Magnetic field around sinusoid shaped wire

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SUMMARY

The discussion focuses on the magnetic field generated by a long straight wire deformed into a sinusoidal shape. The magnetic field is expressed using the Biot-Savart law, with specific equations derived for the components Bx, By, and Bz. The user attempts to simplify the calculations by expanding the denominator in the Biot-Savart law using a Taylor series around a small amplitude 'a'. The user also seeks guidance on using Mathematica for integral calculations and plotting the magnetic field.

PREREQUISITES
  • Understanding of the Biot-Savart law for magnetic fields
  • Familiarity with vector calculus and cross products
  • Knowledge of Taylor series expansions
  • Experience with Mathematica for numerical integration and plotting
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  • Learn how to implement Taylor series expansions in Mathematica
  • Explore advanced applications of the Biot-Savart law in electromagnetic theory
  • Research methods for visualizing vector fields in Mathematica
  • Study the effects of wire deformation on magnetic field patterns
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Students and professionals in physics and engineering, particularly those studying electromagnetism, vector calculus, and computational methods in Mathematica.

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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.
 
Last edited:
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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]?
 
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) you[/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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