Magnetic Fields from Two Elliptical Wires

This boils down to an application of the Biot-Savart Law for current-carrying loops:

$B = \frac{\mu_{0} I}{4\pi} \cdot \oint \frac{1}{R} d\theta$ (i)

For the circular loop, we compute $B_{C}$ as:

$B_{C} = \frac{\mu_{0} I}{4\pi} \cdot \int_{0}^{2\pi} \frac{1}{R} d\theta = \frac{\mu_{0} I \theta}{4\pi R}|_{0}^{2\pi} = \frac{\mu_{0} I}{2R}$ (ii)

and substituting $R = 1$ and $I = 1$ gives $\boxed{B_{C} = 0.5 \mu_{0} }$

For the elliptical loop (with coordinates $(x,y) = (2cos(\theta), sin(\theta)))$ , the radius is now equal to $R = \frac{ab}{\sqrt{a^2 sin^{2}(\theta) + b^2 cos^{2} (\theta)}} = \frac{(2)(1)}{ \sqrt{2^2 sin^{2}(\theta) + 1^2 cos^{2} (\theta)}} = \frac{2}{\sqrt{3sin^{2}(\theta) + 1}}$ . Substituting this expression into (i) now produces:

$B_{E} = \frac{\mu_{0} I}{4\pi} \cdot 4\int_{0}^{ \frac{\pi}{2} } \frac{1}{R} d\theta = \frac{\mu_{0} I}{\pi} \cdot \int_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{3sin^{2}(\theta)+ 1}}{2} d\theta$ (iii)

which the integrand is a complete elliptical integral of the second-kind (with parameter $k^2 = 3$ ). The final computation for (iii) (which I'll use Wolfram Alpha on the elliptical integral) yields:

$B_{E} = \frac{\mu_{0} I}{2\pi} \cdot \int_{0}^{ \frac{\pi}{2} } \sqrt{3sin^{2}(\theta)+ 1} d\theta = \frac{\mu_{0} (1)}{2\pi} \cdot 2.42211 \Rightarrow \boxed{B_{E} = 0.3857\mu_{0} }.$

Thus, the we arrive at the desired ratio of magnetic fields at the origin: $\frac{B_{E}}{B_{C}} = \frac{0.3857\mu_{0}}{0.5\mu_{0}} = \boxed{0.771}.$