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Magnetic field from the displacement current of a charging sphere

  1. Jul 18, 2014 #1
    1. The problem statement, all variables and given/known data

    The problem is #1 given here:
    http://ocw.mit.edu/courses/physics/8-311-electromagnetic-theory-spring-2004/assignments/ps1.pdf
    And there is a solution publicly available here:
    http://ocw.mit.edu/courses/physics/8-311-electromagnetic-theory-spring-2004/assignments/solution1.pdf

    In short, a conducting sphere is being charged through a straight wire with given current I. You are asked to find the displacement current, and finally the magnetic field.

    My question pertains to finding the magnetic field - I don't understand how the displacement current was used to find the magnetic field.

    2. Relevant equations

    The displacement current was found to be:

    [tex] j_d = \frac{I}{4 \pi r^2} \hat{r} [/tex]

    To find the magnetic field, use

    [tex] \int B \cdot dl = \frac{4\pi}{c} \int j \cdot \hat{n} dA[/tex]

    where j is the current density and n is the unit vector normal to the surface.

    3. The attempt at a solution

    For the contour, we consider a loop about the axis of the wire (there is a figure in the solutions). The distance from the center of the sphere to the loop is r. The angle between the loop and the axis of the wire is θ.

    On the left hand side, we thus have

    [tex] \int B \cdot dl = 2 \pi r \sin{(\theta)} B [/tex]

    The surface we are using for the right hand side is a section of a sphere of radius [tex]r[/tex]. So we need to integrate the displacement current (since there is no actual current through this surface, only displacement) over this region:

    [tex] \frac{4\pi}{c} \int j \cdot \hat{n} dA = \frac{4\pi}{c} \int_0^{2\pi} d\phi \int_0^{\theta'} d\theta \sin{\theta} \frac{I}{4 \pi r^2} \hat{r} \cdot \hat{r} [/tex]

    [tex] = \frac{2\pi I}{c r^2} (1-\cos \theta)[/tex]

    Combine this with the left hand side from above:

    [tex] 2 \pi r \sin{(\theta)} B = \frac{2\pi I}{c r^2} (1-\cos \theta)[/tex]
    [tex] B = \frac{I}{c r^3 \sin{\theta}} (1-\cos \theta)[/tex]

    Finally, the question: why is the answer in the solutions given off by a factor [tex] r^2 [/tex]? Why does this factor from the displacement current disappear in the worked solutions?

    Thanks for the help

    Edit: the given solution is
    [tex] B = \frac{(1-\cos \theta) I}{cr \sin \theta} [/tex]
     
  2. jcsd
  3. Jul 18, 2014 #2
    Solved, [tex] j_d = I_d / 4\pi r^2 [/tex] is the displacement current density - the enclosed current is clearly [tex] I_d [/tex]...
     
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