# Determining direction of magnetic field

1. Aug 1, 2012

### s-f

4 Currents of equal magnitude are perpendicular to the paper and arranged at the corners of a square. Currents traveling into the page are denote by an "(x)" and currents flowing out of the page denoted by "(.)". Determine the direction of the magnetic field at the center of the square.

Ok I have like a dozen of these to do so if someone could explain the general concept behind these I would appreciate it. I tried to use the right hand rule but I was wrong. For example if someone could explain why the square below points left that would help a lot, once I get the idea I can probably do the others. Thanks!

(x) (x)

(.) (.)

2. Aug 1, 2012

### DrewD

You need to use the right hand rule, as you already tried, but then you need to properly add the vectors you get from the RHR.

The top right one will create a field circling clockwise. This means that the vector at the center is up and left (it is pointing along the circle formed by letting the top right current be the center and the radius reaches to the center. If that part confuses you, let me know).

If you do the same thing with each corner, you will find that the top left one points down and left, bottom left is up and left, bottom right is down and left. The ups and downs will cancel due to symmetry since the magnitude and distance from the center are both equal. Only the left components will be left.

Does that help? Look for the symmetry when you do these types of problems.

3. Aug 1, 2012

### s-f

I think I'm messing because I don't know how to cancel out the vectors to find the direction of the field...could you explain that part a little more? I don't know much about vectors. Thanks

4. Aug 1, 2012

### DrewD

All you need to know for a question like this is that the field points in some direction that has some amount of left/right and up/down. If there are equal amounts up and down added in from each source, then the final amount of up/down in the direction of the field will be 0. If there is more up than down, then the final direction will be somewhat up. If the questions are all like this one, you don't need to know how much more up than down.

If you want to learn more about vectors, you can think of the direction of the field, the up/down part, and the left/right part as sides of a right triangle (the intersection of left/right and up/down is the right angle). Then using Trig and the Pythagorean Theorem allows you to put a number to the up/down and left/right and the field direction (depending on which ones you are given you may need different information).

This is often drawn on a graph with up/down being the $y-axis$ and left/right being the $x-axis$. So if you have an arrow pointing from $(x,y)=(0,0)$ to $(x,y)=(3,4)$ we know the $x$ component (side of the triangle) is 3 and the $y$ component is 4. Using Pythagorean Theorem, $$\sqrt{x^2+y^2}=\sqrt{3^2+4^2}=5$$ so the length of that arrow is 5.

I need to go cook dinner. Hopefully the first part helps. Let me know if you already knew this and it was a waste of time, or if you have more questions.

5. Aug 1, 2012

### DrewD

The logical leap is probably just recognizing that you can add the perpendicular directions separately. I hope I made that clear.

6. Aug 2, 2012

### s-f

Thanks a lot Drew :) I understand how to find the direction of the magnetic field at the center of the square, but I'm not sure how to find the net magnetic field at the center of the square. Specifically I don't really get the math as I have yet to take trig/precalc.

I'm using this formula:

μ I = B
2pi r

For the square I posted above, all 4 wires have the same current (I). I think the vectors cancel out vertically, and horizontally they point in the same direction (left) but I'm not sure how to add them? I think I would need to find the radius, calculate the magnitude, and then multiply it by 4 since there are 4 wires. Then I know I have to multiply it by Cos theta but is it 45 or 90 degrees?

Most of the problems are similar to this, so if anyone could explain the trig a bit that would really help and hopefully I can figure out the rest. Thanks!

7. Aug 2, 2012

### s-f

Ok I think I have the right answer but someone please tell me if I'm doing it wrong:

I found the radius to = sqrt 1/2 d. Then I plugged that into the formula above as r to get magnitude, and multiplied the whole thing by 4 and by Cos45.

4 x (μ I/2pi sqrt 1/2d) x Cos 45

Then Cos45 = sqrt 1/2 so that cancels out and the whole thing simplifies to:

2 μ I = B
pi d