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Emag magnetism question (EE)

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the net mag field at the points asked for. It is just a matter of getting the distances (r values) correct based on the points asked for relative for each wire.

    http://img4.imageshack.us/img4/5906/picf.th.jpg [Broken]

    2. Relevant equations

    I think the relevant equation is:

    B = (u*I)/(2*pi*r) where u is the permeability of free space.

    3. The attempt at a solution

    Assume Wire 1 is to the right, and Wire 2 is the left one.

    For part (a) -- directly between the point of wires, the net magnetic field is Bnet = B1+B2

    B1 = (u*5A)/(2*pi*5cm) + (u*5A)/(2*pi*5cm)

    For part (b) -- 10cm to the right of the right wire, the net magnetic field is Bnet = B1+B2, where B1 and B2 are as follows:

    B1 = (u*5A)/(2*pi*10cm)
    B2 = (u*5A)/(2*pi*20cm)

    For part (c) -- 20cm left of the wire on the left, the net magnetic field is Bnet = B1+B2, where B1 and B2 are as follows:

    B1 = (u*5A)/(2*pi*30cm)
    B2 = (u*5A)/(2*pi*20cm)


    Does my logic/work seem correct? I believe it is.. looking for your input. Thanks!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 12, 2009 #2
    You need to make sure wheter you're calculating with vectors or magnitudes. If you're using magnitudes here, you're wrong in b and c. Is the point in part a between the wires?
  4. Mar 12, 2009 #3
    Hi, I'm sorry I forgot to put the subquestions! :(

    I placed them in. I am looking for the net magnitude only. Do my answers make sense now? :p
  5. Mar 12, 2009 #4
    Your answer for a is correct. But check the signs in b and c by using the right-hand rule. :wink:
  6. Mar 12, 2009 #5
    I thought the signs were independent on each other? Don't you have to calculate the independent magnetic field produced by each wire first at that particular point then add them two?

    If not, then I must've made a mistake in part a, no? That's what I did for part a.
  7. Mar 12, 2009 #6
    If you use the right-hand rule, you see that the magnetic fields are going into different directions, so you would have wrong answer if you just slapped the numbers into your equations.

    Again, right-hand rule says the magnetic fields are going in the same directions so here you add the magnitudes.
  8. Mar 12, 2009 #7
    I see. So, in the case for b and c, I have to denote one convention to say if the current is going down the wire, it will produce a negative magnetic field, B. So, let's say the current down wire 2 (the wire on the left) will produce negative current.

    Hence, for part (b)

    B1 = (u*5A)/(2*pi*10cm)
    B2 = -(u*5A)/(2*pi*20cm)

    For part (c)

    B1 = (u*5A)/(2*pi*30cm)
    B2 = -(u*5A)/(2*pi*20cm)

    Is that better? :)
  9. Mar 12, 2009 #8
    For parts b and c yes, but that convention would contradict with part a where you now are subrtacting the magnetic fields instead of adding them. Only way to solve these kind of problems are using the http://en.wikipedia.org/wiki/Right_hand_rule" [Broken].
    Last edited by a moderator: May 4, 2017
  10. Mar 12, 2009 #9
    I see. Thanks for the help. I think I will stick to that convention I just mentioned.

    Hence, for part (a)

    Bnet = (u*5A)/(2*pi*5cm) - (u*5A)/(2*pi*5cm) = 0

    I think that looks good right?
  11. Mar 12, 2009 #10
    Nope. For part a Bnet = 2(u*5A)/(2*pi*5cm)
  12. Mar 12, 2009 #11
    Hm. I will have to go back and revise the RHR. Hence, the answers IF ASSUMING if the current going down in wire 2 is negative, I will get:

    Bnet = 2(u*5A)/(2*pi*5cm)

    B1 = (u*5A)/(2*pi*10cm)
    B2 = -(u*5A)/(2*pi*20cm)

    B1 = (u*5A)/(2*pi*30cm)
    B2 = -(u*5A)/(2*pi*20cm)

    I hope this make sense.
  13. Mar 12, 2009 #12
    Yep, that looks right. And the right-hand rule really can save your butt in many situations. :biggrin:
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