Question 275: Magnetism and Matter

Question

A tightly wound long solenoid carries a current of 1.5 A . An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns . The number of turns per metre in the solenoid is _________.
[Take mass of electron $\mathrm{m}_{\mathrm{e}}=9 	imes 10^{-31} \mathrm{~kg}$, charge of electron $\left|\mathrm{q}_{\mathrm{e}}ight|=1.6 	imes 10^{-19} \mathrm{C}$,

$$
\left.\mu_0=4 \pi 	imes 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^2}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}ight]
$$

Accepted Answer

The problem involves an electron executing uniform circular motion inside a solenoid. The objective is to find the number of turns per meter in the solenoid.
Given:
Current in the solenoid, $ I = 1.5 \, 	ext{A} $
Time period of the electron's circular motion, $ T = 75 \, 	ext{ns} = 75 	imes 10^{-9} \, 	ext{s} $
Mass of electron, $ m_e = 9 	imes 10^{-31} \, 	ext{kg} $
Charge of electron, $ |q_e| = 1.6 	imes 10^{-19} \, 	ext{C} $
Permeability of free space, $ \mu_0 = 4\pi 	imes 10^{-7} \, 	ext{N/A}^2 $
Explanation:
The time period $ T $ for a revolving charge in a magnetic field is given by:
$ T = \frac{2\pi m}{qB} $
The magnetic field $ B $ inside a solenoid is:
$ B = \mu_0 n I $
where $ n $ is the number of turns per meter. Thus, substituting $ B $ in the expression for $ T $, we get:
$ T = \frac{2\pi m}{q(\mu_0 n I)} $
Plugging in the known values:
$ 75 	imes 10^{-9} = \frac{(2\pi)(9 	imes 10^{-31})}{1.6 	imes 10^{-19} 	imes 4\pi 	imes 10^{-7} 	imes n 	imes 1.5} $
Solving for $ n $, we find:
$ n = 250 $
Thus, the number of turns per meter in the solenoid is 250.

Magnetism and Matter

Integer / Numeric Answer Type