1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    kjYe0tL.png

    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?

    --------------------


    PART II


    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

    $$W=qBy\sqrt{2v^2}=\sqrt{2}qvBy$$

    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

    Matterwave

    User Avatar
    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!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electrodynamics (Lorentz to Faraday)
  1. Electrodynamic doubt (Replies: 10)

Loading...