(8c28p17) An electron with kinetic energy 1.68 keV circles in a plane perpendicu

Question

(8c28p17) An electron with kinetic energy 1.68 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 29.0 cm.
Find the speed of the electron.

Find the speed of the electron.

Find the magnetic field.

Find the frequency of circling. (Hz)

Find the period of the motion.

Explanation / Answer

1). v = sqrt(2KE/m) => v = sqrt[2*1.68*10^3*1.602*10^-19/(9.11*10^-31)] => v = 2.43 *10^7 m/s w = v/r rad/s => w = 2.43*10^7/0.29 = 8.38 *10^7 rad/s 2. w = qB/m => B = m*w/q => B = 9.11*10^-31*8.38*10^7/(1.602*10^-19) => B = 4.77 *10^-4 T 3). f = w/(2pi) Hz => f = 8.38 *10^7/(2*3.14) = 1.33 *10^7 Hz 4) T = 1/f => T = 7.52 *10^-8 s

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