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 meter in the solenoid.

Answer 1

We will use the formula of magnetic field inside an ideal solenoid. Then we shall use the formula of frequency of a charge particle executing a uniform circular motion under a constant magnetic field. From here, we calculate the magnetic field in terms of the frequency, charge and mass of the electron and compare it with our first equation to get the value of the number of turns per meter.

Complete answer:
Let us first define some terms and their values as they will be of use later.
Let the current in the ideal solenoid be denoted by $I$ . And the value of $I$is equal to:
$\Rightarrow I=2A$
Let the frequency of electrons revolving in the ideal solenoid be denoted by $f$ . And the value of $f$ is equal to:
$\Rightarrow f=1\times {{10}^{8}}rev{{s}^{-1}}$
Also, let the mass of the electron be $m$and its charge be $q$ . So, their values are as follows:
$\Rightarrow m=9.1\times {{10}^{-31}}kg$
$\Rightarrow q=1.6\times {{10}^{-19}}C$
And, we know the permittivity of free space is given by ${{\mu }_{0}}$ . It is equal to:
$\Rightarrow {{\mu }_{0}}=4\pi \times {{10}^{-7}}$
Now that we have defined all the useful terms, we shall proceed with the magnetic field of a solenoid which is given by:
$\Rightarrow B={{\mu }_{0}}nI$ [Let this expression be equation number (1)]
Now, the formula of frequency is given by:
$\Rightarrow f=\dfrac{qB}{2\pi m}$
Therefore, the magnetic field can be written as:
$\Rightarrow B=\dfrac{2\pi mf}{q}$ [Let this expression be equation number (2)]
Now, on equating equation number (1) and equation number (2), we get:
$\Rightarrow {{\mu }_{0}}nI=\dfrac{2\pi mf}{q}$
On simplifying and solving for $n$ , we get;
$\Rightarrow n=\dfrac{2\pi mf}{{{\mu }_{0}}qI}$
Putting the values of all the terms in R.H.S., we get:
$\begin{aligned} & \Rightarrow n=\dfrac{2\pi \times 9.1\times {{10}^{-31}}\times 1\times {{10}^{8}}}{4\pi \times {{10}^{-7}}\times 1.6\times {{10}^{-19}}\times 2} \\\ & \Rightarrow n=1.422\times {{10}^{3}}turns{{m}^{-1}} \\\ & \therefore n=1422turns{{m}^{-1}} \\\ \end{aligned}$
Hence, the number of turns per meter comes out to be 1422 turns per meter.

Note:
This was a question where both, our in-depth knowledge of the topic and our calculational skills were tested. In the solution, we assumed that the solenoid was ideal because it was mentioned that it was tightly wound and very long so there would be no loss.