P29_001 - 1(a We use Eq 29-3 FB = |q |vB sin = 3.2 1019...

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1. (a) We use Eq. 29-3: F B = | q | vB sin φ =(+3 . 2 × 10 19 C)(550 m / s)(0 . 045 T)(sin 52 )=6 . 2 × 10 18 N. (b) a = F B /m =(6 . 2 × 10 18 N) / (6 . 6 × 10 27 kg) = 9 . 5 × 10 8 m / s 2 . (c) Since it is perpendicular to ±v , ± F B does not do any work on the particle. Thus from the work-energy
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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.

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