Question

Two charged particles A and B of masses (m) and ( 2m), charges ( 2q) and ( 3q ) respectively, are moving with the same velocity into a uniform magnetic field such that both particles make the same angle \( \theta (<90^\circ) \)with the direction of the magnetic field. Then the ratio of the pitches of the helical paths of the particles A and B is:

• A. \( 4:3 \)
• B. \( 3:2 \)
• C. \( 3:4 \)
• D. \( 2:3 \)

Hint:

The pitch of a helical path is given by \( P = v \cos\theta \cdot T \).
- The time period of circular motion in a uniform magnetic field is \( T = \frac{2\pi m}{qB} \).

Answer 1

The Correct Option is C

The pitch \( P \) of a helical path in a magnetic field is given by: \[ P = v \cos\theta \cdot T \] where: - \( T \) is the time period of circular motion in the perpendicular plane, - \( v \) is the velocity of the particle. The time period of circular motion is: \[ T = \frac{2\pi m}{qB} \] 1. Pitch for Particle A: \[ P_A = v \cos\theta \cdot \frac{2\pi m}{2qB} = \frac{2\pi m v \cos\theta}{2qB} \] \[ P_A = \frac{\pi m v \cos\theta}{qB} \] 2. Pitch for Particle B: \[ P_B = v \cos\theta \cdot \frac{2\pi (2m)}{3qB} = \frac{4\pi m v \cos\theta}{3qB} \] 3. Ratio of Pitches: \[ \frac{P_A}{P_B} = \frac{\frac{\pi m v \cos\theta}{qB}}{\frac{4\pi m v \cos\theta}{3qB}} \] \[ = \frac{3}{4} \] Thus, the correct answer is \(\boxed{3:4}\).