Question

A tightly-wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00 × 10⁸ rev s⁻¹. Find the number of turns per metre in the solenoid.

Answer 1

Given:
Magnitude of current in the solenoid, i = 2 A
Frequency of the electron, $f=1\times {10}^8\mathrm{rev}/\mathrm{s}$
Mass of the electron, $m=9.1\times {10}^{-31}\mathrm{kg}$
Charge of the electron, $q=1.6\times {10}^{-19}\mathrm{C}$
We know that the magnetic field inside a solenoid is given by
B = µ₀ni
If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by
$$\begin{gathered} f=\frac{q\mathrm{B}}{2\mathrm{\pi }m} \\ \Rightarrow B=\frac{2\mathrm{\pi }mf}{q} \\ \Rightarrow {\mathrm{\mu }}_{0}ni=\frac{2\mathrm{\pi }mf}{q}[\mathrm{Using}(1\left)\right] \\ \Rightarrow n=\frac{2\mathrm{\pi }mf}{{\mathrm{\mu }}_{0}qi} \\ =\frac{2\mathrm{\pi }\times 9.1\times {10}^{-31}\times 1\times {10}^8}{4\mathrm{\pi }\times {10}^{-7}\times 1.6\times {10}^{-19}\times 2} \\ =1420\mathrm{turns}/\mathrm{m} \end{gathered}$$