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Magnetic Fields Question

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Heres the question: http://imgur.com/aFJFxLa
    2. Relevant equations
    B = μ0/2∏r


    3. The attempt at a solution
    μ0 = 4∏*10^-7
    Magnetic field = μ0/2∏r + μ0/2∏r
    = μ0/2∏(0.05) + μ0/2∏(0.05)
    = 4*10^-5

    The answer is 24*10^-6 T. I need some assistance on what i may be doing wrong.

    My 2nd attempt at question:
    B = μ0/4∏*(5+5)Sin 53.1/(0.08)^2 <== this is still wrong.
     
  2. jcsd
  3. Apr 11, 2013 #2
    Well first of all, you are fergetting something in your formula for the magnetic field...
    check it (hint: it will have to depend also on the current, right??)

    Then consider in which point you are computing the field (in order to be 5cm away from each wire) and draw the forces in that point... the apply correctly trigonometry (the fields will not be simply aligned, you will have to compute a sum of vectors) and here is the result
     
  4. Apr 11, 2013 #3
    Opps I forgot to put the current in formula; but I assure you I used it. And can you explain mor about the point, can I pick a point outside the picture I provided, which is 5 cms away from each wire and forms 2 right angle triangles. Also which method should I be using the first one or the 2nd one (1st or 2nd attempt)
     
  5. Apr 11, 2013 #4
    Actually no one of your attempts is correct (as far as I understand at least, it is difficult to see what is denominator and what is numerator but no problem).

    Anyway OK for the fact that you find two right angle triangles. Then you have to compute the value of the field in that point (simply with your formula), and only then you compute (considering that the field vector is tangential to the circumference passing through your point) the components (you will have to compute the angles). Then the components in one direction will cancel while the components in another direction will add up. Do the sum and you're done. By the way, I got really ##24 \mu T##.
     
  6. Apr 11, 2013 #5
    Isn't what I did with the 1st attempt correct, I just need to add angles and is the hypotense the magnetic field?
     
  7. Apr 11, 2013 #6
    Ok, yes, if you compute correctly the components you are actually adding (using angles) and use the current, then yes, it is your first attempt
     
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