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Potential difference in rod rotating in magnetic field

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Q: A rod of length l rotates with a uniform angular velocity ω about the axis passing through its center and perpendicular to its length. A uniform magnetic field B exists with its direction normal to the plane of rotation. The e.m.f. induced between the center and any one end of the rod is:

    (A)[itex]Bl^2ω[/itex] (B)[itex]{\displaystyle\frac{Bl^2ω}{2}}[/itex] (C) [itex]{\displaystyle\frac{Bl^2ω}{4}}[/itex] (D)[itex]{\displaystyle\frac{Bl^2ω}{8}}[/itex]

    2. Relevant equations

    [itex]ΔV = {\displaystyle\frac{\delta\phi}{\delta t}}[/itex], where [itex]\phi[/itex] = magnetic flux = B∙Area

    [itex]ΔV = Blv[/itex], where B is magnetic field, l is length of rod, v is velocity of rod perpendicular to length and magnetic field.

    [itex]ω = {\displaystyle\frac{2\pi}{T}}[/itex]

    [itex]v = rω[/itex]

    3. The attempt at a solution

    The answer provided is (D).

    The given solution goes something like this:

    Area swept in one rotation = [itex]\pi{\displaystyle\frac{l^2}{4}}[/itex]

    [itex]ΔV = B{\displaystyle\frac{ΔA}{ΔT}} = B{\displaystyle\frac{\pi{\displaystyle\frac{l^2}{4}}}{{\displaystyle\frac{2\pi}{ω}}}} = {\displaystyle\frac{Bl^2ω}{8}}[/itex]

    However, the method I tried was:
    [itex]ΔV = Bvl = B(rω)l = B{\displaystyle\frac{lω}{2}}{\displaystyle\frac{l}{2}} = {\displaystyle\frac{Bl^2ω}{4}}[/itex].

    Where did I go wrong?
     
  2. jcsd
  3. Feb 22, 2013 #2

    tiny-tim

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    Homework Helper

    hi kbar1! :smile:

    Bvl only works if every point on the line of length l has speed v

    but this line is stationary at one end, so its average speed is halved :wink:
     
  4. Feb 22, 2013 #3
    Hello! The rod's center is stationary. The two ends are rotating. But I think I got your point. That has to be one of the quickest problem resolutions ever! Thanks a lot!
     
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