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Homework Help: Magnetic Fields of copper wire

  1. Jul 2, 2010 #1
    1. The problem statement, all variables and given/known data

    A 1.0 cm copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate." What are the field's direction and magnitude.

    2. Relevant equations

    B = [(Mu)I]/[2(pi)r]

    3. The attempt at a solution

    I don't really know how to get this one started. Am I supposed to determine the weight of the copper wire by the area and the atomic weight? Can somebody please help me out with this one?
     
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  3. Jul 2, 2010 #2

    kuruman

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    You are supposed to find the mass from the mass density and the volume. Also you need to use the equation for the force on a current-carrying wire in a magnetic field. The equation that you posted gives the magnetic field due to a very long wire. Not the same thing.
     
  4. Jul 2, 2010 #3
    I don't know how to find the mass through the mass density and the volume. I assume the volume would be that of a cylinder?
     
  5. Jul 2, 2010 #4

    kuruman

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    Yes, what expression gives the volume of a cylinder?
     
    Last edited: Jul 2, 2010
  6. Jul 2, 2010 #5
    m = dv
    d = 8.94 g/cm^3
    v = (pi)r^2h
    =(3.14)(0.50 cm)^2(100cm)
    =78.5 cm^3

    Therefore, m = dv
    =(8.94 g/cm^3)(78.5 cm^3)
    =7.01 x 10^2 g
    =0.701 kg?
     
  7. Jul 2, 2010 #6
    F = ma
    =(0.701 kg)(9.81 m/s^2)
    =6.88 N

    B = F/IL
    =(6.88 N)/(50 A)(1.0 m)
    =0.14 T

    Does this answer look correct to you?
     
  8. Jul 2, 2010 #7

    kuruman

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    Your mass calculation is incorrect. You have the wrong radius and the wrong length for the wire. Your calculation for the B field is the correct approach. However, you still need to specify the direction of the field. Hint: Use the right hand rule.
     
  9. Jul 2, 2010 #8
    Oops, I wrote the wrong units for the length of the wire. I apologize, it should read 1.0 m. Let me make another attempt at it.
     
  10. Jul 2, 2010 #9
    v = (pi)r^2h
    =(3.14)(0.050 cm)^2(100 cm)
    =0.785 cm^3

    m = dv
    =(8.94 g/cm^3)(0.785)
    =7.02 g
    =7.02 x 10^-3 kg

    F = ma
    =(7.02 x 10^-3)(9.81 m/s^2)
    =6.88 x 10^-2 N

    Therefore, B would equal 1.4 x 10^-3 T, and the direction of the field would be coming out of the page if it means the force is up in the field of the page.
     
  11. Jul 2, 2010 #10

    kuruman

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    In what direction is the current relative to the page?
     
  12. Jul 2, 2010 #11
    The current is to the east. I wasn't sure if they meant up as in a 3-D view of the ground up or just up in the plane of the page. If they mean that the force is up in the plane of the page, the field would be coming out of the page. If they mean that the force is coming out of the page as if it were a 3-D view of the ground, then the field would be going up in the plane of the page.
     
  13. Jul 2, 2010 #12

    kuruman

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    Suppose you hold the page so that its plane is parallel to the ground. Draw a line right to left and turn the page about a vertical axis so that left-to-right is west to east. Gravity acts perpendicular to the page and down. You want the magnetic force to be up. In what direction should the magnetic field point?
     
  14. Jul 2, 2010 #13

    phyzguy

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    The magnetic force on what? The magnetic force is q VxB, so it depends on the velocity of the charge as well as the magnetic field. In what direction is the charge moving?
     
  15. Jul 2, 2010 #14

    kuruman

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    Please read the statement of the problem carefully to answer this question for yourself.
     
  16. Jul 2, 2010 #15

    phyzguy

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    Sorry, I missed the original statement. Please ignore my post.
     
  17. Jul 2, 2010 #16
    So, the magnetic field should point North.
     
  18. Jul 2, 2010 #17

    kuruman

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    Correct.
     
  19. Jul 2, 2010 #18
    Thanks for your time kuruman!
     
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