# Question about magnitude positive y direction

1. Apr 3, 2015

### davidpotter

Hello everybody.

If I choose the positive y direction to be vertically downwards, and the positive x direction to be to the right, and take the cross product y cross x, then the direction of the resultant is out of the page (if I draw x and y as lines on paper). The magnitude is yx sin(φ), where φ is the angle between them. I do understand that, but it's been put in a context where I can't find φ. An infinite wire carrying current I in the positive y direction generates a field at P, which is a distance a along the x axis. I want to cross y, which is infinite, with x, which is finite but changing. x is the position vector of point P, relative to the infinite wire. I don't know what to use as the angle between them. Is φ=90°? Is sinφ x/(x2+y2)1/2, by Pythagoras theorem and the geometry of the situation, or should that be a y on the numerator? My textbook explains this poorly, and I think it's still more maths than physics, but I'm sorry if I posted in the wrong place.
Thank's a lot!

2. Apr 3, 2015

### Staff: Mentor

Both angle and displacement depend on the point on the conductor (and depending on what you want to calculate, you probably want to use an integral), but if you calculate the cross-product, you'll see that those effects cancel.

What do you want to do?

3. Apr 4, 2015

### HallsofIvy

You seem to be using "x" and "y" in two different ways here. "If I choose the positive y direction to be vertically downwards, and the positive x direction to be to the right, and take the cross product y cross x". If you mean "x" to indicate the x-axis and "y" to indicate the y-axis then the angle between x and y is $\pi/2$ radians.

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