# Magnetic force between conductors

1. Jun 20, 2008

### scholio

1. The problem statement, all variables and given/known data

a straight conductor carrying 5A is separated by 10cm from a second parallel conductor carrying 10A in the same direction. at what distance from the first conductor (along the line connecting the two conductors) is the magnetic field zero?

2. Relevant equations

F_2 = I_2 L X B_1 = [mu_0(I_1)(I_2)L] / 2(pi)d

where F_2 is force on conductor 2, mu_0 is 4(pi)*10^-7 constant, d is distance between conductors, I is current, L is length of conductor

3. The attempt at a solution

since i need to find the distance when B_1 is zero, i set B_1 = zero and attempted to solve for d, the thing is since when B_1 is zero it makes the left hand side equal to zero thus it is impossible to solve for d

F_2 = I_2 L X B_1 = [mu_0(I_1)(I_2)L]/2(pi)d

I_2 L X B_1 = [mu_0(I_1)(I_2)L]/2(pi)d

I_2 L X 0 = [mu_0(I_1)(I_2)L]/2(pi)d

0 = [mu_0(I_1)(I_2)L]/2(pi)d

0 = mu_0(I_1)(I_2)L where mu_0 = 4pi*10^-7, I_1 = 5A, I_2 = 10A, L is constant

i think my approach is wrong because i did not use the 10cm specified in the problem, am i supposed to use I_1, I_2 and d = 10cm =0.1 m to solve for F_2 first?

then do i compare that to something? substitute?

2. Jun 20, 2008

### calef

At the point where the field is zero, each wire's magnetic field is cancelling out the other. And because you know the wires have current running in the same direction, this means that this point has to be between the wires (if that doesn't make sense, think about how each wire's magnetic field vector affects the others not in between the wires).

This means that one wires field will be positive, and the other's will be negative at this point. This gives you two distinct radii. Those radii sum to 10 cm from your problem. From that information, you should be able to make a system.

3. Jun 20, 2008

### scholio

so i assume the magnetic field of each wire has the same magnitude? i am still somewhat confused, do i put the distance d in terms of radius r from 0.10 meters?
---> d = 0.10 - r?

I_2L X B_1 = [mu_0(I_1)(I_2)L]/2(pi)(0.10 - r) + [mu_0(I_1)(I_2)L]/2(pi)r and so letting B_1 = 0, the left hand side goes to zero thus

[mu_0(I_1)(I_2)L]/2(pi)(0.10 - r) = - [mu_0(I_1)(I_2)L]/2(pi)r

2(pi)r(mu_0(I_1)(I_2)L) = - [mu_0(I_1)(I_2)L](2pi(0.10 - r))

2(pi)r = - 2pi(0.10 - r)
2r = - 0.10
r = - 0.05 = 0.05 meters = 5 cm

is that what you meant? i must've done something wrong because i am supposed to get 3.3 cm from the 5ampere wire/conductor

4. Jun 20, 2008

### calef

They have equal, but opposite magnitude at that point, yes, so that their sum is zero.

I think you're using the wrong equations for one thing.

Use (mu * I) / (2 * pi * r) = B

You have two different currents (which are given) and two different radii, which, summed, equal 10 cm.

Edit: Negative and Positive, in this case, are just a convention, because the magnetic field lines propagate in concentric circles about the wires. Just understand that at this point, if you were looking "down" the wires, one vector would be pointing "up" and the other would be "down", each "arrow" being the same length, thus cancelling each other out and equalling zero.

Last edited: Jun 21, 2008
5. Jun 21, 2008

### Gear300

Find 2 separate equations: one for the magnetic field a distance away from one wire, and one for the magnetic field a distance away from the other wire (make the equations relative, meaning that place both wires on a coordinate plane and take into account the distance between them in the equations---one wire could be along the x-axis and the other some value above or below it). Set both equations equal to each other and solve for the distance.

6. Jun 21, 2008

### scholio

gear300, didn't i already find two equations in my second attempt? the distances were in terms of r, where r < 0.10m? or are you describing something else?

in my second attempt, i was able to solve for r, r = 0.05m. i am supposed to get 0.033m. i think my second attempt was close, except i may have made a mistake with how i presented the distances of each wire and thus my final answer was wrong.

calef, am i supposed to use "(mu * I) / (2 * pi * r) = B" for B_1 in " I_2 L X B_1 = [mu_0(I_1)(I_2)L] / 2(pi)d ." but why do i need to solve for magnetic field, B, when the problem states it has to be zero to solve for the distance?

7. Jun 21, 2008

### calef

Don't use the I_2 L X B_1 formula at all. That's an equation for force. And you're setting it equal to a magnetic field.

You only need equations of the form (mu * I) / (2 * pi * r) = B (two of them) and a third equation to link your radius variables.

Another conceptual hint is to think about the proportionality of it. You have one wire that has twice the current of the other. Your answer is 3.33~ cm away from the weaker wire and 6.66~ cm away from the stronger wire. Doesn't it make sense that it would be twice the distance from the stronger wire that it is from the weaker wire?

Last edited: Jun 21, 2008
8. Jun 21, 2008

### scholio

is the third equation, the biot-savart law? B = mu_0/4pi[integral(IdL/r^2)] where mu_0 = constant = 4pi*10^-7, dL is change in L,constant and r is radius?

when i use the "(mu * I) / (2 * pi * r) = B", should the r for one wire be d, and the for (0.10 - d) for the other?

if so i do set the two "(mu * I) / (2 * pi * r) = B" equations equal to each other, solve for r, in my case d, and then substitute the value d for r in the biot-savart eq, let B equal zero, and then solve for r?

9. Jun 22, 2008

### calef

Oh jeez, nonono. It's much more simple than that. In fact, you don't even need the third equation if you're going to use (.1 -d) and d as your radii. I was using r1 and r2 with my third equation being r1 + r2 = .1.