Magnetic force between 3 straight wires

[kiewelb]

Find I2 so that the magnetic force per unit length on each wire is zero. The three parallel wires are equally spaced. Current directions are shown with '<' or '>'.

I1 = I3 = 1 A

< I1 < I1 < I1
---------------
> I2 > I2 > I2
---------------
< I3 < I3 < I3
---------------

I have the formula F = (mu0*I*I')/(2*pi*r) with mu0 = 4*pi*10^-7 but have no clue where to take this to figure out I2 with the 3 wires.

Sorry if this is really easy but I'm not too quick with Physics. Thanks for any help.

 

[gnome]

r in the formula is the distance between 2 wires. So let r = distance between wire 1 and wire 2. Then the distance between wire 1 and wire 3 is ______?

Next, use that formula to find the force exerted on wire 1 by wire 2, and the force exerted on wire 1 by wire 3. If the NET force on wire 1 is 0, those two forces must be equal in magnitude & opposite in direction so set the two F's equal to each other & solve for I₂.

 

[M98Ranger]

To find the value of I2 that will result in a magnetic force per unit length of zero on each wire, we can use the principle of superposition. This means that the total force on a wire is the sum of the forces due to each individual wire. In this case, we have three wires, so we can set up three equations to solve for I2.

First, let's consider the force on the top wire (I1) due to the other two wires (I2 and I3). Since the wires are equally spaced, the distance between I1 and I2 is the same as the distance between I1 and I3. This means that we can use the same value for r in both calculations. The force on I1 due to I2 is given by F12 = (mu0*I1*I2)/(2*pi*r) and the force on I1 due to I3 is given by F13 = (mu0*I1*I3)/(2*pi*r). Since we want the total force on I1 to be zero, we can set F12 + F13 = 0 and solve for I2:

(mu0*I1*I2)/(2*pi*r) + (mu0*I1*I3)/(2*pi*r) = 0
I2 + I3 = 0
I2 = -I3

Next, let's consider the force on the middle wire (I2) due to the other two wires (I1 and I3). Again, using the principle of superposition, we can set up the equation F21 + F23 = 0 and solve for I2:

(mu0*I2*I1)/(2*pi*r) + (mu0*I2*I3)/(2*pi*r) = 0
I1 + I3 = 0
I1 = -I3

Finally, let's consider the force on the bottom wire (I3) due to the other two wires (I1 and I2). Again, using the principle of superposition, we can set up the equation F31 + F32 = 0 and solve for I2:

(mu0*I3*I1)/(2*pi*r) + (mu0*I3*I2)/(2*pi*r) = 0
I1 + I2 = 0
I1 = -I2

We can see that all three equations lead to the same result: I2 = -I3 = -I