Motion of a Charged Particle in a Magnetic Field

When a charged particle moves in a magnetic field, it experiences a force that is always perpendicular to its velocity. Because of this perpendicular nature, the magnetic force does no work on the particle. As a result, the kinetic energy and speed of the particle remain constant.

If a charged particle with charge q moves with velocity v in a magnetic field B, the magnetic force acting on it is given by:

where θ is the angle between the velocity vector (v) and the magnetic field (B).

Motion Based on Initial Conditions

Case 1:

When the charged particle moves parallel (θ = 0°) or anti-parallel (θ = 180°) to the magnetic field, the angle between velocity and magnetic field becomes zero or 180°. Since sin⁡0° = 0 and sin⁡180° = 0, the magnetic force acting on the particle becomes zero:

As no magnetic force acts on the particle, there is no change in its motion. Therefore, the particle continues to move in a straight line at constant velocity parallel to the magnetic field.

Case II:

When a charged particle enters a magnetic field such that its velocity is perpendicular to the magnetic field (θ = 90°), it experiences a maximum magnetic force. This force always acts perpendicular to the velocity, continuously changing its direction but not its speed. The particle moves in a circular path with constant speed. In this case, the magnetic force acts as the centripetal force required for circular motion.

Mathematical Expression

The magnetic force is given by:

This force provides the centripetal force:

On solving, we get the radius of the circular path:

Angular Velocity:

Time Period :The time taken to complete one revolution is:

Frequency

Case III:

When a charged particle enters a magnetic field at an angle other than 0° or 90°, its velocity can be resolved into two components:

• V_||​ → along the direction of the magnetic field
• V⊥​ → perpendicular to the magnetic field

The parallel component V_||​ causes the particle to move in a straight line along the magnetic field, while the perpendicular component V⊥​ produces a circular motion due to the magnetic force. The combined motion is a helical (spiral) path.

Radius of the Helical Path

The circular motion is due to the perpendicular component V⊥ = v sin⁡θ:

Frequency of Revolution

Time Period

Pitch of Helix

Pitch is defined as the distance travelled by the particle along the magnetic field in one complete revolution.

Since V_|| = vcosθ, we get:

Sample Problems

Question 1: A charged particle of charge 2 × 10^-6 C moves with a velocity of 3 × 10⁵ m/s perpendicular to a magnetic field of 0.5 T. Find the magnetic force.

Solution: F = qvB

F = (2 × 10^-6) (3 × 10⁵) (0.5)

F = 0.3 N

Question 2: An electron enters a magnetic field of 0.02 T with velocity 4×10⁶ m/s. Find the radius. (m = 9.1 × 10^-31,  q = 1.6 × 10^-19)

Solution:

Question 3: Find the time period of revolution of a proton in a magnetic field of 0.1 T. (m = 1.67 × 10^-27,  q = 1.6 × 10^-19)

Solution:

Question 4: A particle enters a magnetic field of 0.5 T with velocity 2×10⁶ m/s at an angle of 60 °. Find the radius of the helical path. (m = 1.67 × 10^-27,  q = 1.6 ×10^-19)

Solution:

Unsolved Problems

Question 1: A charged particle of charge 3 × 10^-6 C moves with a velocity of 2 × 10⁵ m/s perpendicular to a magnetic field of 0.4 T. Find the magnetic force acting on it.

Question 2: An electron enters a magnetic field of 0.03 T with a velocity of 5 × 10⁶ m/s perpendicular to the field. Find the radius of its circular path. (m = 9.1 × 10^-31 kg,  q = 1.6 × 10^-19 C)

Question 3: Find the time period of a proton moving in a magnetic field of 0.2 T (m = 1.67 × 10^-27 kg,  q = 1.6 × 10^-19 C)

Question 4: A charged particle enters a magnetic field of 0.6 T with a velocity of 3 × 10⁶ m/s at an angle of 30 °. Find the radius of the helical path. (m = 1.67 × 10^-27 kg,  q = 1.6 × 10^-19 C)