# Parallel conducting wires and zero magnetic field x value

## Homework Statement

Consider two parallel conducting wires along the direction of the z axis as shown below. Wire 1 crosses the x-axis at x = -1.10cm and carries a current of 3.50A out of the xy-plane of the page. Wire 2 (right) crosses the x axis at x = 1.10cm and carries a current of 6.20A into the xy plane.

At which value of x is the magnetic field zero? (Hint: Careful with sign)

B = u0I / 2pi r

## The Attempt at a Solution

I found the B field values for both wires, but I don't understand how to find where they intersect? Can someone help?
Thanks! The diagram is attached

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berkeman
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## Homework Statement

Consider two parallel conducting wires along the direction of the z axis as shown below. Wire 1 crosses the x-axis at x = -1.10cm and carries a current of 3.50A out of the xy-plane of the page. Wire 2 (right) crosses the x axis at x = 1.10cm and carries a current of 6.20A into the xy plane.

At which value of x is the magnetic field zero? (Hint: Careful with sign)

B = u0I / 2pi r

## The Attempt at a Solution

I found the B field values for both wires, but I don't understand how to find where they intersect? Can someone help?
Thanks! The diagram is attached
Use the righthand rule to draw how the two B-fields circle around their respective wires. How do the two B-fields add in the region between the wires...?

The values of the B fields will be additive, because they will both be coming up (y axis) on the page. So if I add the two B fields, I get 1.766E-4, and if I use that to solve for r, I still don't know what to use for I in the equation B = u0I / 2pi r....

berkeman
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The values of the B fields will be additive, because they will both be coming up (y axis) on the page. So if I add the two B fields, I get 1.766E-4, and if I use that to solve for r, I still don't know what to use for I in the equation B = u0I / 2pi r....
Ack, my bad. I missed that the two currents were in opposite directions. So you are right, they will add in the region between the wires. What about outside the wire area?

outside the wire area they will also be additive because they will both be going down the y axis. I guess I still don't know how to solve for the x value where the current equals zero though...

berkeman
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outside the wire area they will also be additive because they will both be going down the y axis.
Nope. Draw the B-field circles, with arrows for direction based on the righthand rule.

I did, and I got them to both be going down. The wire on the right points into the page, (thumb out) and so points down on the outside of it, and the wire on the left points into the page (thumb in) and so points down on the outside of it

Outside being in the -x axis beyond the left wire and +x axis beyond the right wire. Is this what you were asking? It would be the opposite of what the B fields are inside the wires.

berkeman
Mentor
I did, and I got them to both be going down. The wire on the right points into the page, (thumb out) and so points down on the outside of it, and the wire on the left points into the page (thumb in) and so points down on the outside of it

Outside being in the -x axis beyond the left wire and +x axis beyond the right wire. Is this what you were asking? It would be the opposite of what the B fields are inside the wires.
Each B-field circles its respective wire. How can it point "up" no matter which side of the wire (left or right) you are on? Please draw the circulating B-field lines for each wire. Draw arrows along the circles showing from the RHR which way the B-field is circulating.

Maybe this is the confusion -- draw both B-circles to the left of the left wire. Include both the B-field circle from the left wire, and the (larger diameter) circle from the right wire. That's what I mean by adding those two B-fields outside of the wire pair. Now do the same to the right of the right wire.

Oh, sorry I thought you were talking about just the effect of each wire on it's own side (which they do both go down).

If you combine them both they would contradict or cancel (except for different currents) because they point in opposite directions.

But the B field values are different, so how would I calculate where the B field is zero?

berkeman
Mentor
Oh, sorry I thought you were talking about just the effect of each wire on it's own side (which they do both go down).

If you combine them both they would contradict or cancel (except for different currents) because they point in opposite directions.

But the B field values are different, so how would I calculate where the B field is zero?
Use the equation for the B-field strength as a function of radius away from a current conducting wire. You listed it in your original post (OP). It comes from Ampere's Law -- that's what you are applying in this problem.

That's what I thought but I don't know which I value to use, because there are two different ones???

Or can I just do it for whichever current I'm on the side of? But then I will get two different radiuses?

berkeman
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That's what I thought but I don't know which I value to use, because there are two different ones???
Do what I said in post 8. Draw a small circle around the left wire, and a big circle around the right wire -- big enough to touch the left wire's circle on the x axis to the left of the left wire. Now write the equation for the combination (now a subtraction) of the two B-fields in that region to the left of the left wire. Can you find a set of radii from the two wires that makes the combination go to zero?

Now do the same with two new circles to the right of the right wire. Is there a distance out to the right where the two different radii from the two wires, and the two different current magnitudes, lets the sum of the two B-fields go to zero?

I need to bail for a couple hours, so I hope this is making more sense now.

I still don't understand it, sorry. Thanks for trying. I'll have to go get help tomorrow. I got r = .004 m but that was wrong.

berkeman
Mentor
I still don't understand it, sorry. Thanks for trying. I'll have to go get help tomorrow. I got r = .004 m but that was wrong.
You're not solving for an r. You're solving for a value of x where the two raidii to the two wires and the two currents give you two B-field values that are equal and opposite.

Then I have no clue what equation to use. I know the B fields oppose each other on the outside of each wire, but when you say, "Can you find a set of radii from the two wires that makes the combination go to zero" I guess I don't know how to solve for two different variables at once. Don't worry about it, thanks for your help anyway.

berkeman
Mentor
Then I have no clue what equation to use. I know the B fields oppose each other on the outside of each wire, but when you say, "Can you find a set of radii from the two wires that makes the combination go to zero" I guess I don't know how to solve for two different variables at once. Don't worry about it, thanks for your help anyway.
I'll try one last time, (I got stuck at work, but am about to head out).

For the first drawing I mentioned, with the little left circle and the bigger right circle, where they come together at some distance left of the left wire on the x axis, you write an equation that adds up (actually subtracts because of the opposed B vectors) the B fields from the two wires. The r value from the left wire for the left circle is smaller than the r value for the circle around the right wire, correct? Use the equation in your first post for each of the two B fields and add the two and set the sum equal to zero. You will write an expression for each of the two radii that turns it into a value of x (for each x to the left of the left wire, you can write what each of the two radii are, correct?).

You may or may not be able to find an x to the left where they cancel. Do the same for the region to the right of the right wire, and see if there is an x where the two B fields are equal and opposite so they cancel.

Whenever you post again, please post a sketch of the wires and B-field circles I've been describing. That will help us (and any other Homework Helpers) to see where you are having problems.