853_Physics ProblemsTechnical Physics

853_Physics ProblemsTechnical Physics - 193 Chapter 30...

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Chapter 30 193 P30.15 Take the x -direction to the right and the y -direction up in the plane of the paper. Current 1 creates at P a field B I a 1 0 7 2 200 10 300 00500 == ×⋅ µ π .. . Tm A A m ej af bg B 1 12 0 = . T downward and leftward, at angle 67.4 ° below the – x axis. Current 2 contributes B 2 7 0120 = . A m a f clockwise perpendicular to 12.0 cm B 2 500 = to the right and down, at angle –22.6° 5.00 cm 13.0 cm 12.0 cm P I 1 I 2 B 2 B 1 FIG. P30.15 Then, BB B i j i j = + = °− ° + °− ° 12 12 0 67 4 67 4 5 00 22 6 22 6 . ± cos . ± sin . . ± cos . ± sin . T T µµ Bj j j =− = − 111 1 92 13 0 . ± . ± . ± T T b g Section 30.2 The Magnetic Force Between Two Parallel Conductors P30.16 Let both wires carry current in the x direction, the first at y = 0 and the second at y = 10 0 . cm. (a) Bk k 0 7 2 41 0 5 0 0 20 1 0 0 I r ± . . ± TmA A m B 100 10 5 . T out of the page y y = 10.0 cm I 1 = 5.00 A I 2 = 8.00 A z x FIG. P30.16(a)
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