Ampere's Law Exercise (Part 2)

Consider a point on the black circle:

$\vec{r}_{C1} = \cos{\theta} \hat{i} + \sin{\theta} \hat{j}$

An arc length element of this circle is:

$d\vec{r}_{C1} = \left(-\sin{\theta} \hat{i} + \cos{\theta} \hat{j}\right)d\theta$

Consider a point on the blue circle circle:

$\vec{r}_{C2} = \cos{\phi} \hat{i} + \sin{\phi} \hat{j}$

An arc length element of this circle is:

$d\vec{r}_{C2} = \left(-\sin{\phi} \hat{i} + \cos{\phi} \hat{j}\right)d\phi$

Let:

$\vec{r} = \vec{r}_{C2} - \vec{r}_{C1}$

Now, the magnetic field at a point on the blue circle due to an element of the black circle is:

$d\vec{B} = \frac{\mu_o I}{4\pi} \left(\frac{d\vec{r}_{C1} \times \vec{r}}{\lvert \vec{r} \rvert ^3}\right)$

The elementary dot product of this magnetic field along an arc length element of the blue circle is:

$dQ = d\vec{B} \cdot d\vec{r}_{C2}$

Now, it is just a matter of substituting expressions and simplifying a bit, to obtain:

$dQ = \frac{-1}{4\pi} \left(\frac{\cos\left(\mathrm{\theta}\right)-\cos\left(\phi \right)+\cos\left(\phi \right)\,\cos\left(\mathrm{\theta}\right)}{{\left(2\,\cos\left(\phi \right)-2\,\cos\left(\mathrm{\theta}\right)-2\,\cos\left(\phi \right)\,\cos\left(\mathrm{\theta}\right)+3\right)}^{3/2}}\right)d\theta d\phi \implies dQ = f(\theta,\phi) d\theta d\phi$

Now, the following quantities:

$Q_R = \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} f(\theta,\phi) d\theta d\phi$

$Q_L = \int_{\pi/2}^{3\pi/2} \int_{0}^{2\pi} f(\theta,\phi) d\theta d\phi$

From Ampere's Law: $Q_L + Q_R = 1$ and therefore the answer is:

$\boxed{Q_R \approx 0.1787}$