A proton travels in a circle perpendicular to a magnetic field. The field strength is 2T...

Question

Question

A proton travels in a circle perpendicular to a magnetic field. The field strength is 2T and the proton speed is 20m/s. Calculate the following:

radius of circle

period of motion

frequency of motion

Answer 1

Answer #1

Radius of circle $r=\frac{mv}{eB}=\frac{1.6726*10^{-27}*20}{1.6*10^{-19}*2}=1.045*10^{-7}\,m$

Period of motion $T=\frac{2\pi r}{v}=\frac{2\pi *1.045*10^{-7}}{20}=3.28*10^{-8}\,s$

Frequency of motion $f=\frac{1}{T}=\frac{1}{3.28*10^{-8}}=3.04*10^7\,Hz$

Answer 2

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