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- 1. An electron with kinetic energy equal to 1.20 keV moves in a circular path in a plane perpendicular to a uniform magnetic field. The radius of the orbit is 25.0 cm. Find a) the speed of the electron b) the magnitude of the magnetic field c) the frequency f with which it circulates d) the period T of the movement Solution a) KInetic energy of electron K=(1/2)mV 2 1.2*10 3 *1.6*10 -19 =(1/2)(9.11*10 -31 )*V 2 V=2.05*10 7 m/s b) Since centripetal force is balanced by magnetic force qvB=mv 2 /r B=mv/rq=(9.11*10 -31 )(2.05*10 7 )/(0.25)(1.6*10 -19 ) B=4.68*10 -4 T C) frequency is given by f=qB/2pim =(1.6*10 -19 )(4.68*10 -4 )/(2pi*9.11*10 -31 ) f=1.31*10 7 Hz
- 2. d) Period T=1/f =7.66*10 -8 s