A tightly wound long solenoid carries a current of $1.5\ \text{A}$. An electron is executing uniform circular motion inside the solenoid with a time period of $75\ \text{ns}$. The number of turns per metre in the solenoid is ______. [Take mass of electron $m_e = 9\times10^{-31}\ \text{kg}$, charge of electron $|q_e| = 1.6\times10^{-19}\ \text{C}$, $\mu_0 = 4\pi\times10^{-7}\ \dfrac{\text{N}}{\text{A}^2}$, $1\ \text{ns} = 10^{-9}\ \text{s}$]
Numerical value type. Enter your answer.
Answer: 250
Period of circular motion in a magnetic field: $T = \dfrac{2\pi m}{eB}$, so
$$B = \frac{2\pi m}{eT}$$
Inside the solenoid $B = \mu_0 nI$:
$$n = \frac{2\pi m}{eT\mu_0 I} = \frac{2\pi\times9\times10^{-31}}{1.6\times10^{-19}\times75\times10^{-9}\times4\pi\times10^{-7}\times1.5}$$
$$= \frac{18\times10^{-31}}{1.6\times75\times4\times1.5\times10^{-35}} = \frac{18\times10^{4}}{720} = 250\ \text{turns/m}$$
Solution by Sreeraj P, M.Sc Physics