# Magnetic field around sinusoid shaped wire

malganis99

## 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:

Homework Helper
Gold Member
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 $a=0$?

malganis99
$(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) ya$