JEE Main 2025 Jan 22 Shift 2, Physics Q21: Motion of a Charged Particle in a Magnetic Field

In the crossed-field velocity selector, undeflected motion requires $qE=qvB$, so $E=vB$, giving $B=\frac{E}{v}$.

When the electric field is switched off, only the magnetic force acts. The proton moves in a circle of radius:

$r=\frac{mv}{qB}=\frac{m{v}^2}{qE}$.

Solving for $E$:

$E=\frac{m{v}^2}{qr}=\frac{(1.6\times 10^{-27})(2\times 10^5{)}^2}{(1.6\times 10^{-19})(2\times 10^{-2})}$

$=\frac{1.6\times 10^{-27}\times 4\times 10^{10}}{3.2\times 10^{-21}}=\frac{6.4\times 10^{-17}}{3.2\times 10^{-21}}=2\times 10^4\text{N/C}$.

$\therefore x=2$.