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Unformatted text preview: 21. So that the magnetic ﬁeld has an eﬀect on the moving electrons, we need a nonnegligible component of
B to be perpendicular to v (the electron velocity). It is most eﬃcient, therefore, to orient the magnetic
ﬁeld so it is perpendicular to the plane of the page. The magnetic force on an electron has magnitude
FB = evB , and the acceleration of the electron has magnitude a = v 2 /r. Newton’s second law yields
evB = me v 2 /r, so the radius of the circle is given by r = me v/eB in agreement with Eq. 2916. The
kinetic energy of the electron is K = 1 me v 2 , so v = 2K/me . Thus,
2
r= me
eB 2K
=
me 2m e K
.
e2 B 2 This must be less than d, so
2m e K
≤d
e2 B 2
or
2m e K
.
e2 d2
If the electrons are to travel as shown in Fig. 2933, the magnetic ﬁeld must be out of the page. Then
the magnetic force is toward the center of the circular path, as it must be (in order to make the circular
motion possible).
B≥ ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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