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Magnitism Question-help!

  1. May 21, 2008 #1
    1. The problem statement, all variables and given/known data

    The magnitude of the magnetic force between two conductors 4.00m long, separated by 8.00cm is, 2.85X10^-5 N. The current in one conductor is twice the current in the other. Calculate both currents.

    Given: FM = 2.85X10^-5 N
    l = 4.00m
    Mo = 4pie X10^-7
    d = 0.08m
    I2 = 2(I1)
    I1 = ?

    2. Relevant equations
    F2/l = Mo(I1)(I2)/2pie(d)



    3. The attempt at a solution
    Couldnt really understand how I am supposed to find one of the currents when i dont know either of them, if i knew one i could find it easily.
     
  2. jcsd
  3. May 21, 2008 #2

    rock.freak667

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    Call one current [itex]I_1[/itex] and the other current [itex]I_2[/itex]

    They directly told you how to relate one current to the other.
     
  4. May 21, 2008 #3
    Um yes they told me but im not sure how im supposed to use that in the equation when i still dont know either of the actual currents. How am I supposed to set up the equation knowing one current is twice the other current?
     
  5. May 21, 2008 #4

    rock.freak667

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    If

    [tex]F=\frac{\mu_0 I_1 I_2 l}{2 \pi d}[/tex]

    and that [itex]I_1=2I_2[/itex]

    can you substitute that into the equation and then solve?
     
  6. May 21, 2008 #5
    Well all I can get from that is that it would be Mo2(I2)(I2)(l)/2pie(d) and the I2's would cancel out so ur left with F = Mo(2)(l)/2pie(d) but then how can i solve for I? I put in the values into that equation and did not end up with the correct answer
     
  7. May 21, 2008 #6

    rock.freak667

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    if [itex]I_1=2I_2[/itex] and you put that into the formula, [itex]I_2 \times I_2 =(I_2)^2[/itex]
     
  8. May 21, 2008 #7
    Ok so then F=Mo(2)(I2)^2(l)/2pie(d) plug in 2.85X10^-5 for F, plug in 4pieX10^-7 for Mo, plug in 4 for l and plug in 0.08 for d. I calculated and ended up with 1.19 X10^-7, the answer is supposed to be 1.18 and 2.36
     
  9. May 21, 2008 #8

    rock.freak667

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    Then you plugged in the values wrongly or calculated wrong. I got close to those answers.
    Make [itex](I_2)^2[/itex] the subject and then plug in the values.
     
  10. May 21, 2008 #9
    Ok I tried that so its (I2)^2 on one side so it would be = to F(2pie x d)/Mo(2)(l) which still equalled 1.19X10^-7 so i dont know how or what way u used to get it
     
  11. May 21, 2008 #10

    rock.freak667

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    [tex]\frac{2\pi Fd}{2 \mu l}= \frac{2.85 \times 10^{-5} \times 2\pi \times 0.08}{2\times 4\pi \times 10^{-7} \times 4}[/tex]

    Did you find that?
     
    Last edited: May 21, 2008
  12. May 21, 2008 #11
    That equals 2.85
     
  13. May 22, 2008 #12

    rock.freak667

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    No I get 1.425
     
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