Question

An electron of kinetic energy 1.77 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 28.2 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Answer 1

r = 28.2 cm , K = 1.77 keV

m = 9.1*10^-31 C, q = 1.6*10^-19 C

(a) K = (1/2)mv^2

1.77*1000*1.6*10^-19 = 0.5*9.1*10^-31*v^2

v = 2.5*10^7 m/s

(b) centripetal force = magnetic force

mv^2/r = qvB

mv = qBr

9.1*10^-31*2.5*10^7 = 1.6*10^-19*B*0.282

B = 0.0005 T = 0.5 mT

(c) f = qB/2pi*m

f = 1.6*10^-19*0.0005/(2*3.14*9.1*10^-31)

f = 1.4*10^7 Hz

(d) T = 1/f = 1/(1.4*10^7)

T = 7.14*10^-8 s