A positively charged particle of mass 6.70 Times 10^-8 kg is traveling due east

Question

A positively charged particle of mass 6.70 Times 10^-8 kg is traveling due east with a speed of 78.6 m/s and enters a 0.259-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 3.91 Times 10^-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field, (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge. Number units Number units

Explanation / Answer

While the particle is in the field it travels distance d = 78.6x3.91x10^-3 = 0.307326 m

radius r = 2d/pi =2x0.307326/pi = 0.1956 m.
so,the magnitude of the magnetic force acting on the particle

F = mv^2/r = 6.70x10^-8 x 78.6^2/0.1956 = 2.116x10^-3 N ............................ans a)

omega = v/r =78.6/0.1956 = 401.84 rad/s
the magnitude of the charge of the particleq

= m*omega/B = 6.70x10^-8 x401.84/0.259 = 1.0395x10^-4 C ........................ans b)

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