Ask your own question, for FREE!
Physics 13 Online
OpenStudy (anonymous):

Two very long straight conducting wires are perpendicular to the x-y plane...

OpenStudy (anonymous):

OpenStudy (anonymous):

Calculate the magnitude of the current in wire 1 and 2

OpenStudy (anonymous):

The equation \[B= \mu I/2\pi r \] should be helpful I think

OpenStudy (anonymous):

Not sure what to use for B

OpenStudy (fwizbang):

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?

OpenStudy (anonymous):

to the left?

OpenStudy (anonymous):

as shown by the line at point p. It makes circles around the wires

OpenStudy (fwizbang):

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....

OpenStudy (anonymous):

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

OpenStudy (fwizbang):

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.)

OpenStudy (anonymous):

Yeah that's the part I'm trying to figure out lol.

OpenStudy (fwizbang):

Can you get the field from wire 1?

OpenStudy (anonymous):

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

OpenStudy (fwizbang):

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.

OpenStudy (anonymous):

But we don't know the field produced by only wire 1

OpenStudy (fwizbang):

You said the field from wire 1 pointed to the left, so B1x = - mu I/(2 pi 2cm) and B1y= 0

OpenStudy (anonymous):

So its the vector sum from the two wires?

OpenStudy (anonymous):

so I just get each magnitude into its x and y components add them and then do sqrt(Bx2 + By2).

OpenStudy (fwizbang):

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

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!