Magnetic field around sinusoid shaped wire

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The discussion revolves around the effects of sinusoidal deformation on the magnetic field generated by a long straight wire. The user applies the Biot-Savart law to derive the magnetic field expressions while attempting to account for the deformation's amplitude. They explore the possibility of simplifying the calculations by expanding terms in a Taylor series due to the small amplitude assumption. The user encounters difficulties with Mathematica in solving the integrals and considers using NIntegrate for numerical solutions. Additionally, they express uncertainty about how to visualize the magnetic field, suggesting a potential need for further assistance in a different forum.
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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

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

The Attempt at a Solution



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

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

d\vec{l} = \vec{\xi}(t)dl

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

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

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

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

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

B_{z}=\frac{μ_{0}*I}{4\Pi}\int\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}}dt

When I expanded denominator in Biot_savart law I ignored the terms with a^{2} 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 a is small, why not expand \left[ (x-t)^2 + (y-a\sin t)^2 + z^2 \right]^{-\frac{3}{2}} in a Taylor series around a=0?
 
Thanks for the reply.

I expanded that to the first order of a.

(t^2 - 2 t x + x^2 + y^2 + z^2)^{3/2} - 3\sqrt{t^2 - 2 t x + x^2 + y^2 + z^2}Sin(t) you

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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