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 $\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}$, charge of electron $\left|\mathrm{q}_{\mathrm{e}}\right|=1.6 \times 10^{-19} \mathrm{C}$, $$ \left.\mu_0=4 \pi \times 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^2}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right] $$

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