Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance betw

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 of 1.40 A. To the right, wire 2 carries a downward current of 4.80 A. 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 (b) the magnitude and direction of the current in wire 3 (a) the position of wire 3 (b) the magnitude and direction of the current in wire 3 (a) the position of wire 3 (b) the magnitude and direction of the current in wire 3

Explanation / Answer

The force(per unit length) on wire 1 by 2 is to the left and

= u*I1*I2/2pir = 2x10^-7*1.4* 4.8/0.20 = 6.72 x10^-6N/m

From third law we have the force on wire2 by wire 1 is 6.72 x10^-6N/m and to the right

So the magnitude of the force on each wire by wire 3 must be 6.72 x10^-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 ?*I1*I3/2pix = ?*I3*I2/2pi(x + 0.20)

so solving for x we get

I1/x = I2/(x+0.20)

or x +0.20 = 4x

so x = 0.082 m

Now ?*I1*I3/2pix = 6.72 x10^-6

so I3 = 1.968 A and its direction is down

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