19_Ch 19 College Physics ProblemCH19 Magnetism

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

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Magnetism 159 19.42 Since the proton moves with constant velocity, the net force acting on it is zero. Thus, the magnetic force due to the current in the wire must be counterbalancing the weight of the proton, or where qvB mg = 0 2 BId µ π = . This gives 0 2 qv I d mg = , or the distance the proton is above the wire must be ( ) ( ) ( )( ) () 19 4 7 6 27 2 10 C 2.30 10 m s 4 10 T m A 1.20 10 A 2 21 . 6 71 0 k g9 . 8 0 m s qv d −− ××× × × 0 1.60 I mg == 2 10 m 5.40 5.40 cm d = 19.43 (a) From 0 2 BIr = , observe that the field is inversely proportional to the distance from the conductor. Thus, the field will have one-tenth its original value if the distance is increased by a factor of 10. The required distance is then 10 10 0.400 m 4.00 m rr = (b) A point in the plane of the conductors and 40.0 cm from the center of the cord is located 39.85 cm from the nearer wire and 40.15 cm from the far wire. Since the currents are in opposite directions, so are their contributions to the net field.
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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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