Question

Find the expression of time period, velocity, frequency of charged particle when it enters perpendicular to the direction of magnetic field. (Circular path).

Answer 1

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. This is typical of uniform circular motion. The simplest case occurs when a charged particle moves perpendicular to a uniform B-field, such as shown in. (If this takes place in a vacuum, the magnetic field is the dominant factor determining the motion. ) Here, the magnetic force (Lorentz force) supplies the centripetal force

Fc = mv^2/r

where m= mass
v= velocity
r= radius
Fc = centripetal force

Noting that
sinθ=1sinθ=1
we see that

F=qvB.F=qvB.
The Lorentz magnetic force supplies the centripetal force, so these terms are equal:
qvB= mv^2/r

.
solving for r yields
r=mv/qB
Here, r, called the gyroradius or cyclotron radius, is the radius of curvature of the path of a charged particle with mass m and charge q, moving at a speed v perpendicular to a magnetic field of strength B. In other words, it is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field. If the velocity is not perpendicular to the magnetic field, then v is the component of the velocity perpendicular to the field.

f=frequency = v/2πr

becomes
f =qB/2πm

The cyclotron frequency is trivially given in radians per second by
ω=qB/m

A moving charge in a cyclotron will move in a circular path under the influence of a constant magnetic field. If the time to complete one orbit is calculated:

T= 2πr/v

=2πmv/qBv

=2πm/qB