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Force to move a bar in a magnetic field

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data
    I made this image to illustrate:
    r26d5g.jpg

    A rod with resistance R lies across frictionless conducting rails in a constant uniform magnetic field B. Assume the rails have negligible resistance. The magnitude of the force that must be applied by a person to pull the rod to the right at a constant speed v is:


    2. Relevant equations
    [tex]E = -\frac{d \phi}{dt}[/tex]
    Plus any of maxwells equations et al.

    3. The attempt at a solution

    From what I figure, as the bar moves to the right, the magnetic flux will be increasing, and so, the current in the loop will produce a field opposing the change. This B field would be coming out of the screen/page, so the current must be going counter clockwise.

    Now,
    [tex]
    \Phi = B \dot A\\
    A = Lx\\
    \Phi = BLxcos(\theta)\\
    [/tex]

    But theta (angle between area vector and B is 0:

    [tex]
    \Phi = BLx
    [/tex]

    The velocity of the bar is basically in x per seconds, so I think I can say that basically x is time. That way, I can differentiate the flux as so:

    [tex]
    \Phi = BLx\\
    \frac{d\Phi}{dt} = BL\\
    [/tex]

    Now, E in the loop will equal negative BL since:
    [tex]E = -\frac{d\Phi}{dt}[/tex]

    and so, via Ohms law, the current in the loop is:
    [tex]
    V=IR\\
    I=\frac{V}{R}\\
    I = \frac{-BL}{R}\\
    [/tex]

    Then, by the equation of the Lorentz force:
    [tex]
    F_{B} =I(L×B)\\
    F_{B} = ILBsin(\theta)\\
    [/tex]

    But theta here is 90 degrees, and sin(90) = 1,

    So I think the "opposing" force on the bar due to it moving and increasing the flux through the loop is equal to ILB.

    So at least a force of ILB is required to move the bar to the right, and with a force of ILB, the bar will remain still. So now I am lost.

    What's the next step? I know the solution but I need to figure this out myself.
     
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2
    I forgot this part:

    Since:
    [tex]I = \frac{-BL}{R}\\
    [/tex]

    and

    [tex]F = ILB[/tex]

    Then:
    [tex]F = -\frac{B^{2}L^{2}}{R}[/tex]

    Which is almost the answer I need, it's definitely the force to the left. So to pull it to the right I need to put that force.

    There should be a v in that equation though, and I don't see how.
     
  4. Apr 15, 2012 #3

    WannabeNewton

    User Avatar
    Science Advisor

    Well you have [itex]\Phi = BLx[/itex] so [itex]\frac{\mathrm{d}\Phi }{\mathrm{d} t} = BL\frac{\mathrm{d} x}{\mathrm{d} t} = Blv[/itex]. So there's your v.
     
  5. Apr 15, 2012 #4
    Ah, I see. Idk what I was thinking with this sentence:

    "The velocity of the bar is basically in x per seconds, so I think I can say that basically x is time. That way, I can differentiate the flux as so:"


    Thanks
     
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