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 meter in the solenoid is ………… 

Hint:

The magnetic field in a solenoid is directly proportional to the current and the number of turns per unit length. Use the magnetic force on the electron to relate the field and the motion.

Answer 1

The magnetic field \( B \) inside a solenoid is given by: \[ B = \mu_0 n I, \] where: - \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \) is the permeability of free space, - \( n \) is the number of turns per meter, - \( I = 1.5 \, \text{A} \) is the current. The electron moves in a circular path under the influence of the magnetic field. The force acting on the electron is the Lorentz force, which provides the centripetal force: \[ e v B = \frac{m v^2}{r}, \] where \( e \) is the charge of the electron, \( m \) is the mass of the electron, \( v \) is the velocity of the electron, and \( r \) is the radius of the circular path. The time period \( T \) of the motion is related to the frequency \( f \) and the radius \( r \). We can use the relation: \[ T = \frac{2\pi m}{e B}. \] We are given the time period \( T = 75 \, \text{ns} = 75 \times 10^{-9} \, \text{s} \). Solving for \( B \) and then using the formula for \( B \) in the solenoid, we can calculate \( n \). After substituting known values and solving, we find: \[ n = 10^5 \, \text{turns per meter}. \] Final Answer: \( 10^5 \).