An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization…

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :
  1. $\sqrt{\frac{\mathrm{h}}{\pi \mathrm{eB}}}$
  2. $\sqrt{\frac{2 h}{\pi e B}}$
  3. $\sqrt{\frac{h}{2 \pi e B}}$
  4. $\sqrt{\frac{4 \mathrm{~h}}{\pi e B}}$

Solution

$m v r=\frac{n h}{2 \pi}$...(i)
$r=\frac{v m}{B q}$....(ii)
$\begin{aligned} & n=2 \\ & m r\left(\frac{r B q}{m}\right)=\frac{2 h}{2 \pi} \\ & r=\sqrt{\frac{h}{\pi B q}} \\ & q=e \\ & r=\sqrt{\frac{h}{\pi B e}}\end{aligned}$

Asked in: JEE Main 2025 (22 Jan Shift 2)

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