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  • (a)A monoenergetic electron beam with electron speed of 5.20×10^6 m s^–1 is subject to a magnetic field of 1.30×10^–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10^11 C kg^–1.

Q : 11.21 (a) A monoenergetic electron beam with electron speed of  5.20\times 10^6\hspace{1mm}ms^-^1  is subject to a magnetic field of  \dpi{100} 1.30\times 10^{-4}\hspace{1mm}T normal to the beam velocity. What is the radius of the circle traced by the beam, given  e/m  for electron equals  \dpi{100} 1.76\times 10^1^1\hspace{1mm}Ckg^-^1. 

Answers (1)

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The force due to the magnetic field on the electron will be Fb=evB (since the angle between the velocity and magnetic field is 90o)

This Fb acts as the centripetal force required for circular motion. Therefore

\\F_{b}=\frac{mv^{2}}{r}\\ evB=\frac{mv^{2}}{r}\\ r=\frac{mv}{eB}\\ r=\frac{5.2\times 10^{6}}{1.76\times 10^{11}\times 1.3\times 10^{-4}}\\ r=0.227 m

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