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Current Density

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Capture.JPG

    2. Relevant equations

    J=i/a

    3. The attempt at a solution

    Studying for a test. This sample exam went pretty well. I am having trouble with 3 out of the 20. I can't figure this one out. I can't seem to find how to get a denominator value.

    The answer is B.

    I know that current density is equal to the current over the area.
    [tex]J=\frac{i}{A}[/tex]
    [tex]J_o(r/R)=\frac{i}{\pi (\frac{r}{R}^2)}[/tex]

    I know I'm doing something wrong. Anyone help?
     
  2. jcsd
  3. Oct 31, 2009 #2
    You are right at this formula:

    [tex]
    J = \frac{i}{A}
    [/tex]

    But it is just the special case of this:

    i = [tex]\int[/tex] JdA
    and dA = 2 * pi * r * dr.
     
  4. Oct 31, 2009 #3
    Oh. Right.

    I still don't know how to get a correct answer out of that.

    [tex]i = \int J \cdot da [/tex]

    [tex]i = \int_0^r \frac{J_or}{R} \cdot 2\pi r dr [/tex]

    pulling out the constants:

    [tex]i = \frac{J_o2\pi}{R} \int_0^r r^2 dr [/tex]

    [tex]i = \frac{J_o2\pi}{R}\frac{r^3}{3}|_0^r[/tex]

    [tex]i = \frac{J_o2\pi}{R}\frac{r^3}{3}[/tex]

    Where's the extra r in the numerator coming from?
     
  5. Oct 31, 2009 #4
    When you set r = R, you will get what you want (because problem asks you to find total current, not current at radius r, i think).
     
    Last edited: Oct 31, 2009
  6. Oct 31, 2009 #5
    Oh. because I'm integrating from a radius of 0 to a radius of R. That makes sense.

    [tex]
    i = \frac{J_o2\pi}{R}\frac{r^3}{3}|_0^R
    [/tex]

    [tex]
    i = \frac{J_o2\pi}{R}\frac{R^3}{3}
    [/tex]

    [tex]
    i = \frac{J_o2\pi R^2}{3}
    [/tex]

    Thanks :)
     
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