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By the end of this section, you will be able to:
  • Explain how a charged particle in an external magnetic field undergoes circular motion
  • Describe how to determine the radius of the circular motion of a charged particle in a magnetic field

A charged particle experiences a force when moving through a magnetic field. What happens if this field is uniform over the motion of the charged particle? What path does the particle follow? In this section, we discuss the circular motion of the charged particle as well as other motion that results from a charged particle entering a magnetic field.

The simplest case occurs when a charged particle moves perpendicular to a uniform B -field ( [link] ). If the field is in a vacuum, the magnetic field is the dominant factor determining the motion. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The particle continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.

An illustration of the motion of a charged particle in a uniform magnetic field. The magnetic field points into the page. The particle is negative and moves in a clockwise circle. Its velocity is tangent to the circle, and the force points toward the center of the circle at all times.
A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper (represented by the small × ’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion. (Note that because the charge is negative, the force is opposite in direction to the prediction of the right-hand rule.)

In this situation, the magnetic force supplies the centripetal force F c = m v 2 r . Noting that the velocity is perpendicular to the magnetic field, the magnitude of the magnetic force is reduced to F = q v B . Because the magnetic force F supplies the centripetal force F c , we have

q v B = m v 2 r .

Solving for r yields

r = m v q B .

Here, r is the radius of curvature of the path of a charged particle with mass m and charge q , moving at a speed v that is perpendicular to a magnetic field of strength B . The time for the charged particle to go around the circular path is defined as the period, which is the same as the distance traveled (the circumference) divided by the speed. Based on this and [link] , we can derive the period of motion as

T = 2 π r v = 2 π v m v q B = 2 π m q B .

If the velocity is not perpendicular to the magnetic field, then we can compare each component of the velocity separately with the magnetic field. The component of the velocity perpendicular to the magnetic field produces a magnetic force perpendicular to both this velocity and the field:

v perp = v sin θ , v para = v cos θ .

where θ is the angle between v and B . The component parallel to the magnetic field creates constant motion along the same direction as the magnetic field, also shown in [link] . The parallel motion determines the pitch p of the helix, which is the distance between adjacent turns. This distance equals the parallel component of the velocity times the period:

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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