Line Integral of Magnetic Field due to Infinitely Long Wire

To find the magnitude of the line integral $\int \vec{B} \cdot d\vec{l}$ along the straight line path from $P_1(-\sqrt{3}a, a, 0)$ to $P_2(a, a, 0)$, we analyze the magnetic field produced by an infinitely long straight wire carrying current $I$ along the $+z$-axis.

1. Expression for Magnetic Field

The magnetic field $\vec{B}$ at any point $(x, y, z)$ in Cartesian coordinates due to a current $I$ along the $+z$-axis is given by: $\vec{B} = \frac{\mu_0 I}{2\pi (x^2 + y^2)} \left( -y \hat{i} + x \hat{j} \right)$

2. Parameterization of the Path

The straight line connects $P_1(-\sqrt{3}a, a, 0)$ to $P_2(a, a, 0)$. Along this straight line segment:

• $y = a$ (constant, so $dy = 0$)
• $z = 0$ (constant, so $dz = 0$)
• $x$ varies from $-\sqrt{3}a$ to $a$

Thus, the differential displacement vector along the path is: $d\vec{l} = dx \hat{i}$

3. Evaluating the Line Integral

Substituting $\vec{B}$ and $d\vec{l}$ with $y = a$: $\vec{B} \cdot d\vec{l} = \left[ \frac{\mu_0 I}{2\pi (x^2 + a^2)} \left( -a \hat{i} + x \hat{j} \right) \right] \cdot (dx \hat{i}) = -\frac{\mu_0 I a}{2\pi (x^2 + a^2)} dx$

Integrating from $x = -\sqrt{3}a$ to $x = a$: $\int \vec{B} \cdot d\vec{l} = \int_{-\sqrt{3}a}^{a} -\frac{\mu_0 I a}{2\pi (x^2 + a^2)} dx$

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

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

Using the standard inverse trigonometric values: $\tan^{-1}(1) = \frac{\pi}{4}$ $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$

Substituting these back into the integral: $\int \vec{B} \cdot d\vec{l} = -\frac{\mu_0 I}{2\pi} \left[ \frac{\pi}{4} - \left(-\frac{\pi}{3}\right) \right] = -\frac{\mu_0 I}{2\pi} \left( \frac{7\pi}{12} \right) = -\frac{7\mu_0 I}{24}$

4. Magnitude of the Line Integral

Taking the magnitude of the result: $\left| \int \vec{B} \cdot d\vec{l} \right| = \frac{7\mu_0 I}{24}$

Thus, the correct option is (A).