How Do Magnetic Fields Interact in a System of Three Parallel Wires?

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Homework Statement

Three long parallel wires are 4.0 cm from one another. (Looking along them, they are at three corners of an equilateral triangle.) The current in each wire is 8.00 A, but its direction in wire M is opposite to that in wires N and P. By convention we can say that M is at the top of our equilateral triangle and N on the lower left hand corner.

Determine the magnitude and direction of the magnetic field at the midpoint of the side of the triangle between wire M and wire N.

I'm totally lost; any input would be much appreciated.

Homework Equations

B = \frac{\mu_0 I}{2 \pi r}

The Attempt at a Solution

B = \frac{\mu_0 8}{2 \pi (.0447213595)} since the field from M and from N cancel each other out we're left with only the field from P which is \sqrt{.04^2 + .02^2} m from our midpoint of MN. This would give us B = \frac{2(10^{-7})(8)}{.0447213595} = 3.57770877*10^{-5} T.

I'm not sure how to calculate the angle although if I do right hand rule for r to B giving me I as out of the page I think that it should be along the plane of the page.

 

[LowlyPion]

since the field from M and from N cancel each other out we're left with only the field from P ...

Welcome to PF.

But not so fast there Tonto. These currents M,N are in opposite directions.

Complicating it a little is that the B from P is a vector pointing in a different direction. (Happily it is at least perpendicular to M,N at that mid-point.)

 

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Welcome to PF.

But not so fast there Tonto. These currents M,N are in opposite directions.

Complicating it a little is that the B from P is a vector pointing in a different direction. (Happily it is at least perpendicular to M,N at that mid-point.)

Since the currents are in opposite directions and the distance from the midpoint of MN is the same doesn't the field from M cancel out the field from N?

I'm not sure quite what you meant in that last bit. Sorry.

 

[LowlyPion]

Since the currents are in opposite directions and the distance from the midpoint of MN is the same doesn't the field from M cancel out the field from N?

I'm not sure quite what you meant in that last bit. Sorry.

The B-field is the right hand rule. You are on opposite sides of two inversely related currents.

 

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The B-field is the right hand rule. You are on opposite sides of two inversely related currents.

Oh duh, go me...

So B = 2*B_M+B_P where B_P is what I came up with in the first part and B_M = \frac{2(10^{-7})(8)}{.02}?

 

[LowlyPion]

Does this help?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html#c1

 

[LowlyPion]

Oh duh, go me...

One way to think about the field that I find useful is to use the right hand, thumb up for positive and left hand with thumb up for negative currents.

Aside from looking like McCain on the campaign trail, you can see your little pinkies pointing together.

 

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One way to think about the field that I find useful is to use the right hand, thumb up for positive and left hand with thumb up for negative currents.

Aside from looking like McCain on the campaign trail, you can see your little pinkies pointing together.

Haha which means instead of canceling out they replicate each other so I have what I said above (B = 2B_M + B_P)?

 

[LowlyPion]

Haha which means instead of canceling out they replicate each other so I have what I said above (B = 2B_M + B_P)?

As vectors of course.

 

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As vectors of course.

Of course. For magnitude was my initial concern.

I guess where I get stuck adding magnetic fields as vectors in finding direction is the curvature thing; straight vectors are easy. I have the field from M and N going towards P and from P going perpendicular to these vectors along the line MN.

I'd just need to find the angle trigonometrically with the right triangle with sides of the magnitude above?

 

[LowlyPion]

I'd just need to find the angle trigonometrically with the right triangle with sides of the magnitude above?

Exactamente. Let Pythagoras be your guide for magnitude and tan^-1 gives the angle.

 

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Exactamente. Let Pythagoras be your guide for magnitude and tan^-1 gives the angle.

Thank you so much!