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