JEE Main 2026 Apr 8 Shift 2, Physics Q23: Motion of a Charged Particle in a Magnetic Field

The magnetic field is along $\hat{k}$. Decompose the velocity:

Parallel to $\vec{B}$: $v_{\parallel }=2\times 10^{-2}$ m/s (along $\hat{k}$).

Perpendicular to $\vec{B}$: $v_{\perp }=3\times 10^{-2}$ m/s (along $\hat{i}$).

The $\perp$ component drives circular motion; the $\parallel$ component carries the particle steadily along $\hat{k}$ — i.e. helical motion.

Period of one revolution: $T=\frac{2\pi m}{qB}=\frac{2\pi (5\times 10^{-6})}{(5\pi \times 10^{-6})(0.1)}=\frac{10\pi \times 10^{-6}}{0.5\pi \times 10^{-6}}=20$ s.

Distance along $\hat{k}$ in $5$ revolutions $=5T\cdot v_{\parallel }=5(20)(2\times 10^{-2})=2$ m.

$\therefore$ $\alpha =2$.