Motion of a Charged Particle in a Magnetic Field

Last Updated : 26 May, 2026

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:

\text{F = qv B sinθ}

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

2056958175

Motion Based on Initial Conditions

Case 1:

2056958172

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:

F = qvB \sin 0° = 0

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:

2056958173

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: \text{F = qvB}

This force provides the centripetal force:

\frac{mv^2}{R} = qvB

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

R = \frac{mv}{qB}

Angular Velocity: \omega = \frac{v}{R} = \frac{qB}{m}

Time Period :The time taken to complete one revolution is: T = \frac{2\pi}{\omega}= \frac{2\pi m}{qB}

Frequency f = \frac{1}{T} = \frac{qB}{2\pi m}

Case III:

2056958174

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:

r = \frac{mv_{\perp}}{qB}= \frac{mv\sin\theta}{qB}

Frequency of Revolution

f = \frac{qB}{2\pi m}

Time Period

T = \frac{2\pi m}{qB}

Pitch of Helix

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

\text{Pitch} = v_{\parallel} \cdot T

Since V|| = vcosθ, we get:

\text{Pitch} = v\cos\theta \cdot \frac{2\pi m}{qB}

2056958171

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: R = \frac{mv}{qB}

R = \frac{9.1 \times 10^{-31} \times 4 \times 10^6}{1.6 \times 10^{-19} \times 0.02}

R = \frac{36.4 \times 10^{-25}}{3.2 \times 10^{-21}}

R = 1.14 \times 10^{-3} \, m

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:

T = \frac{2\pi m}{qB}

T = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.1}

T = \frac{10.48 \times 10^{-27}}{1.6 \times 10^{-20}}

T = 6.55 \times 10^{-7} \, s

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: v_{\perp} = v \sin\theta = 2 \times 10^6 \times \frac{\sqrt{3}}{2} = 1.732 \times 10^6

r = \frac{mv_{\perp}}{qB}

r = \frac{1.67 \times 10^{-27} \times 1.732 \times 10^6}{1.6 \times 10^{-19} \times 0.5}

r = \frac{2.89 \times 10^{-21}}{0.8 \times 10^{-19}}

r = 3.61 \times 10^{-2} \, m

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)

Comment

Explore