JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Magnetic Effects of Current

Take the wire along the z-axis carrying current I, so the field at perpendicular distance r is $B={\mu }_{0}I/(2\pi r)$, directed azimuthally. A particle of charge q and mass m moving with velocity $\vec{v}$ feels $\vec{F}=q\vec{v}\times \vec{B}$; since this force is always perpendicular to $\vec{v}$, it does no work, so the particle's speed stays fixed at $v_0$ throughout the motion — only its direction rotates.

Resolving the equations of motion for the component of velocity toward the wire ($v_x$) and along the wire's length ($v_y$) gives, for the phase in which the particle approaches the wire from distance a down to its closest distance $x_1$ (where the velocity has become entirely radial):

Integrating from $v_x=0$ at $r=a$ to $v_x=v_0$ at $r=x_1$:


By the same reasoning, the particle then rebounds and swings back toward the wire under the identical rotational field, so the same relation carries it from $x_1$ in to the final turning point x:


Combining the two legs of the motion,