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malganis99
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Homework Statement
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?
Homework 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]
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.
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