Unformatted text preview: 28. We consider the point at which it enters the ﬁeld-ﬁlled region, velocity vector pointing downward. The
ﬁeld points out of the page so that v × B points leftward, which indeed seems to be the direction it is
“pushed”; therefore, q > 0 (it is a proton).
(a) Eq. 29-17 becomes
2 130 × 10−9
which yields |B | = 0.252 T. =
e |B |
2π 1.67 × 10−27
(1.60 × 10−19 ) |B | √
(b) Doubling the kinetic energy implies multiplying the speed by 2. Since the period T does√
depend on speed, then it remains the same (even though the radius increases by a factor of 2).
Thus, t = T /2 = 130 ns, again. ...
View Full Document
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