Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance betw
ID: 2185800 • Letter: T
Question
Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance between wire 1 and wire 2 is 20.0 cm. On the left, wire 1 carries an upward current of1.80A. To the right, wire 2 carries a downward current of4.80A. Wire 3 is located such that when it carries a certain current, each wire experiences no net force. Find the following
(a) the position of wire 3 cm ---Position---left of wire 1right of wire 1 (b) the magnitude and direction of the current in wire 3 A ---Direction---updownExplanation / Answer
The force(per unit length) on wire 1 by 2 is to the left and = ?o*I1*I2/2?r = 2x10^-7*1.8*4.8/0.20 = 8.64 x10^-6N/m From third law we have the force on wire2 by wire 1 is 8.64x10^-6N/m and to the right So the magnitude of the force on each wire by wire 3 must be 8.64x10^-6N/m Since the current in wire 2 is > wire 1 then wire 3 must be closer to wire 1 than 2 and wire 3 must be to the left of wire 1 Let x be the distance from wire 3 to wire 1 so x + 0.20 = distance from wire 3 to wire 2 Now ?o*I1*I3/2?x = ?o*I3*I2/2?(x + 0.20) so solving for x we get I1/x = I2/(x+0.20) or x +0.20 = x*4.8/1.8 x+0.2=2.67x so x = 0.20/1.67 = 0.1197 m Now ?o*I1*I3/2?x = 8.64x10^-6 so I3 = 8.64x10^-6*0.1197/(2.0x10^-7*1.8) = 2.8728 A and its direction is down
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