Two insulated current-carrying wires (wire 1 and wire 2) are bound together with

Question

Two insulated current-carrying wires (wire 1 and wire 2) are bound together with wire ties to form a two-wire unit. The wires are 2.59 m long and are stretched out horizontally parallel to each other. Wire 1 carries a current of I1 = 8.00 A and the other wire carries a current I2 in the opposite direction. The two-wire unit is placed in a uniform magnetic field of magnitude0.400 T such that the angle between the direction of I1 and the magnetic field is 68.5°. While we don't know the current in wire 2, we do know that it is smaller than the current in wire 1. If the magnitude of the net force experienced by the two-wire unit is 3.50 N, determine the current in wire 2.

Explanation / Answer

Use F = I*L*B*sin(?)
so I (net) = F/(L*B*sin(?)) = 3.50/(2.59*0.400*sin(68.5) = 3.631A
So I = 8 - 3.631 = 4.369A
Since you didn't state the direction of the force the other current could be 8.00 + 3.631 = 11.631A
Basically the difference in currents must be 3.631A. So with either 4.369A or 11.631A the net current would be 4.369A

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