Question 2

The magnetic field $\vec{B}$ at a point $(x, y, 0)$ due to an infinite wire along the $z$-axis is given by:

$\vec{B} = \frac{\mu_0 I}{2\pi (x^2 + y^2)} (-y \hat{i} + x \hat{j})$

The path of integration is a straight line from $(-\sqrt{3}a, a, 0)$ to $(a, a, 0)$. Along this path, $y = a$ (constant), so $dy = 0$ and $dz = 0$. The differential length vector is $d\vec{l} = dx \hat{i}$.

The dot product is $\vec{B} \cdot d\vec{l} = B_x dx = -\frac{\mu_0 I a}{2\pi (x^2 + a^2)} dx$.

The line integral is:

$\int_{-\sqrt{3}a}^{a} -\frac{\mu_0 I a}{2\pi (x^2 + a^2)} dx = -\frac{\mu_0 I a}{2\pi} \int_{-\sqrt{3}a}^{a} \frac{1}{x^2 + a^2} dx$

$= -\frac{\mu_0 I a}{2\pi} \left[ \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \right]_{-\sqrt{3}a}^{a} = -\frac{\mu_0 I}{2\pi} \left[ \tan^{-1}(1) - \tan^{-1}(-\sqrt{3}) \right]$

$= -\frac{\mu_0 I}{2\pi} \left[ \frac{\pi}{4} - \left(-\frac{\pi}{3}\right) \right] = -\frac{\mu_0 I}{2\pi} \left( \frac{7\pi}{12} \right) = -\frac{7\mu_0 I}{24}$

The magnitude of the integral is $\frac{7\mu_0 I}{24}$.