JEE Main 2025 Jan 22 Shift 2, Physics Q7: Motion of a Charged Particle in a Magnetic Field

For circular motion in a magnetic field, the magnetic force provides centripetal force:

$evB=\frac{m{v}^2}{r}\Rightarrow r=\frac{mv}{eB}$, i.e. $mv=eBr$.

Bohr's quantization rule: $mvr=\frac{nh}{2\pi }$.

Substituting $mv=eBr$:

$(eBr)r=\frac{nh}{2\pi }\Rightarrow r^2=\frac{nh}{2\pi eB}\Rightarrow r=\sqrt{\frac{nh}{2\pi eB}}$.

First excited state corresponds to $n=2$:

$r=\sqrt{\frac{2h}{2\pi eB}}=\sqrt{\frac{h}{\pi eB}}$.