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B inside a cylinder of radius R

  1. Mar 13, 2005 #1
    hey all, this is confusing me a lot:

    consider an infinitely long cylinder of cross-section radius R. we choose symmetry axis of the cylinder as the z-axis. The cylinder carries a uniform current density J in the +z direction throughout it's cross section. what is B at r inside of the cylinder? Express you answer in the component form B = Bx i + By j + Bz k


    ...what i'm confused about is whether or not I can use ampere's law with an amperian loop inside the cylinder or if i have to use Biot-Savart.....

    i did it using ampere's law and i got |B| = uJs/2 ....and the vector B is in the phi direction.....wrapping around the z-axis.......did i do that right? how do i get it into cartesian coords? Do i have to use Biot-Savart in to be able to get it in cartesian coords regardless of whether or not ampere's law can be used?
     
    Last edited: Mar 13, 2005
  2. jcsd
  3. Mar 13, 2005 #2

    Andrew Mason

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    It is a fairly simple Ampere's law problem. The line integral of the magnetic field around a circle at radius r is just:

    [tex]\oint \vec B\cdot ds = \mu_0I_{encl}= \mu_0AJ = \mu_0\pi r^2J[/tex]

    due to symmetry, |B| is constant and always in the direction of ds so:

    [tex]\oint \vec B\cdot ds = B2\pi r[/tex]

    [tex]B2\pi r = \mu_0\pi r^2J[/tex]

    [tex]B = \mu_0rJ/2[/tex]

    At a given point [itex]\vec r = x\hat i + y\hat j[/itex], the magnetic field vector is perpendicular to the radial vector. So By/Bx = x/y. Divide by r to get the unit vectors.

    [tex]\vec B = B\frac{y}{r}\hat i + B\frac{x}{r}\hat j[/tex]

    where [tex]B = \mu_0rJ/2[/tex]

    AM
     
  4. Mar 13, 2005 #3
    ok thanks man...that makes sense....ok so thaty's B for inside the cylinder....now whatabout outside?
     
  5. Mar 14, 2005 #4

    Andrew Mason

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    Outside the enclosed current is the entire current in the cylinder. So:

    [tex]B = \mu_0J\pi R^2/2\pi D[/tex] where D is the distance from the centre.

    AM
     
  6. Mar 14, 2005 #5
    IN a right handed coordinate system, Should that be

    [tex]\vec B = -B\frac{y}{r}\hat i + B\frac{x}{r}\hat j[/tex]
     
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