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Faraday's law problem

  1. May 22, 2013 #1
    I just did an exercise from the book "a student's guide to Maxwell's Equation". I like the book, because the author has detailed solutions on his website. However, I suspect that his solution is wrong on this one.

    Problem statement:

    "A conducting bar descends with speed v down conducting rails in the presence of a constant, uniform magnetic field pointing into the page, as shown in the figure.
    Write an expression for the emf induced in the loop"

    I've attached both my solution and the solution on the website. I thought that since the area is decreasing, the sign of the flux change should be negative. Since speed must be a positive scalar, I just put a minus sign there.

    Thanks for any help!

    Attached Files:

    • ex3.jpg
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    • ex32.jpg
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  2. jcsd
  3. May 22, 2013 #2


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    I think it may be where you write $$\frac{d}{dt} BLw = BL \frac{dw}{dt} = -BLv$$ ##v## here is the speed at which the loop is being pulled out the field, so dw/dt = v.
  4. May 22, 2013 #3
    Thank you for your reply. But w is getting smaller, so how can dw/dt be a positive number?
  5. May 22, 2013 #4

    rude man

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    The way to figure out the direction of the emf is to imagine the direction of force on free charges in the bar.

    In this example, if we set up a coordinate system xyz with x sideways, y up and z out of the page, with unit vectors i , j and k & I use bold face for all vectors:

    F = q v x B
    with v = -|v| j ( v down)
    B = -|B| k (B into the page)
    so F = |B||v| i (i.e. along the +x axis).

    If the charges are pushed to the + end along the x axis, that end is + and the -x end is -. So the right end of the bar is + and the left is - with |emf| = BLv.

    BTW it's not only easier but sometimes mandatory to use the "BLv" law rather than the -d(phi)/dt law. I know of an example where the
    d(phi)/dt law fails to give the right answer. This is true wherever there are moving media. Maxwell's law of del x E = -∂B/∂t does not always hold in moving media.

    EDIT: in general, with moving media, del x E = -∂B/∂t + del x (v x B).
    Last edited: May 22, 2013
  6. May 22, 2013 #5
    Thank you for your input! I allready figured out the direction of the current, as being clockwise to oppose the change in flux. So it means from left to right in the bar. I still didn't understand if the emf should be + or - though...
  7. May 22, 2013 #6
    I agree that the induced current will flow from left to right.
    The current is produced by an induced EMF in the wire. You have to see the wire as a source of EMF and current flows from - to + through an EMF (think about current through a battery !!!) so the right hand end of the wire is + and the left hand end is -
  8. May 22, 2013 #7
    But will this emf have a positive or a negative sign? Haha, I must be really slow....
  9. May 22, 2013 #8

    rude man

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    If the source is an emf then current travels from - to +! Like in a battery.
  10. May 22, 2013 #9
    The emf is shown with a + sign on the right hand end of the wire and a - sign on the left hand end.
    It is doing the same job as a battery (EMF) connected this way
  11. May 22, 2013 #10
    I thought the emf in an induced electric field measured the work done per unit charge to complete a circle in the wire.
  12. May 22, 2013 #11
    What I meant is, when a battery is connected to a wire, you have a point where charges end up and get transported from - to +. In a circuit like this, the emf would work on the entire circuit so even though charges move from left to right at the bar on top, they move from right to left at the lower wire. I just don't get how to deterimine the sign of the emf on the entire wire.
  13. May 22, 2013 #12

    rude man

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    The emf is the work done ON a unit positive charge in going from - to +. Same as the potential.
  14. May 22, 2013 #13

    rude man

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    Why would they move right to left in the lower wire? Why is the lower wire any different in any respect than the bar?

    Your picture shows both the bar and the lower wire moving thru the B field.
    The upper bar will generate |BLv| volts with polarity from left to right.
    The bottom wire will do exactly the same thing.
    So the two cancel each other and the net emf around the loop is zero. As is the current.
  15. May 22, 2013 #14
    The bottom bar is not moving through the magnetic field. There is no emf generated in the bottom bar. The emf is generated in the moving top bar.
    (the bottom bar is connected to the 2 side bars to make a U shaped frame down which the top bar slides,)
    The current flows clockwise with the emf being generated in the top bar.
    If anything you should say that the voltage across the bottom bar is a potential difference (energy is dissipated)
    Last edited: May 22, 2013
  16. May 22, 2013 #15

    rude man

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    Take another look at the drawing.

    Obviously, if the bottom wire is outside the B field it will not develop any emf. The current would then be clockwise.
    Last edited: May 22, 2013
  17. May 22, 2013 #16
    Take another look at his drawing.
    The 2 sides and the bottom bar form a U shape.
    Obviously the bottom bar is not moving
    The top bar slides down the 2 sides. The velocity arrows are attached to the ends of the top bar
    The whole lot is in a magnetic field but only one conductor is moving.
  18. May 23, 2013 #17
    Technician is right, the U-shaped wire does not move, and is fully emerged in the magnetic field. Just read the problem statement.

    "The bottom bar is not moving through the magnetic field. There is no emf generated in the bottom bar. The emf is generated in the moving top bar."

    The magnetic flux through a surface containing the lower bar is changing, so I thought that meant a circulating electric field would be created throughout the loop. But, seeing it from the perspective of the Lorentz force, what you said makes more sense. Perhaps the circulating electric field according to Faraday's law need not be uniform?

    Still, my original question was not which way the current would flow, as I knew that all along. My question was, using Faraday's law, should one get a positive or negative emf?
  19. May 23, 2013 #18
    If this:
    \varepsilon = -\frac{\mathrm{d}}{\mathrm{d}t} \int_S \mathbf{B} \cdot \mathrm{d}\mathbf{A}
    is what you want to calcuate, then, for instance, if [itex]S[/itex] lies in the plane of your paper, you have to choose if [itex]\mathrm{d}\mathbf{A}[/itex] is directed into, or out from, the paper. One choice gives you a positive number for ε, the other a negative number and they're both correct if you show your choice of orientation of [itex]\mathrm{d}\mathbf{A}[/itex].

    This does not mean, however, that the sign of ε is always arbitrary. If you had a cut in your lower bar and wanted to assign a reference polarity to the open-circuit voltage present there, you'd have make sure that the sign of ε doesn't violate Lenz's law, which in turn constricts your choice of orientation of [itex]\mathrm{d}\mathbf{A}[/itex].
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