# Calculate B due to current

1. Jun 22, 2010

### Neophyte

1. The problem statement, all variables and given/known data
Figure below shows two very long straight wires (in cross section) that each carry a current of 4.26 A directly out of the page. Distance d1 = 6.01 m and distance d2 = 4.18 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?

2. Relevant equations
This is a bad picture of it.

0
|d1 is all of the |
|
d2 is the --'s
------------P
|
|d1/2

0

0 is long current out of screen

3. The attempt at a solution

Is failure. I screw up right from the start

μ0i / 2πR

Anyways I always get the direction of b wrong I want both of them of them to point up, I know that they don't but I cant seem to figure out how to do this.

I thought it was suppose to go in a concentric circle around the current and so it would be going counterclockwise and at point p it would go up on both of them eventually, but this method only seems to allow it to go up or down and idk anymore its becoming a problem. I mean I am missing something probably everything.

2. Jun 22, 2010

### quantum13

both wires do contribute to an 'upwards' magnetic field (up the page, not out of the plane of the computer screen)

why do you know that they can't point 'up'?

3. Jun 22, 2010

### collinsmark

Hello Neophyte,
After adding the B fields from each wire together (as vectors), the resulting vector will point up. But when taken individually, the B field from each wire has an upward component, but also a horizontal component too (either left or right, depending on the wire).
Once you find the magnitude of the B vector at point P, for a given wire (just one of them), you need to break that vector into its respective x and y components.
Yes, so far so good...
Yes, but that won't be the case until after you add all the respective components together of both Bs. When working with the B vector associated with just one of the wires, you can expect an upward component; and then a left or right component depending on which wire. So an important step is to figure out how to break up these two B vectors into their respective x and y components such that you can add them together.
Don't get discouraged, you're doing okay. Redraw the diagram on a piece of paper. This step is important because you're going to drawing a few angles and vectors and stuff on it.

Draw the line from one of the wires to point P. The length of this line is R (as in the 'R' in the relevant equation that you gave). Notice that R, d1 and d2 form a right triangle, where R is the hypotenuse. You can use the Pythagorean theorem to calculate the length of R.

This business of using right triangles is just getting started. Geometry/Trigonometry will mostly get your through the rest of this problem. I'll help out a little more and then leave the rest up to you.

Also note that you can calculate the angle θ between R and d2 at point P. The relationship

$$\mathrm{tan} \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}}$$

might help here.

From the right hand rule (or whichever rule you are using for the cross product), you know that the direction of B is perpendicular to the direction of R. So the direction of B is the direction of R, offset by 90o.

Now, once you know the direction of B, you can find its x and y components.

Next move on to the other wire (you can repeat the above again, or use symmetry to figure out the B direction for the other wire).

4. Jun 23, 2010

### Neophyte

I think I love you.:!!) It was the perpendicular to R part I had forgotten, thank you for your help.