Let $I$ be an electric current inside an infinite helix $H$ which is given by the following parametrization:
$$H(t)=(\cos(t),\sin(t),t)\\t\in(-\infty,\infty)$$
Find the magnetic field $\vec{B}$ that is generated by the current $I$ at the origin $(0,0,0).$
My Attempt
We know by Biot-Savart Law that the magnetic field $\vec{B}$ generated by the current $I$ in a curve $\Gamma(t):[a,b]\to\mathbb{R}$ is given by:
$$\vec{B}=\frac{\mu_0I}{4\pi}\int_a^b\frac{d\vec{l}\times(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3}dt$$
When $d\vec{l}$ is a length element of the curve, $\vec{r}$ is a vector from the origin to the point where we calculate the field, and $\vec{r}'$ is a vector from the origin to a general point of the curve.
Thus, in our case:
$$\vec{r}=(0,0,0)\\ \vec{r}'=(\cos(t),\sin(t),t)\\ d\vec{l}=(-\sin(t),\cos(t),1)\ dt$$
After performing the cross product, we receive the following integral:
$$\vec{B}=\frac{\mu_0I}{4\pi}\int_{-\infty}^\infty\frac{(\sin(t)-t\cos(t))\hat{x}+(-\cos(t)-t\sin(t))\hat{y}+\hat{z}}{(1+t^2)^{3/2}}dt$$
Now, I have two problems with this integral:
- I don't know how to calculate the $x$ and $y$ components.
- I did notice, though, that the $x$ component is a-symmetric; thus it means the $x$ component should be $0$. That seems physically weird to me, since I would expect either both of the $x$ and $y$ components to cancel, or both of them not to. I don't have symmetry here, so I think I might've done something wrong.
Another thing that I noticed, is the $z$ component, which I can actually compute, is identical to the $z$ component I would get if instead of the helix, the current would flow through a circle laying in the plane $z=0$. I found that rather interesting.
Since this problem is actually mathematical (rather than physical), I came here for your help. Thank you!
You are correct that the $x$ integral vanishes by asymmetry. Since you don't seem to have a problem with the $z$ integral, that leaves the $y$ integral: $$ B_y = -\frac{\mu_0 I}{4\pi}\int_{-\infty}^\infty \frac{\cos(t)+t\sin(t)}{(1+t^2)^{3/2}}dt $$
To do this integral, first integrate by parts on the $t\sin(t)$ term, then use this integral representation of the Bessel $K$ function: $$ K_\nu(z) = \frac{2^\nu}{z^\nu}\frac{1}{\sqrt{\pi}}\int_0^\infty \frac{\cos(z t)}{(1+t^2)^{\nu+1/2}}dt. $$
You should also try working this out for the helix $$ H(t) = \{R\cos(t),R\sin(t),ht/(2\pi)\}, $$ that is, the helix of radius $R$ and spacing $h$ between the coils. Then show in the limit that $h/R\rightarrow 0$, you recover the expression for the field of an ideal solenoid.