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Explain the Motion of a charged particle in a uniform magnetic field. |
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Answer» Solution :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 aloneacts 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 = (2pi 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 velocityor the radius of the circular path . |
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