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Derivation of Faraday's Law from Lorentz's

  1. Aug 7, 2014 #1
    $$\vec{F}=q\vec{v}\times\vec{B}$$

    $$\frac{d\vec{F}}{dq}=\vec{v}\times\vec{B}$$

    $$\int\frac{d\vec{F}}{dq} \cdot ds=\int(\frac{d\vec{s}}{dt}\times\vec{B}) \cdot ds$$

    from here, I went about it two different ways:

    1.) Here I assumed everything was at right angles and got rid of all the vectors and vector products

    $$\varepsilon=\int \frac{ds}{dt}B ds=\int \frac{ds}{dt}B \frac{ds}{dt}dt$$


    By u substitution

    $$u=\frac{ds}{dt}, du=dt$$
    $$\varepsilon=\int B(u^2)du=\frac{Bv^3}{3}$$

    where v = ds/dt


    That was the first way i went about it, but i didn't feel any closer to Faraday's law.

    2.) Here I left the vectors alone on the RHS; I figured since [itex]\hat{v}[/itex] and d[itex]\hat{s}[/itex] were perpendicular, the quantity ([itex]\vec{v}[/itex]s) would be a time derivative of the area formed

    $$\varepsilon=\int\frac{ds}{dt}B ds=\int(\vec{v}\times\vec{B}) \cdot d\vec{s}=\dot{A}B$$

    $$\varepsilon=\frac{BA}{dt}$$

    don't know where the minus sign is; probably was supposed to do something with the cross product, but didn't know what.





    Well I got alot further with the second "method," but is this a valid derivation? and what went wrong with the first method?
     
    Last edited: Aug 7, 2014
  2. jcsd
  3. Aug 9, 2014 #2
    Oh! PS, well, more like pre-script... my goal is to derive faraday's induction law from lorentz force law
     
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