Ampere's Law Exercise (Part 3) - "A bit" less calculus

Key steps of the problem are highlighted. Consider the position vector of a point $P$ on the circular loop:

$\vec{r}_p = \cos{\theta} \hat{i} + \sin{\theta} \hat{j}$ $d\vec{r}_p = \left(-\sin{\theta} \hat{i} + \cos{\theta} \hat{j}\right)d\theta$

The position vector of the location of the wire in the X-Y plane (Named $C$ ) is:

$\vec{r}_p = 0.5 \hat{i}$

A vector joining point $C$ and $P$ and directed towards $P$ is:

$\vec{r} = \vec{r}_p-\vec{r}_c$

The magnetic field at point $P$ due to the infinite wire is (assuming current flows into the plane of the paper):

$\vec{B} = \frac{\mu_o I}{2\pi \lvert \vec{r} \rvert} \left(-\hat{k} \times \hat{r}\right)$

Having obtained $\vec{B}$ , the next step is to calculate the following dot product:

$dQ = \vec{B} \cdot d\vec{r}_p$

Finally, all expressions are substituted and the resulting expression is:

$dQ = \frac{\mu_o I}{\pi}\left(\frac{\cos{\theta}-2}{5-4\cos{\theta}}\right)d\theta$

The working is left out in this solution. Finally,

$Q_R = \frac{\mu_o I}{\pi}\int_{-\pi/2}^{\pi/2} \left(\frac{\cos{\theta}-2}{5-4\cos{\theta}}\right)d\theta$

$Q_L+Q_R = \frac{\mu_o I}{\pi}\int_{0}^{2\pi} \left(\frac{\cos{\theta}-2}{5-4\cos{\theta}}\right)d\theta = -\mu_o I$

The second integral verifies Ampere's law. The closed-form expressions are evaluated (working left out) and the required ratio evaluates to:

$\boxed{\frac{Q_R}{Q_R+Q_L}= \frac{4\arctan{3} + \pi}{4\pi} \approx 0.6475836}$