Question

An electron enters the region of \( 0.3 \, \text{T} \) magnetic field at an angle of 60° with the speed of \( 4 \times 10^5 \, \text{ms}^{-1} \). Find the radius of the helical path and the pitch (distance between two consecutive spirals) of the electron beam.

Hint:

The radius of the helical path of a charged particle moving in a magnetic field depends on the velocity, magnetic field strength, and the angle between the velocity and the magnetic field.

Answer 1

Step 1: Magnetic Force on the Electron.
The magnetic force \( F \) acting on the electron is given by the formula: \[ F = qvB \sin\theta \] Where:
- \( q = 1.6 \times 10^{-19} \, \text{C} \) is the charge of the electron,
- \( v = 4 \times 10^5 \, \text{ms}^{-1} \) is the velocity of the electron,
- \( B = 0.3 \, \text{T} \) is the magnetic field,
- \( \theta = 60^\circ \) is the angle between the velocity and the magnetic field.
Step 2: Centripetal Force and Radius.
For circular motion, the magnetic force provides the centripetal force, so: \[ qvB \sin\theta = \frac{mv^2}{r} \] Where \( r \) is the radius of the helical path and \( m = 9.11 \times 10^{-31} \, \text{kg} \) is the mass of the electron. Substitute the known values: \[ 1.6 \times 10^{-19} \times 4 \times 10^5 \times 0.3 \times \sin(60^\cir = \frac{9.11 \times 10^{-31} \times (4 \times 10^5)^2}{r} \] \[ r = \frac{9.11 \times 10^{-31} \times (4 \times 10^5)^2}{1.6 \times 10^{-19} \times 4 \times 10^5 \times 0.3 \times \sin(60^\cir} \] Solving for \( r \), we get: \[ r \approx 32 \times 10^{-2} \, \text{m} \]
Step 3: Pitch of the Helical Path.
The pitch of the helical path is the distance the electron moves along the direction of the magnetic field during one complete revolution. The pitch \( p \) is given by: \[ p = v \cos\theta \times T \] Where \( T \) is the time period of the electron’s circular motion.
The time period \( T \) is: \[ T = \frac{2\pi r}{v \sin\theta} \] Substitute the values: \[ p = v \cos\theta \times \frac{2\pi r}{v \sin\theta} \] \[ p = \frac{2\pi r \cos\theta}{\sin\theta} \] Substituting the known values: \[ p = \frac{2\pi \times 32 \times 10^{-2} \times \cos(60^\cir}{\sin(60^\cir} \approx 1.5 \times 10^{-2} \, \text{m} \]
Final Answer:
The radius of the helical path is \( \boxed{32 \times 10^{-2} \, \text{m}} \) and the pitch of the electron’s path is \( \boxed{1.5 \times 10^{-2} \, \text{m}} \).