symmetry argument instead (as well?)
    But your approach should get the same result.

    Merry Xmas.
     
  4. Dec 25, 2013 #3

    rude man

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    What is your R and how did you come up with your formula for B?

    The dissection must be along the z axis, otherwise the stated currents could not exist.

    If you draw an amperian circle of arbitrarily small radius centered on the z axis you can readily determine what the B field along the axis must be. Even in light of the asymmetry of the current density around the axis.
     
  5. Dec 25, 2013 #4

    TSny

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    I don't see how you can use Ampere's law for this problem - there just isn't enough symmetry. But it would be nice if I'm mistaken!

    Brute force :cry: - break the current distribution into elements such that each element of current produces a magnetic field on the axis that you can easily calculate. (You could use Ampere's law to get the field of an individual element). Then integrate the contributions from all of the elements.
     
  6. Dec 25, 2013 #5

    rude man

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    EDIT (again!)

    I changed my mind - again. The best way to solve it that I can think of is by mag. potential.
     
    Last edited: Dec 25, 2013
  7. Dec 25, 2013 #6

    rude man

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    If you used mag. potential I think you could eliminate the current in half the wire completely and integrate the other half with current I1 - I2. Also of course you only have to integrate half the infinite wire & multiply by 2.

    You could choose a d-shaped slice of wire of thickness dz, divide it into half-annuli of radial width dr, and compute the potential due to one half-annulus and then to the total slice, then integrate z from 0 to infinity. Use cylindrical coordinates.
     
  8. Dec 25, 2013 #7

    rude man

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    Also, I don't think this problem belongs in the introductory physics forum unless we are all missing something.
     
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