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 dvec{l} ) along a straight line from the point ( (-sqrt{3}a, a, 0) ) to ( (a, a, 0) ) is given by: [text{[} mu_0 text{ is the magnetic permeability of free space.] }][latexpage]

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Problem:

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:

$\text{[} \mu_0 \text{ is the magnetic permeability of free space.] }$

(A) $\frac{7 \mu_0 I}{24}}$ (B) $\frac{7 \mu_0 I}{12}}$

A straight infinitely long wire lies along the z-axis and carries a steady current $I$ in the +z-direction. This current creates a magnetic field $\vec{B}$ in the surrounding space. The magnetic field at a point in the xy-plane due to such a wire is tangential (circular around the wire) and its magnitude depends inversely on the radial distance $r$ from the wire.

The problem asks to compute the magnitude of the line integral $\int \vec{B} \cdot d\vec{l}$ along a straight line path from point $A(-\sqrt{3}a, a, 0)$ to point $B(a, a, 0)$ , both of which lie in the xy-plane. The integral represents the component of magnetic field along the path due to the wire.

Explanation:

The magnetic field due to an infinite straight current-carrying wire is given by:

$\vec{B} = \frac{\mu_0 I}{2\pi r} \hat{\phi}$

where $r$ is the perpendicular distance from the wire, and $\hat{\phi}$ is the azimuthal direction (tangential to the circle centered on the wire). The field circles around the wire in planes perpendicular to it.

For a line integral $\int \vec{B} \cdot d\vec{l}$ , the result depends on how the path traverses the circular magnetic field lines. The wire lies on the z-axis, so the path from $A$ to $B$ is in the xy-plane, not enclosing the wire completely, but subtending some angle at the wire.

In this case, we need the angle subtended at the z-axis (the location of the wire) by the straight line joining the two points. This is the same as the angular sweep $\theta$ from vector $\vec{r}_A$ to $\vec{r}_B$ as seen from the origin.