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18_Ch 19 College Physics ProblemCH19 Magnetism

18_Ch 19 College Physics ProblemCH19 Magnetism - 158...

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158 CHAPTER 19 (b) At point P , ( ) 1 0.200 m 2 r = and B is directed at 1 1 135 θ = + ° . The magnitude of is 1 B ( ) ( ) ( ) 7 0 1 1 1 0 T m A .00 A 2.12 T 2 2 0.200 2 I B r π µ 4 1 3 m µ π π × = = = The contribution from wire 2 is in the – x direction and has magnitude ( ) ( ) ( ) 7 0 2 2 2 0 T m A .00 A 5.00 T 2 2 0 0 m I B r π µ 4 1 5 .20 µ π π × = = = ( ) cos135 cos180 2.12 T c 135 5.00 T cos180 6. x B B B Therefore, the components of the net field at point P are: ( ) 1 2 os 50 T µ µ µ ° = − = ° + = ° ° + and ( ) 1 2 s 35 sin 2.12 T s y B B B in1 180 in135 0 1.50 T µ µ + = + = ° + ° = ° Therefore, 2 2 6.67 T net x y B B B µ = + = at 1 1 6.50 T tan 1.50 T x y B B µ µ = = tan 77.0 θ = ° or 6.67 T at 77.0 to the left of vert net µ = ° B JG ical 19.41
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