Net force on 3 parallel wires carrying current.

[Motorscooter]

Homework Statement

You are given 3 wires parallel to each other
(1)------> I
(2)<-------I
(3)------->I
http://www.webassign.net/knight/p32-38alt1.gif (not sure it this link works)
I1=I2=I3 = 16A
the distance between the wires is d=2cm

length of the wires is l= 50cm

Homework Equations

I assume I need to use the μ0 * I1*I2*l/(2∏*d)

The Attempt at a Solution

I calculated that the middle wires force is 0
but for the first wire I tried solving for the force for 1 on 2 and then 1 on 3 and subtracting them because wire 2 repels wire 1 and wire 3 attracts wire 1 I ended up with:
Force 1 on 2 = .0128 N
Force 1 on 3 = .0064 N
I am pretty sure I converted the units correctly but the answer is wrong. Please help.

 

[anathema]

Hi there,

Thank you for sharing your attempt at solving this problem. Your approach is correct, but there are a few things that need to be adjusted.

Firstly, the equation you have used is correct, but you need to use the absolute value of the current for the calculation. This is because the direction of the current will not affect the magnitude of the force, only the direction.

Secondly, the distance between the wires should be converted to meters (m), as the units for the permeability of free space (μ0) are in N/A^2. So, the distance should be 0.02m.

Lastly, when calculating the force exerted by wire 2 on wire 1, you need to use the distance between wire 1 and wire 2, which is 0.04m (since they are 2cm apart and there are 2 wires between them). Similarly, for the force exerted by wire 3 on wire 1, you need to use the distance of 0.06m (since there are 3 wires between them).

Using these adjustments, the force exerted by wire 2 on wire 1 would be 0.008N and the force exerted by wire 3 on wire 1 would be 0.004N. Subtracting these two forces, you would get a net force of 0.004N exerted on wire 1 by the other two wires.

I hope this helps. Keep up the good work!