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Find magnetic flux density(B) via vector potential(A)

  1. Apr 22, 2014 #1
    Hello
    As you know we can find the flux density (B) via vector potential with this equation
    gif.gif
    So with this equation and solution at first i try to find vector potential completely then use above equation

    Problem:
    7085326300_1398155382.jpg
    Solution:
    5925891300_1398155383.jpg

    my approach:
    gif.gif

    gif.gif

    gif.gif

    now we use curl A to find B
    5024359900_1398156076.jpg
    we find that
    gif.gif

    but it is different form the book soltuion

    what is my problem and wrong?

    Thanks
     
    Last edited: Apr 22, 2014
  2. jcsd
  3. Apr 22, 2014 #2

    TSny

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    Note how you wrote R in terms of x, y, z. Is the 3/2 power correct?
     
  4. Apr 25, 2014 #3
    Thanks Dear Tsny.you'r correct
    as i want to solve the new integral
    Cvec%7Bay%7Ddxdy%7D%7B%5Cpi%20%28x%5E2+y%5E2+z%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D.gif
    a new problem appears

    for example if i want to calculate the inner integral the integral going to be infinite.
    8967079300_1398439030.jpg

    %7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%3Dln%28+%5Cinfty%20%29-ln%28%5Csqrt%7By%5E2+z%5E2%7D%29.gif

    Ln(infinite)=infinite

    now how can solve this problem?
    Thanks
     
  5. Apr 25, 2014 #4

    TSny

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    Yes, the integral is infinite. It's similar to trying to finding the potential V of an infinite line charge by integrating dV = k dq/r. Here you will also get a divergent integral. This is due to the charge distribution extending to infinity while at the same time trying to take the potential to be zero at infinity. (Note kdq/r goes to zero at infinity.)

    You can avoid the problem by not demanding that the potential be zero at infinity. For example, in the infinite line charge example, you can pick an arbitrary point in space to be zero potential. The potential at other points is then defined relative to this chosen point.

    You can try a similar trick for the vector potential. Thus define A as the following (with constants left out)

    $$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\sqrt{x^2 +y^2+z^2}} dx dy - \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\sqrt{x^2 +y^2+z_0^2}} dx dy$$

    ##z_0## is just an arbitrarily chosen point on the z axis where you are taking A to be 0.

    Each integral individually diverges. But if you combine the integrals before letting the upper limits go to infinity, you get a finite result that you can use for A.
     
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