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 $$\int \vec{B} \cdot d\vec{l}$$ 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.]

The z-axis carries an infinitely long straight wire with current $$I$$ along $$+z$$. At a point with position vector $$\vec{r}=x\,\hat{i}+y\,\hat{j}$$ in the $$xy$$-plane its magnetic field is

$$\vec{B}=\frac{\mu_0 I}{2\pi r}\,\hat{\phi} ,\qquad r=\sqrt{x^{2}+y^{2}}$$

For current in $$+z$$, $$\hat{\phi}= -\sin\phi\,\hat{i}+\cos\phi\,\hat{j}$$, where $$\tan\phi=\dfrac{y}{x}$$. With $$x$$ variable and constant $$y=a$$ (our path), we have

$$\sin\phi=\frac{a}{r},\qquad \cos\phi=\frac{x}{r}$$

Hence

$$\vec{B}= \frac{\mu_0 I}{2\pi r}\Bigl(-\frac{a}{r}\,\hat{i} +\frac{x}{r}\,\hat{j}\Bigr) =\frac{\mu_0 I}{2\pi r^{2}}\bigl(-a\,\hat{i}+x\,\hat{j}\bigr)$$

The required line integral is taken along the straight segment from $$P_1(-\sqrt{3}a,\,a,\,0)$$ to $$P_2(a,\,a,\,0)$$. Along this segment $$y=a,\;z=0$$ and $$x$$ varies, so

$$d\vec{l}=dx\,\hat{i}$$

Therefore

$$\vec{B}\cdot d\vec{l}= \frac{\mu_0 I}{2\pi r^{2}}\bigl(-a\,\hat{i}+x\,\hat{j}\bigr)\!\cdot\!(dx\,\hat{i}) =-\frac{\mu_0 I\,a}{2\pi (x^{2}+a^{2})}\,dx$$

Integrate from $$x=-\sqrt{3}a$$ to $$x=a$$:

$$\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l} =-\frac{\mu_0 I\,a}{2\pi}\int_{-\sqrt{3}a}^{a}\frac{dx}{x^{2}+a^{2}}$$

Use the standard integral $$\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\!\Bigl(\frac{x}{a}\Bigr)$$:

$$\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l} =-\frac{\mu_0 I}{2\pi}\Bigl[\tan^{-1}\!\Bigl(\frac{x}{a}\Bigr)\Bigr]_{x=-\sqrt{3}a}^{x=a}$$

Evaluate the limits:

$$\tan^{-1}(1)=\frac{\pi}{4},\qquad \tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}$$

Thus

$$\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l} =-\frac{\mu_0 I}{2\pi}\!\Bigl(\frac{\pi}{4}-\bigl(-\frac{\pi}{3}\bigr)\Bigr) =-\frac{\mu_0 I}{2\pi}\!\Bigl(\frac{7\pi}{12}\Bigr) =-\frac{7\mu_0 I}{24}$$

The negative sign only indicates that $$\vec{B}$$ and $$d\vec{l}$$ are oppositely directed. The question asks for the magnitude, hence

$$\Bigl|\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l}\Bigr| =\frac{7\mu_0 I}{24}$$

Option A which is: $$\dfrac{7\mu_0 I}{24}$$