Chapter_28_Solutions - 28.1 For a charge with velocity v...

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28.1: For a charge with velocity G v = (8.00 × 10 6 m s ) ˆ j , the magnetic field produced at a position r away from the particle is G B = µ 0 4 π q G v × ˆ r r 2 . So for the cases below: a) G r = ( + 0.500 m) ˆ i ˆ v × ˆ r = − ˆ k, r 0 2 = 1 4 G B = − µ 0 4 π qv r 0 2 ˆ k = − µ 0 4 π (6.0 × 10 6 C)(8.0 × 10 6 m s ) (0.50 m) 2 ˆ k = − (1.92 × 10 5 T) ˆ k ≡ − B 0 ˆ k. b) G r = ( 0.500 m) ˆ j ˆ v × ˆ r = 0 G B = 0. c) G r = (0.500 m) ˆ k ˆ v × ˆ r = + ˆ i , r 0 2 = 1 4 . G B = + µ 0 4 π qv r 0 2 ˆ i = B 0 ˆ i . d) G r = − (0.500 m) ˆ j + (0.500 m) ˆ k ˆ v × ˆ r = − ˆ i , r 2 = 1 2 = 2r 0 G B = + µ 0 4 π qv r 2 ˆ i 2 = + B 0 2 ˆ i 2 = + B 0 ˆ i 2 2
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