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Electrodynamics (Lorentz to Faraday)

  1. Jul 26, 2014 #1
    Electrodynamics (Lorentz Force & Voltage) [edited]

    Two part inquiry

    PART I

    Situation: I have a B-field and i'm pulling a straight piece of wire through that B-field


    You pull the wire perpendicularly through the B-field at velocity v; charges between points a and b will experience the Lorentz Force and will begin to move along the wire with velocity u, and thus will experience a force opposite to the direction which you are pulling the wire.

    Goal: Trying to derive the work done on a unit charge the classical way (by integrating along the path it takes)

    Let us define the force per unit q to pull the wire through the field: f(pull)[itex]\equiv[/itex]uB

    $$\int f_{pull} \cdot dl = (uB) \frac{y}{cosθ}sinθ= vBhy$$

    Problem: The vector f(pull) should be pointing outward (opposite the direction of wire pulling). so then shouldn't there be a negative sign somewhere?



    I decided to try and derive it my own way and i ran into a problem somewhere but i can't seem to spot where it is.

    I said that, there is a vertical work component and a horizontal work component.

    vertical work component

    (From the initial pull)

    $$W_y= \int (q\vec{v} \times \vec{B}) \cdot dy = (qvB)y= qvBy$$

    horizontal work component

    $$W_x= \int (q\vec{u} \times \vec{B}) \cdot dx = quBx =quBytanθ$$

    vectorial sum

    (yes i know work is a scalar quantity. but there is a work in x direction and work in the y direction so i thought i'd try this approach)

    $$W= \sqrt{(qvBy)^2+(quBytanθ)2}= qBy\sqrt{v^2+u^2tan^2θ}$$

    but since utanθ=v


    i'm off by a factor of [itex]\sqrt{2}[/itex]

    Could someone please help me find the mistake in doing it this way?

    help is much appreciated. ty all!
    Last edited: Jul 26, 2014
  2. jcsd
  3. Jul 26, 2014 #2


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    Science Advisor
    Gold Member

    Notice that on the left hand side is ##u\tan(\theta)## and on the right hand side is ##v##. So it must be that ##u\tan(\theta)=v##. Look at figure 7.11 to see why this might be
  4. Jul 26, 2014 #3

    yeah i actually caught that right after i posted it. I'm actually now stuck with a different set of questions.

    but thanks!!
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