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Finding current density given B-field

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the current density as a function of distance ##r## from the axis of a radially symmetrical parallel stream of electrons if the magnetic induction inside the stream varies as ##B=br^a##, where ##b## and ##a## are positive constants.


    2. Relevant equations



    3. The attempt at a solution
    Honestly, I don't have any idea for this problem. I don't see any equations in my notes which relates the B-field and current density.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. May 25, 2013 #2

    haruspex

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    No, but I'm sure you have an equation for the field based on the current in a wire. Won't it just be a matter of integrating the density, viewed as many wire elements?
     
  4. May 25, 2013 #3
    Ampere's circuital law?

    [tex]\int \textbf{B}\cdot d\textbf{l} = \mu_o\int \textbf{J} \cdot d\textbf{S}[/tex]
    What should I replace dl with? ##2\pi r##?
     
  5. May 25, 2013 #4

    ehild

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    Because of radial symmetry, the vector B is tangential to the circles around the axis and its magnitude depends only on the distance from the axis. dl=rdθ and yes, the line integral is B(2πr).

    ehild
     
  6. May 25, 2013 #5
    That gives me
    [tex]2\pi b r^{a+1}=\mu_o \int \textbf{J} \cdot d\textbf{S}[/tex]
    Is ##dS=2\pi rdr##?

    And how am I supposed to solve the above equation? :confused:
     
  7. May 25, 2013 #6

    ehild

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    Plug in the expression for dS. The integral is valid for all r. You get an equation of form f(r)=0r∫g(r')dr'. How do you get g(r)?

    ehild
     
  8. May 25, 2013 #7
    Thank you ehild! :smile:

    I get
    [tex]br^{a+1}=\int_0^{r}\textbf{J}rdr[/tex]
    Differentiating both the sides,
    [tex]\Rightarrow (a+1)br^{a}=\textbf{J}r[/tex]
    [tex]\Rightarrow \textbf{J}=(a+1)br^{a-1}[/tex]
     
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