(1) v = ?2 K/m e where K is the electron\'s kinetic energy and m e is the electr
ID: 1262623 • Letter: #
Question
(1) v = ?2K/me
where K is the electron's kinetic energy and me is the electron mass. The component of velocity along the x direction is then:
(2) vx = v cos? = ?2K/me cos? = _____ m/s
The magnitude of the velocity component in the yz plane, however, is:
(3) vyz = v sin? = ?2K/me sin? = ______ m/s
An electron with kinetic energy K = 8.0 An electron with kinetic energy K = 10-18 J enters a region where there is a uniform magnetic field of magnitude 1.1 T directed along the positive x axis. Its initial direction is at an angle of 19 A degree with respect to the x axis. Find how far the electron moves along the x axis in the time it takes to complete one cycle of its circular motion as seen in the yz plane. (1) v = ?2K/me where K is the electron's kinetic energy and me is the electron mass. The component of velocity along the x direction is then: (2) vx = v cos? = ?2K/me cos? = _____ m/s The magnitude of the velocity component in the yz plane, however, is: (3) vyz = v sin? = ?2K/me sin? = ______ m/sExplanation / Answer
centrepetal force provided by electron=force on moving charge
mV2/R =Q*vBsin(theta)
KE=0.5mV2
mV2=2KE
2KE=Q*vRBsin(theta)
2*8*10-18=1.6*10-19*4.192*R*sin(19)*1.1*106
16/2.402 *10-5=R
R=66.6*10-5m
2)
VX=Vcos(theta)
=4..192*106*cos(19)
=3.963*106m/s
3)
vyz = v sin(theta)
=4..192*106*sin(19)
=1.364m/s
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