ch28-p020 - 20. Using Eq. 28-16, the radius of the circular...

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2 mv mK r qB qB == where 2 /2 Km v = is the kinetic energy of the particle. Thus, we see that K = ( rqB ) 2 /2 m q 2 m –1 . (a) () 2 2 21 4 1 . 0 M e V ; pp p p p Kq qm m K K K αα α = = (b) 2 2 1 1 2 1.0 MeV 2 0.50MeV.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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