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Magnetic field around sinusoid shaped wire

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a long straight wire which is slightly deformed into sinusoid shape. How does the magnetic field change with deformation? Can you express the magnetic field change linearly with sinusoid amplitude?


    2. Relevant equations

    parametrized sinuosid equation

    x = t
    y = a*sin(t)
    z=0

    [itex]\vec{B}[/itex]=[itex]\frac{μ_{0}*I}{4\Pi}[/itex][itex]\int[/itex][itex]\frac{d\vec{l}×(\vec{r}'(t)-\vec{r}(t))}{|\vec{r}'(t)-\vec{r}(t)|^{3}}[/itex]

    3. The attempt at a solution

    [itex]\vec{r}[/itex](t)=(t, a*sin(t),0)
    [itex]\dot{\vec{r}}[/itex](t)=(1,a*cos(t),0)

    [itex]\vec{\xi}[/itex](t)=[itex]\frac{\dot{\vec{r}}(t)}{|\dot{\vec{r}}(t)|}[/itex]=[itex]\frac{(1,a*cos(t),0)}{\sqrt{1+a^{2}*cos^{2}(t)}}[/itex]

    d[itex]\vec{l}[/itex] = [itex]\vec{\xi}[/itex](t)dl

    dl=|[itex]\frac{d\vec{r}}{dt}[/itex]|*dt=[itex]\sqrt{1+a^{2}*cos^{2}(t)}[/itex]*dt

    The cross product in Biot-Savart law
    d[itex]\vec{l}[/itex]×([itex]\vec{r}[/itex]'(t)-[itex]\vec{r}[/itex](t))=((1,a*cos(t),0)×(x-t,y-a*sin(t),z))*dt where [itex]\vec{r}[/itex]'(t) is a point in space (x,y,z)

    [itex]|\vec{r}[/itex]'(t)-[itex]\vec{r}[/itex](t)|[itex]^{3}[/itex]= [itex]\sqrt{((x-t)^{2}+(y-a*sin(t))^{2}+z^{2})}[/itex][itex]^{3}[/itex]

    [itex]\vec{B}[/itex]=[itex]\frac{μ_{0}*I}{4\Pi}[/itex][itex]\int[/itex][itex]\frac{(z*a*cos(t)\hat{i}-z\hat{j}+(y+a(t*cos(t)-x*cos(t)-sin(t))\hat{k})}{\sqrt{(x-t)^{2}+(y-a*sin(t))^{2}+z^{2})}^{3}}[/itex]dt and integral goes from -∞ to +∞


    B[itex]_{x}[/itex]=[itex]\frac{μ_{0}*I}{4\Pi}[/itex][itex]\int[/itex][itex]\frac{z*a*cos(t)\hat{i}}{(x^{2}+y^{2}+z^{2}-2xt+t^{2}-2y*a*sin(t))^{3/2}}[/itex]dt

    B[itex]_{y}[/itex]=[itex]\frac{μ_{0}*I}{4\Pi}[/itex][itex]\int[/itex][itex]\frac{-z\hat{j}}{(x^{2}+y^{2}+z^{2}-2xt+t^{2}-2y*a*sin(t))^{3/2}}[/itex]dt

    B[itex]_{z}[/itex]=[itex]\frac{μ_{0}*I}{4\Pi}[/itex][itex]\int[/itex][itex]\frac{(y+a(t*cos(t)-x*cos(t)-sin(t)))\hat{k}}{(x^{2}+y^{2}+z^{2}-2xt+t^{2}-2y*a*sin(t))^{3/2}}[/itex]dt

    When I expanded denominator in Biot_savart law I ignored the terms with a[itex]^{2}[/itex] because they are negligible if a is small.

    When I put these integrals in Mathematica it can't solve them. Should I approach the problem differently? Are my calculations correct?

    Thanks for the help.
     
    Last edited: Sep 18, 2012
  2. jcsd
  3. Sep 22, 2012 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    As [itex]a[/itex] is small, why not expand [itex]\left[ (x-t)^2 + (y-a\sin t)^2 + z^2 \right]^{-\frac{3}{2}}[/itex] in a Taylor series around [itex]a=0[/itex]?
     
  4. Sep 23, 2012 #3
    Thanks for the reply.

    I expanded that to the first order of a.

    [itex](t^2 - 2 t x + x^2 + y^2 + z^2)[/itex][itex]^{3/2}[/itex] - [itex]3\sqrt{t^2 - 2 t x + x^2 + y^2 + z^2}[/itex][itex]Sin(t) ya [/itex]

    I can calculate the integrals with mathematica using NIntegrate. I don't know how to plot the field though. As that is mathematica question I think it's better if I post it in another forum.
     
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