An infinitely long wire, located on the z-axis, carries a current I along the +z-direction and produces the magnetic field B . The magnitude of the line integral ∫ B ⋅.dl ⃗⃗ along a straight line from the point (−√3a, a, 0) to (a, a, 0) is given by [u0 is the magnetic permeability of free space.]

 To determine the magnitude of the line integral 

$\int \mathbf{B}\cdot d\mathbf{l}$ along the straight path from $(-\text{√}3a,a,0)$ to $(a,a,0)$ due to an infinite wire on the $z$-axis carrying current $I$, we proceed as follows:

Key Steps:

1. Magnetic Field of the Wire:

   • The magnetic field at a distance $r$ from the wire is given by Ampère's Law:

     $$B=\frac{{\mu }_{0}I}{2\pi r},$$

     where $r=\sqrt{x^2+y^2}$. For the path $y=a$, $r=\sqrt{x^2+a^2}$.

2. Direction of $\mathbf{B}$:

   • The field circulates counterclockwise (right-hand rule). At a point $(x,a)$, the tangential direction of $\mathbf{B}$ is $(-a,x)$, normalized by $r^2$.

3. Line Integral Setup:

   • The path is horizontal ($dl=dx\widehat{\mathbf{i}}$) at $y=a$. The dot product $\mathbf{B}\cdot d\mathbf{l}$ simplifies to:

     $$\mathbf{B}\cdot d\mathbf{l}=\left(\frac{{\mu }_{0}I}{2\pi (x^2+a^2)}\right)(-a)dx\mathrm{.}$$

4. Integration:

   • Integrate from $x=-\text{√}3a$ to $x=a$:

     $$\int \mathbf{B}\cdot d\mathbf{l}=-\frac{{\mu }_{0}Ia}{2\pi }{\int }_{-\text{√}3a}^a\frac{dx}{x^2+a^2}\mathrm{.}$$

   • Using $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$:

     $$\text{Result}=-\frac{{\mu }_{0}I}{2\pi }[{\tan }^{-1}(1)-{\tan }^{-1}(-\text{√}3)]\mathrm{.}$$

   • Evaluate arctangents: ${\tan }^{-1}(1)=\frac{\pi }{4}$ and ${\tan }^{-1}(-\text{√}3)=-\frac{\pi }{3}$.

     $$\text{Result}=-\frac{{\mu }_{0}I}{2\pi }[\frac{\pi }{4}+\frac{\pi }{3}]=-\frac{{\mu }_{0}I}{2\pi }\cdot \frac{7\pi }{12}=-\frac{7{\mu }_{0}I}{24}\mathrm{.}$$

5. Magnitude:

   • The magnitude of the integral is $\frac{7{\mu }_{0}I}{24}$.

Final Answer:

$$\boxed{{\displaystyle \frac{7{\mu }_{0}I}{24}}}$$

Correct Option: (A)