Potential difference in rod rotating in magnetic field

  • Thread starter kbar1
  • Start date
  • #1
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Homework Statement



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]

Homework 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]

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?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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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:
 
  • #3
15
0
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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