Parabola Magnetic Field

The magnetic field due to a charge $q$ moving with velocity $\mathbf{v}$ at a point with displacement $\mathbf{r}$ from the particle is $\frac{\mu_0}{4\pi} \frac{q\mathbf{v} \wedge \mathbf{r}}{|\mathbf{r}|^3}$ Thus the magnetic field due to a current $I$ running along a curve $C$ at a point $P$ with displacement $\mathbf{r}$ is $\mathbf{B}_P \; = \; \frac{\mu_0I}{4\pi}\int_C \frac{d\mathbf{s} \wedge (\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3}$ Thus there is no magnetic field at the point $(0,1)$ due to the semi-infinite straight segments of wire. To integrate along the parabolic curve, we adopt the parametrization $\mathbf{s} = (t,t^2)$ for $-1 \le t \le 1$ , and obtain the magnetic field $\mathbf{B}_{(0,1)} \; = \; \frac{\mu_0I}{4\pi} \int_{-1}^1 \frac{1 + t^2}{\big(t^2 + (1-t^2)^2\big)^{\frac32}}\,dt \;\mathbf{k}$ where $\mathbf{k}$ is a unit vector perpendicular to the plane. Thus $\alpha \; = \; \int_{-1}^1 \frac{1 + t^2}{\big(t^2 + (1-t^2)^2\big)^{\frac32}}\,dt \; = \; \boxed{3.42211}$ after a numerical evaluation of the integral.

This solution is written by me, a person who knows very little about this sort of physics. I just wanted to show a way of solving this question that did not require any knowledge of magnetism or vectors, and simply involved me looking up a formula and doing a lot of probably unnecessary problem solving.

Searching around online, I found this website , which states this formula: $dB=\frac{\mu I\sin\theta}{4\pi R^2}dL$ $dL$ is the length of an arbitrarily small part of the wire, $R$ is the distance of that part of the wire from the point we are calculating the magnetic flux of, and $\theta$ is the angle from the part of the wire to the point. Integrating this for the length of the wire, we will get the answer.

It is clear to see that for the straight section of the wire, $\theta=0$ , and therefore the straight section will have no impact on the magnetic flux.

Now let us get everything in terms of x. $dL$ is a length of wire, so it can be found by the Pythagorean theorem: $dL=\sqrt{(dx)^2+(dy)^2}\\ \text{But since }y=x^2,dy=2xdx\\ \text{So }dL=\sqrt{(dx)^2+(2xdx)^2}\\ dL=\sqrt{1+4x^2}dx$ R will be the distance from a point $(x,y)$ on the curve to the point $(0,1)$ , which equals: $\begin{aligned}R&=\sqrt{x^2+(y-1)^2}\\ &=\sqrt{x^2+(x^2-1)^2}\\ &=\sqrt{x^4-x^2+1}\end{aligned}$ Finding $\theta$ in terms of x is more tedious. We know that for a point $(x,y)$ on the curve, the slope of the tangent line is $2x$ , and the slope of the line from the point to $(0,1)$ is $\frac{y-1}{x-0}=\frac{x^2-1}{x}$ . Therefore $\theta$ is just the difference between these two angles: $\begin{aligned}\theta&=\tan^{-1}\left(2x\right)-\tan^{-1}\left(\frac{x^2-1}{x}\right)\\ \sin\theta&=\sin\tan^{-1}\left(2x\right)\cos\tan^{-1}\left(\frac{x^2-1}{x}\right)-\sin\tan^{-1}\left(2x\right)\cos\tan^{-1}\left(\frac{x^2-1}{x}\right)\\ &=\frac{2x}{\sqrt{1+4x^2}}\frac{x}{\sqrt{x^2+(x^2-1)^2}}-\frac{x^2-1}{\sqrt{x^2+(x^2-1)^2}}\frac{1}{\sqrt{1+4x^2}}\\ &=\frac{x^2+1}{\sqrt{1+4x^2}\sqrt{x^4-x^2+1}}\end{aligned}$ Putting these together, we can integrate and find a value for B: $\begin{aligned}dB&=\frac{\mu I\sin\theta}{4\pi R^2}dL\\ &=\frac{\mu I}{4\pi (x^4-x^2+1)}\frac{x^2+1}{\sqrt{1+4x^2}\sqrt{x^4-x^2+1}}\sqrt{1+4x^2}dx\\ &=\frac{\mu I(x^2+1)}{4\pi (x^4-x^2+1)^\frac{3}{2}}dx\\ B&=\frac{\mu I}{4\pi}\int_{-1}^1\frac{x^2+1}{(x^4-x^2+1)^\frac{3}{2}}dx\\ &\approx\frac{\mu I}{4\pi}\times3.4221\end{aligned}$ Therefore $\alpha=\boxed{3.4221}$