Explain the Motion of a charged particle in a uniform magnetic field.

Explanation

Consider a charged particle of charge q having mass m enters into a region of uniform magnetic field `vecB ` with velocity `vecv` such that velocity is perpendicular to the magnetic field . As soon as the particle enters into the field , Lorentz force acts on it in a direction perpendicular to both magnetic field `vecB` and velocity `vecv`.
`vecB` in

As a result, the charged particle moves in a circular orbit as shown in Figure .
The Lorentz force on the charged particle is given by
` vecf = q ( vecv xx vecB) `
Since Lorentz force alone acts on the particle , the magnitude of the net force on the particle is
` sum_(i) F_(i) = F_(m) = q v B `
This Lorentz force acts as centripetal force for the particle to execute cirtular motion. Therefore,
` qvB = m v^(2)/r`
The radius of the circular path is
` r = (mv)/(qB) = p/(qB) ` .......(1)
where ` p = mv` is the magnitude of the linear momentum of the particle. Let T be the time taken by the particle to finish one complete circular motion then
` T = (2 pi r)/ v` ......(2)
Hence substituting (1) in (2) , we get
` T = (2 pi m ) / ( q B ) ` .....(3)
Equation (3) is called the cyclotron period .
The reciprocal of time period is the frequency f, which is
` f = 1/T`
` f = (qB)/(2 pi m) ` ......(4)
In terms of angular frequency `omega`,
` omega = 2 pi f = q/m B` ......(5)
Equation (4) and equation (5) are called as cyclotron frequency or gyrofrequency .
From equations, we infer that time period and frequency depend only on charge - to - mass ratio ( specific charge ) but not velocity or the radius of the circular path .