825_Physics ProblemsTechnical Physics

825_Physics ProblemsTechnical Physics - Chapter 29 P29.9 FB...

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Chapter 29 165 P29.9 F v B B q vB ij k ijk i j k Fv B ×= + − + ++− =−+ ++ + =+ + + + = = × = × −− ±±± ±± ± ± ± ± . .. . 241 123 12 2 1 6 4 4 10 7 8 1 0781 4 6 160 10 146 234 10 222 19 18 a f a f a f ej bg Tms C T m s N B q P29.10 q Ek k =− × × 200 320 10 19 18 ± . ± N C N FEvB a ki B k B k iB k =+×= −× × × = × × × ×⋅ × = × qq m 3 20 10 1 60 10 9 11 10 2 00 10 3 20 10 1 92 10 1 82 10 1 92 10 5 02 10 18 19 31 12 18 15 18 15 18 . ± . ± ± . ± . ± . ± . ± . ± N C 1.20 10 m s m s N C m s N Cms N 42 e j e j e j e j The magnetic field may have any -component x . B z = 0 and B y 262 . m T . Section 29.2 Magnetic Force Acting on a Current-Carrying Conductor P29.11 FI L B B = sin θ with FFm g Bg == mg ILB = sin so m L gI B = sin I = . A and m L = F H G I K J 0500 100 1000 500 10 2 gcm cm m gkg kg m . Thus 5 00 10 9 80 2 00 90 0 2 . s i n . ° af B FIG. P29.11 B = 0245 T e s l a with the direction given by right-hand rule:
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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