Magnetic field in the left focus of an ellipse current loop

Question

given a 2D ellipse in 3D space parametrized as $x(t)=a\cos{t}$, $y(t)=b\sin{t}$, I would like to find the magnetic field at point $(-c,0,0)$, $c=\sqrt{a^2-b^2}$. (The current $I$ is running counter-clockwise)

Using the Biot-Savart law: $dB=\frac{\mu_0I}{4\pi}\frac{|d\vec{l}\times \vec{r}|}{r^3}$ I make use of $|d\vec{l}\times \vec{r}|=dl \cdot r \sin{\alpha}$ and then I further relate $dl$ with $dt$ as $dl=\sqrt{x'^2+y'^2}\ dt$ (calculating the length of the ellipse). Now I get something like $dB=\frac{\mu_0I}{4\pi}\int_{-\pi}^{\pi}\frac{\sin{\alpha}\sqrt{a^2-c^2\cos{t}}}{(c\cos{t}+a)^2}dt$ which may or may not be correct, but mostly I'd like to know how to relate the sine of the angle $\alpha$ between the two vectors with the parameter $t$ so I can complete the integral. I can't quite see it.

Answer 1

• Referring to another post in Math SE, we have

\begin{align} \mathbf B (-c,0,0) &= \frac{\mu_0 I}{4\pi} \oint_C \frac {d\boldsymbol s \times (-c\, \mathbf i-\mathbf r)} {|-c\, \mathbf i-\mathbf r|^3} \\[5pt] &= \frac{\mu_0 I}{4\pi} \oint_C \frac {\mathbf r'(t) \times (-c\, \mathbf i-\mathbf r)} {|-c\, \mathbf i-\mathbf r|^3} \, dt \\[5pt] &= \frac{\mu_0 I}{4\pi} \, \mathbf k \int_0^{2\pi} \frac{b(a+c\cos t) \, dt}{[(c+a\cos t)^2+b^2\sin^2 t]^{3/2}} \\[5pt] &= \frac{\mu_0 I}{4\pi} \, \mathbf k \int_0^{2\pi} \frac{b(a+c\cos t)}{[(a+c\cos t)^2]^{3/2}} \, dt \\[5pt] &= \frac{\mu_0 I}{4\pi} \, \mathbf k \int_0^{2\pi} \frac{b}{(a+c\cos t)^2}\, dt \\[5pt] &= \frac{\mu_0 I}{2\pi} \, \mathbf k \left[ \frac{2a}{b^2} \tan^{-1} \left( \sqrt{\frac{a-c}{a+c}} \tan \frac{t}{2} \right)- \frac{c\sin t}{b(a+c\cos t)} \right]_0^\pi \\[5pt] &= \frac{\mu_0 aI}{2b^2} \,\mathbf k \end{align}

• Note that $\dfrac{2b^2}{a}$ is the latus rectum of the ellipse.

• Also refer to Am J Phys for your further interests.

• Compare another integral in Math SE.