Question

A charged particle enters a magnetic field with velocity \( v \) at an angle of \( 60^\circ \) with the field. If the time period of the helical path is 2 seconds, the pitch of the helical path is:

• A. \( 2v \)
• B. \( \frac{v}{2} \)
• C. \( v \)
• D. \( \frac{v}{8} \)

Hint:

For a charged particle moving in a magnetic field, the pitch of the helical path is calculated by multiplying the velocity component along the field direction with the time period of the circular motion.

Answer 1

The Correct Option is C

The motion of a charged particle in a magnetic field forms a helical path due to the combination of circular motion in the plane perpendicular to the magnetic field and uniform motion along the direction of the field. - Let the velocity of the particle be \( v \). - The velocity has two components: - One component parallel to the magnetic field, \( v_{\parallel} = v \cos \theta \), where \( \theta = 60^\circ \). - The other component perpendicular to the magnetic field, \( v_{\perp} = v \sin \theta \). In this case, we are given that the time period \( T = 2 \, \text{sec} \) is the time taken for one complete revolution of the charged particle in the magnetic field. The pitch of the helical path refers to the distance traveled along the magnetic field in one complete revolution. This is given by the distance traveled in the direction parallel to the field, which is: \[ \text{Pitch} = v_{\parallel} \times T \] Substituting the value of \( v_{\parallel} \): \[ \text{Pitch} = v \cos 60^\circ \times T \] Since \( \cos 60^\circ = \frac{1}{2} \), we get: \[ \text{Pitch} = \frac{v}{2} \times T \] Given that \( T = 2 \, \text{sec} \), we get: \[ \text{Pitch} = \frac{v}{2} \times 2 = v \] Thus, the pitch of the helical path is \( v \).