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 01 1 1 0 T m A. 0 0 A 2.12 T 2 20 . 2 0 02 I B r π µ 41 3 m ×⋅ == = The contribution from wire 2 is in the – x direction and has magnitude ( )( ) 7 2 2 0 T m 0 0 A 5.00 T 22 0 0 m I B r 5 . 2 0 ππ = cos135 cos180 2.12 T c 135 5.00 T cos180 6. x BB B Therefore, the components of the net field at point P are: 12 os 50 T µµ ° = + ° + and ( ) s 35 sin 2.12 T s y B in1 180 in135 0 1.50 T + = + = ° Therefore, 6.67 T net x y B =+ = at 11 6.50 T tan 1.50 T x y B B −−    tan 77.0 = ° or 6.67 T at 77.0 to the left of vert net B JG ical ± ± ± ² ± ³ ² ² ² ³ ² ± ³ ±² ± ² ± ´µ¶ ³ ± ·¸·¹º»¼½· ± ¾ ³ ´ ·¸·²»¼½· ± ¾ ² 19.41 Call the wire along the x axis wire 1 and the other wire 2. Also, choose the positive
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## This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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