19_Chapter 19 HomeworkCH19 Magnetism

# 19_Chapter 19 HomeworkCH19 Magnetism - Magnetism 19.42 159...

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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 document was uploaded on 10/20/2011.

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