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

i = 2a, f = 108rev/sec, n= ?, me = 9.1 × 10–31kg, Read more on Sarthaks.com - https://www.sarthaks.com/67872/a-tightly-wound-long-solenoid-carries-a-current-of-2-00a

Answer 2

The number of turns per metre in the solenoid is 1421 turns/m

Explanation:

Step 1:

Given data in the question  :

Magnitude of Solenoid current, i = 2 A

Electron Frequency,  $f=1 \times 10^{8} \mathrm{rev} / \mathrm{s}$

Electron mass

Electron Charge, $q=1.6 \times 10^{-19} \mathrm{C}$

We know that inside a solenoid the magnetic field is given by

$B=\frac{\mu_{0} n i}{2 a}$

Step 2:

When a particle executes uniform circular motion within a magnetic field, the particle frequency is given by

    $f=\frac{q B}{2 \pi m}$

   $B=\frac{2 \pi m f}{q}$

$u_{0} n i=\frac{2 \pi m f}{q}$

    $n=\frac{2 \pi m f}{\mu_{0} q i}$

      $=\frac{2 \pi \times 9.1 \times 10^{-31} \times 1 \times 10^{8}}{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19} \times 2}$

      $=\frac{2 \pi \times 9.1 \times 10^{-23}}{8 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}$

      $=\frac{9.1 \times 10^{-4}}{4 \times 10^{-7} \times 1.6}$

      $=\frac{9.1 \times 10^{-4}}{4 \times 10^{-7} \times 1.6}$

      $=\frac{9.1 \times 10^{3}}{6.4}$

      $=\frac{9.1 \times 10^{3}}{6.4}$

      $=1.421 \times 10^{3}$

      $=1421 turns/m$

Thus the number of turns per metre in the solenoid is 1421 turns/m .