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Magnetic Field Integral

  1. Feb 11, 2009 #1
    I need to do the following integral:

    [itex]B_z=\frac{\mu_0 I N a^2}{2} \int_{0}^{L} (a^2+(z-z')^2)^{-\frac{3}{2}} dz'[/itex]

    so far i have set [itex]x=z-z' \Rightarrow dx=-dz'[/itex] and the limits also change to give

    [itex]B_z = - \frac{\mu_0 I N a^2}{2} \int_{z}^{z-L} \frac{dx}{(a^2+x^2)^{\frac{3}{2}}}=\frac{\mu_0 I N a^2}{2a^3} \int_{z-L}^{z} \frac{dx}{(1+(\frac{x}{a})^2)^{\frac{3}{2}}}[/itex]

    im having trouble with the power of 3/2 as otherwise its just a standard integral to arctan. any advice?
     
  2. jcsd
  3. Feb 11, 2009 #2

    Hootenanny

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    You can evaluate this integral using exactly the same substitution. Try letting

    [tex]\frac{x}{a} = \tan\theta[/tex]

    and see where it takes you.
     
  4. Feb 11, 2009 #3
    cheers. this gives

    [itex]\frac{\mu_0 I N}{2a} \int_{\arctan{(\frac{z-L}{a})}}^{\arctan{(\frac{z}{a})}} \frac{1}{\sec{\theta}} d \theta =\frac{\mu_0 I N}{2a} \int_{\arctan{(\frac{z-L}{a})}}^{\arctan{(\frac{z}{a})}} \cos{\theta} d \theta =\frac{\mu_0 I N}{2a} [\sin{\theta}]_{\theta=\arctan{(\frac{z-L}{a})}}^{\arctan{(\frac{z}{a})}}[/itex]

    how do i evaluate sin theta at these points though?
     
  5. Feb 11, 2009 #4

    nicksauce

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    Draw a right triangle, such than one of the angles, theta, has arctan z/a. Then find the sin of that angle.
     
  6. Feb 12, 2009 #5
    i can draw it if tan theta = z/a but what would it look like if arctan theta = z/a?
     
  7. Feb 12, 2009 #6
    You have something like sin(arctan z/a). The term inside the bracket is an angle (let's call it alpha). As it is an angle, you can express it as arcsin something. So you need to know sine alpha, right? And you already know tan alpha.
     
  8. Feb 12, 2009 #7
    so now i get:

    [itex]B_z=\frac{\mu_0 I N}{2a} [\frac{z}{\sqrt{z^2+a^2}}-\frac{z-L}{\sqrt{(z-L)^2+a^2}}][/itex]

    is there anyway to further simplify this or am i done?
     
  9. Feb 12, 2009 #8

    gabbagabbahey

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    You seem to have an extra factor of [tex]\frac{1}{a}[/tex]....It probably arose when you did your substitution: [tex]dx=a\sec^2\theta d \theta[/tex]
     
  10. Feb 12, 2009 #9
    cheers. but the rest of it is correct?
     
  11. Feb 13, 2009 #10

    gabbagabbahey

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