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, \[f = 1 \times {10}^8\]  rev/s

Mass of the electron, \[\text{ m  }= 9 . 1 \times {10}^{- 31}\]  kg

Charge of the electron, \[q = 1 . 6 \times {10}^{- 19} C\]

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

\[f = \frac{qB}{2\pi m}\]
\[ \Rightarrow B = \frac{2\pi mf}{q}\]
\[ \Rightarrow \mu_0 ni = \frac{2\pi mf}{q} [\text{ Using } (1)]\]
\[ \Rightarrow n = \frac{2\pi mf}{\mu_0 qi}\]
\[ = \frac{2\pi \times 9 . 1 \times {10}^{- 31} \times 1 \times {10}^8}{4\pi \times {10}^{- 7} \times 1 . 6 \times {10}^{- 19} \times 2}\]
\[ = 1420 \] turns/m