# Point between two wires at which the flux density is zero

1. Sep 25, 2016

### moenste

1. The problem statement, all variables and given/known data
A long wire (X) carrying a current of 30 A is placed parallel to, and 3.0 cm away from, a similar wire (Y) carrying a current of 6.0 A. What is the flux density midway between the wires: (a) when the currents are in the same direction, (b) when they are in opposite directions? (c) When the currents are in the same direction there is a point somewhere between X and Y at which the flux density is zero. How far from X is this point? (μ0 = 4 π * 10 -7 H m-1.)

Answers: (a) 3.2 * 10-4 T, (b) 4.8 * 10-4 T, (c) 2.5 cm

2. The attempt at a solution
(a) B = μ0 I / 2 π a = [4 π * 10 -7 * (30 - 6)] / 2 π * (0.03 / 2) = 3.2 * 10-4 T. I is (30 - 6) since the currents are in the same direction and 0.03 is divided by two because we need the flux density midway between the wires.

(b) Same formula but 30 + 6, since the currents are in opposite directions = 4.8 * 10-4 T.

(c) Regarding this part I don't know. I used the abovementioned formula and got: [4 π * 10 -7 * (30 - 6)] / 2 π * a = 0 → 7.2 * 10-6 a-1 = 0. And that's as far as I got.

2. Sep 25, 2016

### The Buttered Cat

The way you have it written it looks like you expect the point to be equidistant from both wires. Does that make sense to you, knowing the wires have different currents?

3. Sep 26, 2016

### moenste

Hm, I think I got it:
μ0 IX / 2 π aX = μ0 IY / 2 π aY
μ0 IX 2 π aY = μ0 IY 2 π aX
30 aY = 6 aX
5 aY = aX

We know that aX + aY = 0.03 m

Since we need aX: aY = 0.2 aX → aX + 0.2 aX = 0.03 → 1.2 aX = 0.03 → aX = 0.025 cm or 2.5 m.

4. Sep 26, 2016

### The Buttered Cat

Be careful with your units there, but otherwise you are exactly right.

5. Sep 26, 2016

### moenste

Oh yes, it's the other way around :). 0.025 m and 2.5 cm.