Ampère's law with ellipse

Question

Background information first: Using Ampère's law we can find the magnetic field due to a current, I. We do this with a line integral around a closed path encompassing the current (e.g. long straight wire).

$\oint_c \vec B \cdot \vec dl$ = $\mu_0 I_{enc}$

If we choose a circle centered on the wire (like the images below) we greatly simplify things because the magnetic field (right-hand-rule) is perpendicular to the current flow. With the selected path (circle) the B field is colinear with the circle so the dot product between $\vec B$ and $\vec dl$ reduces to |B||dl| since cosine $0^\circ$ is 1. Right-hand figure of image below is looking down from above with wire in center of circle.

This choice of using a circle for the closed path around the current conveniently reduces the problem to that of finding the circumference of the circle, resulting in the following equation for the magnetic field.

$B = \frac{\mu_0 I_{enc}}{2\pi r}$

So, i wanted to see if i could derive an equation for B using an ellipse instead.

In this case, since the magnetic field is perpendicular to the flow of current the incremental length $\vec dl$ and $\vec B$ will only be colinear at 4 points (on the axes). EDIT - i redrew the right-hand image above to clarify the relationship between the direction of $\vec B$ and the direction of $\vec dl$ at a point on the ellipse. The dashed-red circle is just used to find the direction of $\vec B$ which is tangential to the circle.

However, we know the slope of the ellipse at each (x,y) is $\frac{-xb^2}{ya^2}$. We know that the slope of $\vec B$ at (x,y) is the slope of a circle that has the same origin as the ellipse and passes through (x,y). So, the slope of $\vec B$ is $\frac{-x}{y}$...and now we know the slope of both the ellipse and the $\vec B$ at (x,y).

For the dot product we need $\cos(θ)$. We know that $\tan(\theta) = \dfrac{m_2-m_1}{1+m_1m_2}$ so knowing the 2 slopes we have $\tan(\theta)$. Knowing that $\cos^2 = \frac{1}{1+\tan^2}$ we end up with the following:

$\cos^2(\theta) = \frac{1}{1+|\frac{-xya^2+xyb^2}{a^2y^2+x^2b^2}|^2}$

Knowing the circumference of the ellipse, P, I believe that Ampère's integral is now as follows.

$BP\oint_c \cos(\theta)$ = $\mu_0 I_{enc}$

which reduces to,

$B = \frac{\mu_0 I_{enc}}{P\oint_c \cos(\theta)}$

Does this appear correct to this point? How can i work this into a form that is comparable to the circle case, $B = \frac{\mu_0 I_{enc}}{2\pi r}$, but for an ellipse? The idea is to be able to calculate the B at any (x,y) around the wire. Since $r = \sqrt{x^2+y^2}$ this is easy with the formula derived from the circle. We also know that at any point (x,y) on the ellipse i should get the same value of B as from $B = \frac{\mu_0 I_{enc}}{2\pi \sqrt{x^2+y^2}}$.

Your help/input/corrections appreciated.

Addendum: This edit below made after reading Andreas and user's comments just to make few comments that i'll dwell on further (not sure if will be useful).

We know that $B = \frac{\mu_0 I_{enc}}{2\pi r}$ which is same as $B = \frac{\mu_0 I_{enc}}{2\pi \sqrt{x^2+y^2}}$.

So, $\oint_c \vec B \cdot \vec dl = \oint_c |\frac{\mu_0 I_{enc}}{2\pi \sqrt{x^2+y^2}}||dl|\cos(\theta)$, where $\theta$ is the angle between $\vec B$ and $\vec dl$ at each (x,y).

Knowing that $\oint_c dl = P$ (perimeter of ellipse) we can simplify the r.h.s,

$= P\frac{\mu_0 I_{enc}}{2\pi} \oint_c |\frac{1}{\sqrt{x^2+y^2}}|\cos(\theta)$

Which, since $\oint_c \vec B \cdot \vec dl$ = $\mu_0 I_{enc}$, we can now write

$\oint_c |\frac{1}{\sqrt{x^2+y^2}}|\cos(\theta) = \frac{2\pi}{P} $

Knowing formula for $\cos^2(\theta)$ this becomes,

$\oint_c |\frac{1}{\sqrt{x^2+y^2}}| \sqrt{\frac{1}{1+|\frac{-xya^2+xyb^2}{a^2y^2+x^2b^2}|^2}} = \frac{2\pi}{P} $

Answer 1

The calculation using the circular path depended on two things working out very nicely: (1) The direction of the magnetic field is tangent to the circle at every point of the circle. (2) The magnitude of the magnetic field is the same at all points of the circle. When you modified the calculation to use an ellipse, you correctly noted that (1) is no longer true, and you worked out the relationship between the field's direction and the ellipse's tangent. But you seem to have neglected the fact that (2) is also no longer true.

Fact (2) was true in the circular case because of the symmetry of the set-up. The magnitude of the magnetic field depends only of the distance from the current that causes the field, so it's constant along your circle. But it's not constant along your ellipse. So you can't just factor $B$ out of the integral as if it were constant.