An infinitely long wire, located on the \(z\)-axis, carries a current \(I\) along the \(+z\)-direction and produces the magnetic field \(\vec{B}\). The magnitude of the line integral \(\int \vec{B} \cdot \overrightarrow{d l}\) along a straight line from the point \((-\sqrt{3} a, a, 0)\) to \((a, a, 0)\) is given by
[ \(\mu_0\) is the magnetic permeability of free space.]

\(\begin{aligned} & \Rightarrow|\overrightarrow{\mathrm{d} \ell}|=\mathrm{rd} \theta \\ & \Rightarrow|\overrightarrow{\mathrm{B}}|=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\int|\overrightarrow{\mathrm{B}}||\overrightarrow{\mathrm{d} \ell}| \cos 0^{\circ} \\ & =\int\left(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}\right) \times(\operatorname{rd} \theta) \\ & =\int_{\theta_1}^{\theta_2} \frac{\mu_0 \mathrm{I}}{2 \pi} \mathrm{d} \theta=\frac{\mu_0 \mathrm{I}}{2 \pi}\left[\theta_2-\left(-\theta_1\right)\right]\end{aligned}\)
[ \(\theta_1\) is anticlockwise hence taken negative]

\(\begin{aligned} & \Rightarrow \tan \theta_1=\frac{\mathrm{a} \sqrt{3}}{\mathrm{a}}=\sqrt{3} \\ & \theta_1=\frac{\pi}{3} \\ & \Rightarrow \tan \theta_2=\frac{\mathrm{a}}{\mathrm{a}}=1 \\ & \theta_2=\frac{\pi}{4} \\ & \Rightarrow \int \mathrm{B} \cdot \mathrm{d} \ell=\frac{\mu_0 \mathrm{I}}{2 \pi}\left[\frac{\pi}{3}+\frac{\pi}{4}\right] \\ & =\frac{7 \mu_0 \mathrm{I}}{24} \\ & \Rightarrow \text { Ans. Option (1) }\end{aligned}\)