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Current in a straight wire

  1. Nov 5, 2008 #1
    A wire of length 2m and mass 20g is help horizontally. It has a current running through it. You have a second identical 2m long wire that you want to "float" 1cm below this wire.

    a. How much current would you need in each wire (assume you have the same current through both wires) and in what direction?

    b. If you wanted to float the second wire ABOVE the first one, what would you change?

    a. I am thinking I would use the equation B = (m_0*I)/2pi*r but I do not know what B is.
    I can get force from F=ma but I can not move on from that either.
     
  2. jcsd
  3. Nov 5, 2008 #2
    B is a term for magnetic field. Look up "right-hand rule"...
     
  4. Nov 5, 2008 #3

    gabbagabbahey

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    Hint 1: What two forces are acting on the second wire, and in which direction does each force point?

    Hint 2: If the first wire creates a magnetic field with strength B = (m_0*I)/2pi*r how much magnetic force does it exert on the second wire, if theey are both carrying a current I, in the same direction?

    Hint 3: If the second wire is "floating", what is its acceleration? What does that mean the net force on the wire has to be?
     
  5. Nov 5, 2008 #4
    Okay, since the second wire is floating the acceleration is 9.8m/s2 which means the net force would be F=ma=20*9.8=196N

    Would I set B1 = B2?
     
  6. Nov 5, 2008 #5

    gabbagabbahey

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    If its acceleration were 9.8m/s2 then it would be accelerating....doesn't floating mean stationary? :wink:
     
  7. Nov 5, 2008 #6
    Okay, so that would make F=0.... so how do I get current?
     
  8. Nov 5, 2008 #7

    gabbagabbahey

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    Yes, the net force on the second wire is zero...so what forces are acting on the second wire?
     
  9. Nov 5, 2008 #8
    the magnetic force?
     
  10. Nov 5, 2008 #9

    Redbelly98

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    If the net force is zero, there are either no forces or at least two forces.

    If there were only the magnetic force, Fnet wouldn't be zero.

    So, what else besides the magnetic force acts on the wire?
     
  11. Nov 5, 2008 #10
    I have no idea
     
  12. Nov 5, 2008 #11
    gravity? Im just not sure where this is going.
     
  13. Nov 5, 2008 #12

    Redbelly98

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    Yes, gravity.

    Next step is to set up an equation for Fnet, given that the forces are are gravity, and the magnetic force due to the other wire.

    Hint: after setting up the equation, think about how the wire current enters into it.
     
  14. Nov 5, 2008 #13
    Im completely lost, I thought we said F=0
     
  15. Nov 5, 2008 #14

    gabbagabbahey

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    The net force is zero, but that doesn't mean that each of the individual forces acting on the second wire are zero; it just means that they add up to zero.

    Do you follow this?
     
  16. Nov 5, 2008 #15
    I need to see an equation, I dont understand with just talking about it
     
  17. Nov 5, 2008 #16

    gabbagabbahey

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    Well, you know that the individual forces acting on the second wire are gravity and the magnetic force from the first wire.... correct?

    That means that [itex]F_{net}=F_{grav}+F_{mag}=0[/itex]...But what are the equations for [itex]F_{grav}[/itex] and [itex]F_{mag}[/itex]?
     
  18. Nov 5, 2008 #17
    Fnet = ma + ILBsintheta

    but I dont know B
     
  19. Nov 5, 2008 #18

    gabbagabbahey

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    You must mean: Fnet = -mg + ILBsin(theta) where g is the acelleration to to gravity (9.8m/s^2) right? :wink:

    Remember, gravity is pulling downward, so you need the negative sign in front of mg.

    As for B, it is the magnetic field that is created by the first wire, at the location of the second wire... The magnetic field created by the first wire at a distance 'r' from the wire is just B = (m_0*I)/2pi*r isn't it?....but what is 'r' in this case?

    Also, you want the wires to be attracted to each other rather then repulsed, so does that mean the currents should be in the same direction, or in opposite directions?

    And what is the angle 'theta' between the magnetic field of the first wire and the direction of the current in the second wire in this case?
     
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