JEE Advanced 2024 · Paper 1, Physics question 2

For a long straight wire on the $z$-axis, $\vec B = \dfrac{\mu_0 I}{2\pi r}\hat\phi$, so $\vec B\cdot d\vec l = \dfrac{\mu_0 I}{2\pi}\,d\phi$. The line integral depends only on the change in azimuthal angle.

The start $(-\sqrt3 a, a)$ is at $\phi = 150^\circ$ and the end $(a, a)$ at $\phi = 45^\circ$. The straight segment $y = a$ does not encircle the wire, so $|\Delta\phi| = 105^\circ = \dfrac{7\pi}{12}$.

$$\left|\int\vec B\cdot d\vec l\right| = \frac{\mu_0 I}{2\pi}\cdot\frac{7\pi}{12} = \frac{7\mu_0 I}{24}.$$