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Infinite straight line charg problem

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data


    Supopose you have two infinite straight line charges lambda, a distance apart , moving at a constant speed v. How great would v have to be in order for the magnetic attraction to balance the electrical repulsion ? Work out the actual number... Is this a reasonable sort of speed.
    2. Relevant equations



    3. The attempt at a solution

    F_mag= I (dl x B)=(lambda*v)(lambda*v)*dl/(2*pi*d)
    F_elec=(q_1*q_2/r^2)*(1/4*pi*epilison_0)

    Since the problem says that the magnetic attraction balances out the electrical repulsion, does that mean F_mag = F_elec?
     
  2. jcsd
  3. Nov 21, 2009 #2
    Thought I would get a response by now. I am only asking about the easier part of the problem. Am I not getting an immediate response because I spelled the word charge wrong?
     
  4. Nov 21, 2009 #3
    Seems reasonable, but I have a note and a question:

    The Note: If this is the correct way (which, as I said, seems reasonable), I would suggest writing the charges [itex]q_1,\,q_2[/itex] in terms of [itex]\lambda[/itex].

    The Question: What is moving and how is it moving?
     
  5. Nov 21, 2009 #4
    The two seemingly infinite wires are what is moving. If I right q_1 and q_2 in terms of lambda_1*dl and lambda_2*dl, then the dl and lambda's will cancel each other out right when I set F_comb=F_mag right?
     
  6. Nov 21, 2009 #5
    Correct, the [itex]\lambda[/itex] terms should cancel if you are to express the velocity as an actual number. Now are these wires moving away from each other, towards each other, or what? (I only ask this because it might introduce a negative sign in the formula)
     
  7. Nov 21, 2009 #6
    both wires are parrallel to each other and they are moving in the direction that they are positioned in.
     
  8. Nov 21, 2009 #7

    gabbagabbahey

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    The total electric and magnetic forces on each wire will be infinite (because the wires are infinite and the fields do not taper off along the length of the wires), so in order to compare the two forces, you will need to look at the force per unit length...
     
  9. Nov 21, 2009 #8

    ideasrule

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    It's not easy to use F_elec=kq1q1/r^2 for this type of problem. Instead, try using Gauss' law to figure out the electric field induced by one line of charge in the location of the other, then use F=qE to find force. Your equation for F_mag seems right, except you're missing μ0.
     
  10. Nov 21, 2009 #9
    thats easy. E*da=Q/e_0 , Q=lambda*(2*pi*d)=> E=lambda*(2*pi*d)/(pi*d^2)=F_elec=(lambda*d )*(lambda*(2*pi*d)/(pi*d^2))
     
  11. Nov 22, 2009 #10

    ideasrule

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    What Gaussian surface are you considering? If you're considering a cylinder surrounding one of the line charges and extending to the other, why would Q be lambda*2pi*d? It should be lambda*L. Why would area be pi*d^2? The integral of E(dot)dA around the cylinder should be E*area of cylinder's "sides".
     
  12. Nov 22, 2009 #11
    well the area and length of the ends of a cylinder are respectively pi*d^2 and ,2* pi*d
     
  13. Nov 22, 2009 #12

    gabbagabbahey

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    Sure, but the ends of the cylinder are perpendicular to the field (do you see why?:wink:) and the flux through them is thus zero....What about the curved surface of the cylinder?
     
  14. Nov 22, 2009 #13
    Not , really, but I will give my best explanation . the ends of the cylinder are perpendicular to the field because the end intersects the wire that is traveling at a 90 degree angle with respect to the ends of the cylinder that invisibly encircling the wire. If the flux is zero, does that mean E must be zero and therefore F_elec is zero?
     
  15. Nov 22, 2009 #14

    gabbagabbahey

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    Huh?:confused:

    Let's think through this a little more carefully....appeal to the cylindrical symmetry of the wire....can the field it creates have an axial component?...Can it have an azimuthal component?...What variables can it's radial component depend on?

    But the flux isn't zero...only the flux through the encaps is zero....your Gaussian surface is the entire cylindrical surface (endcaps and curved surface).
     
  16. Nov 22, 2009 #15
    for cylindrical symmetry, I should focus on the r, phi and z components. I don't see why all 3 components should not be take n into serious consideration. then dl is:
    .
    dl=dr r-hat + r*dphi phi-hat + dz z-hat. maybe phi=0 since the wire travel in the straight lines?

    That makes much more sense . E(2*pi*z*r)=flux of the surface of the cylinder.
     
  17. Nov 22, 2009 #16

    gabbagabbahey

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    All three components should be taken into consideration....but what is the point of using [itex]d\textbf{l}[/itex] here?


    Instead, realize that if [itex]\textbf{E}[/itex] has a non-zero [itex]\phi[/itex]-component (azimuthal component), then a test charge placed near the wire will have an azimuthal force on it which will tend to push it towards some other value of [itex]\phi[/itex]....but why would there be such a force?...what would makes [itex]\phi=0[/itex] any more/less preferable a location than say [itex]\phi=\pi/4[/itex] for example?...Doesn't the test charge "see" exactly the same charge distribution at all values of [itex]\phi[/itex]?
     
  18. Nov 22, 2009 #17
    The point of used [itex]d\textbf{l}[/itex] because in order to find Q, I need to set Q= lambda* [itex]d\textbf{l}[/itex] for a line charge right?



    yes , since the figure itself is symmetrical right? For the electrical force, I don't see why the d in F= E*d will have an azimuthal component and not just a radial component.
     
    Last edited: Nov 22, 2009
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