JEE Main 2025 Jan 24 Shift 2, Physics Q21: Motion of a Charged Particle in a Magnetic Field

The electron moves in a circle perpendicular to the solenoid axis (which carries the magnetic field). The period of cyclotron motion is

$T=\frac{2\pi m_e}{|q_e|B}$

Inside a long solenoid, $B={\mu }_{0}nI$, where $n$ is turns per metre. Combining:

$T=\frac{2\pi m_e}{|q_e|{\mu }_{0}nI}\Rightarrow n=\frac{2\pi m_e}{|q_e|{\mu }_{0}IT}$

Substituting numerical values:

$n=\frac{2\pi (9\times 10^{-31})}{(1.6\times 10^{-19})(4\pi \times 10^{-7})(1.5)(75\times 10^{-9})}$

$n=\frac{2\cdot 9\times 10^{-31}}{1.6\times 10^{-19}\cdot 4\times 10^{-7}\cdot 1.5\cdot 75\times 10^{-9}}=\frac{18\times 10^{-31}}{72\times 10^{-34}}$

$n=\frac{18\times 10^{-31}}{72\times 10^{-34}}=\frac{18}{72}\times 10^3=0.25\times 10^3=250$

Answer: 250