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Would the induced EMF & current change?

  1. Dec 3, 2015 #1
    From this diagram:
    etND47v.png
    If a conductor of length(##L##) is moving with a velocity(##v##) inside constant magnetic field(##B##), there is an induced EMF as indicated at the top of copper slab, and connected to a load and current will flow.

    I've been curious with the way the wires are connected to the conductor, what if the bottom wire has been changed from it's position to this:
    buKdlUm.png
    In the calculations for ##\epsilon##, would I just focus on the length ##L_2## or ##L##?
    My initial analysis,is the induced EMF on the conductor regardless of where the connection of the circuit wire is remains unchanged, while as the current... I'm not sure it's the same. What has changed?
     
  2. jcsd
  3. Dec 3, 2015 #2
    I think, for the second part it's as if I'm connected it half way like so:
    YV350cV.png H
    However, that portion at the bottom is still existent, and moving in the magnetic field.
    My initial guess, would be: ## \epsilon = vBL_2##
     
  4. Dec 4, 2015 #3
    Good old @jim hardy can you give me your thoughts on this?
    I know it's similar to previous post we discussed about, however, do we define "##L##" as the distance between the two connection wires or the length that is perpendicular to the magnetic field alone? Or both? I'm curious to know, if I had a large conductive slab passing it through a magnetic field and instead of connecting it by the ends, I'd connected it like the diagrams. I think about the separation of charge like so:


    G5z6g7d.png
    It seems that there still would be negative charges at the bottom, would the geometric change in connection change things?
     
  5. Dec 4, 2015 #4

    NascentOxygen

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    Staff: Mentor

    I expect you'd see just the voltage across that portion you are tapping, even though there will be voltage induced in the full length.

    But you'll need to be careful with those wires to the slab: when they cut across flux lines there will be an induced voltage in the wires. The way you have shown the lower one it will cut flux lines, so arrange these wires horizontally to avoid an induced voltage in the connecting wires.
     
  6. Dec 4, 2015 #5
    Very interesting, by "the portion I'm tapping" that means the induced EMF has changed to a smaller value with respect to the length? Or is it the same voltage o_O? I'm confused with the voltage induced in the full length part.
    About the wires, I've made the bottom one "somewhat" perpendicular to indicate how it's connected, but it will most likely be parallel to the magnetic field.
     
  7. Dec 4, 2015 #6
    The motion induced electric field established inside the conducting bar is uniform, directed from top to bottom according to your diagram. With a uniform electric field, the potential difference is proportional to the length of the bar that you tap.
     
  8. Dec 4, 2015 #7
    Got it, making the length of the wire be ##L_2## for any calculation
     
  9. Dec 4, 2015 #8

    jim hardy

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    Science Advisor
    Gold Member
    2016 Award

    others have answered it above.

    Every individual charge moving in the field experiences force QVcrossB, and they're lined up
    that's why the voltage is the integral along the path
    start with that thought and it becomes intuitive

    Your diagram is made of straight segments.

    Imagine yourself very small and inside the wire where each atom is the size of a basketball, every electron the size of a grain of fine sand.
    You are holding a unit of charge.
    You measure the force exerted on that charge at every point in the wire. , or calculate it using vector multiplication QVcrossB
    You multiply that force by the length of each straight segment .
    You add those force-distance products along the whole wire length of interest.
    If you used newtons, meters, and coulombs your result is volts. (Basics - a volt is a Joule per Coulomb)

    If it's a curved wire you have to figure out its formula and solve the integral.

    Figure things out from the basics...

    i hope i did that right - unsure of thinker lately.

    old jim
     
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