An electron of kinetic energy 40 keV movesin a circular orbit perpendicular to a

Question

An electron of kinetic energy 40 keV movesin a circular orbit perpendicular to a magnetic field of0.300 T. (a) Find the radius of the orbit.
1 mm

(b) Find the frequency of the motion.
2 GHz
Find the period.
3 ns (a) Find the radius of the orbit.
1 mm

(b) Find the frequency of the motion.
2 GHz
Find the period.
3 ns

Explanation / Answer

As we know,the magnetic force is the centripetal force => F =BeV = mV^2/R => R = mV/Be on the other hand : the kinetic energy : K = (1/2)mV^2 => V =(2K/m) => the radius : R = m*(2K/m) /(Be) = 2Km /(Be) =(2*40x10^3*1.6x10^-19*9.1x10^-31)/(0.3*1.6x10^-19) =2.25x10^-3 m the period : T = 2R/V =2m/(Be) => the frequency of motion : f = 1/T = Be/(2m) =8.395x10^9 Hz period : T = 1/f = 1.19x10^-10 s

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