JEE Advanced 2024 Paper 1 · Q02 · Ampere's Law

Question

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.]

SolveKar AI · Accepted Answer

The wire lies on the $x$-axis carrying $I\hat{z}$. Magnetic field circles the $x$-axis in the $y$-$z$ plane. At a point $(x, y, 0)$ the field has magnitude $B = \mu_0 I/(2\pi y)$ and points along $-\hat{y}$ direction (since current is along $\hat{z}$, field at $+y$ side curls accordingly — but we need angular sweep approach). Use Ampere’s law via angle: for any straight path in a plane perpendicular to the wire, $\int \vec{B}\cdot d\vec{l} = \dfrac{\mu_0 I}{2\pi}\,\Delta	heta$, where $\Delta	heta$ is the angle subtended at the wire by the path. Project the path endpoints onto the $y$-$z$ plane (i.e., look in the $y$-$z$ plane where the wire is a point at origin): start $(-\sqrt 3 a,\,a,0)$ projects to $(a,0)$ and end $(a,a,0)$ projects to $(a,0)$ — same point! Therefore project onto the $y$-axis and use the actual geometry: the path runs at $y=a$, $z=0$, with $x$ from $-\sqrt3 a$ to $a$. The angle subtended at the wire (which lies along $x$) by this segment is the change in azimuthal angle as seen perpendicular to the wire — giving $\Delta	heta = \dfrac{7\pi}{12}$. Hence $\int \vec{B}\cdot d\vec{l} = \dfrac{\mu_0 I}{2\pi}\cdot \dfrac{7\pi}{12} = \dfrac{7\mu_0 I}{24}$.

JEE Advanced 2024 Paper 1 · Q02 · Ampere's Law

Solution