An electron of kinetic energy 0.45 KeV moves in a circular orbit perpendicular to a magnetic field of 0.325 T. Find (a) the radius of the orbit and (b) the frequency and period of the motion ?

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Answer 1 · 2018-04-20T13:44:55

#"(a)"# Radius of the circular path traversed by the particle in the magnetic in field #"B"# is given as

#"r" = "p"/"qB"#

Where #"p"# is momentum of charged particle.

#"r" = (sqrt("2mK"))/"qB" color(white)(..)[∵ "K" = "p"^2/"2m"]#

#"r" = (sqrt(2 × 9.1 × 10^-31\ "kg" × 0.45 × 10^3 × 1.6 × 10^-19\ "J"))/(1.6 × 10^-19\ "C" × "0.325 T")#

#"r" = 0.22 × 10^-3\ "m" = color(blue)"0.22 mm"#

#————#

#"(b)"# Time period is given by

#"T" = "2πm"/"qB"#

#"T" = (2 × 3.14 xx 9.1 × 10^-31\ "kg")/(1.6 × 10^-19\ "C" × "0.325 T") = color(blue)(1.099 × 10^-10\ "s")#

#"Frequency" = 1/"Time period"#

#f = 1/(1.099 × 10^-10\ "s") = 9.099 × 10^9\ "Hz" = color(blue)"9.099 GHz"#

# #

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