Question

A tightly-wound, long solenoid carries a current of $2.00A$. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of $1.00\times {10}^{8}rev{s}^{-1}$.Find the number of turns per metre in the solenoid.

Answer 1

Magnitude of current in the solenoid, $i=2A$

Frequency of the electron, $f=1\times {10}^{8}rps$

Mass of the electron, $m=9.1\times {10}^{-31}Kg$

Charge of the electron, $q=1.6\times {10}^{-19}C$

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

$B={\mu }_{0}ni$

If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by

$f=\frac{qB}{2\pi m}$

$\Rightarrow B=\frac{2\pi mf}{q}$

$\Rightarrow {\mu }_{0}ni=\frac{2\pi mf}{q}$ [Using (1)]

$\Rightarrow n=\frac{2\pi mf}{{\mu }_{0}qi}$

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

$=1420turns/m$