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Parallel conductors

  • Thread starter BunDa4Th
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  • #1
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Homework Statement



The two wires shown in Figure P19.38 carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field at the following locations.

http://www.webassign.net/sf5/p19_38.gif

(a) at a point midway between the wires

(b) at point P1, 10.0 cm to the right of the wire on the right

(c) at point P2, 20.0 cm to the left of the wire on the left

Homework Equations



B = UoI/2pi(d)


The Attempt at a Solution



I tried using this formula and finding each one but it does not work. I am not sure if I am doing this correct but this is what i have tried for b and c

B = (4pi x 10^-7)(5)/(2pi)(.10)

B = .00001 T which is incorrect.

then i tried

B = (4pi x 10^-7)(5)(5)/(2pi)(.10)

B = .00005 T
which is also incorrect.

not sure what i am doing wrong but i decided to input 5 x 10^-6 it said it was correct for part b.

any help would be great.
 

Answers and Replies

  • #2
berkeman
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I'm not sure what you are doing, but are you adding up the contributions from each wire? I don't see the sums.

-a- What is the sum of the two B field contributions midway between the wires? What are the two magnitudes and directions? Remember to use the right-hand rule to determine the directions.

-b,c- Show us the sums.
 
  • #3
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Okay, I figure out how to do this. For A i was suppose to find B1 and B2 then add them and it is going out of the page i believe.

For B it is going into the page and C is out of the page.

I didn't realize that I had to find the two B field. I was only thinking of one bar at a time when doing the problem.
 
  • #4
berkeman
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56,842
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Not exactly, but closer.

You need to consider the magnitude and direction of the two contributions. Use the right hand rule to tell you the direction of each contribution at each point, and use the equation that you showed to calculate the magnitude of each contribution at each point in the question.
 
  • #5
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Well this is what I did but not sure if it is correct.

B1 = UoI1/2pid = 4pi x 10^-7/2pi(.1) = 2 x 10^-5
B2 = 2 x 10^-5

B1 + B2 = Bp = 4 x 10^-5

b) B1 = 1 x 10^-5 and B2 = 1.5 x 10^-6

B1 distance is .10 m and B2 distance is .20 m

then i did B1 - B2 = Bp1 = 5 x 10^-6

c) B1 = UoI/2pi.30m = 3.33 x 10^-6
B2 = 5 x 10^-6 ( i believe this is it)

Bp2 = 1.67 x 10^-6

I am not sure how to use the right hand rule and still learning it. Also i did that with a classmate and left my notebook with her so i am trying to remember what i did by memory.

I think i need an easier explanation on the right hand rule.
 
  • #6
berkeman
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56,842
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Look at the way the right hand in the third figure determines the direction of the B-field circulation around a wire:

http://en.wikipedia.org/wiki/Right_hand_rule

Now use that mental picture to help you with question -a-. You probably have the B1 and B2 magnitudes correct (I didn't check the math), but you need to add B1 and B2 as vectors, where you get their direction correct. So grab the left wire in your figure to determine if the B at the midpoint between the wires comes out of the page or down into it. Then do the same with the other wire. Do you see why it is okay for you to add the magnitudes to get the result? What is the direction of the result?

Now for the other questions, will the two B components add or subtract? Why? For each of your answers, give whether the result points up out of the page or down into it.
 
  • #7
Help

I have this same problem right now and I am having a lot of trouble with it. I don't even know where to start. What equations will I be needing? The forces will repel if the currents are opposite right?
 

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