Two very long straight conducting wires are perpendicular to the x-y plane...
Calculate the magnitude of the current in wire 1 and 2
The equation \[B= \mu I/2\pi r \] should be helpful I think
Not sure what to use for B
First think only about the mag field produced at point p by wire 1. If the current comes out of the paper, which direction would this mag field point?
to the left?
as shown by the line at point p. It makes circles around the wires
yes. Now what field do I need to add to that to get the total field shown in the picture. You just need a qualitative answer, like up and to the righth....
The field from the second wire, which points it downwards and to the left. The program requires a number for the current in each wire
Ok. to get the numbers, use the fact that the field is tangential to the circles and B= mu I/ 2 pi r to calculate the components of the two fields at point p. then add them together.(The y-component of the total field should all come from wire 2.)
Yeah that's the part I'm trying to figure out lol.
Can you get the field from wire 1?
I don't know which B to use for wire 1, since wire 2 affects both the x and y directions. I guess you would have to start with wire 2 because it affects the y direction
the field produced by wire 1 only depends on the current I1 in wire 1, so first we just need to write down an expression for its x and y components.
But we don't know the field produced by only wire 1
You said the field from wire 1 pointed to the left, so B1x = - mu I/(2 pi 2cm) and B1y= 0
So its the vector sum from the two wires?
so I just get each magnitude into its x and y components add them and then do sqrt(Bx2 + By2).
For wire 2, we need to find the distance and angle in the right triangle made by wires 1 and 2 and p.|dw:1343585881940:dw| Use the tringle to find the distance d and angle A. From d, you can find the magnitude of B2 = mu I2 / (2 Pi d), and then using A you can find the components of B2. Add the x components of B2 and B1 to get the x-component of Btot and set it equal to the given Bx
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