Current induced in rotating conductor in Magnetic Field

  • Thread starter Jimmeh
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  • #1
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Homework Statement


attachment.php?attachmentid=18020&stc=1&d=1237310867.jpg



Homework Equations


[itex]l = 0.02 m[/itex]

[itex]r = 0.3 m[/itex]

[itex]E = \frac{-d\phi}{dt}[/itex]

[itex]\phi = BA[/itex]


The Attempt at a Solution


[itex]A = \pi 0.32^2 - \pi 0.3^2 = 0.0124\pi [/itex] .....Area "cut" by conductor

[itex]\phi = 1.5a_r0.0124\pi = 0.0186\pi a_r [/itex] ....change in flux in one revolution

[itex]1600 rev/min = 1600 rev/ 60 seconds = 1 rev / \frac{60}{1600} seconds[/itex]

[itex] E = \frac{-0.0186\pi a_r}{\frac{60}{1600}}[/itex]

[itex] E = -0.496\pi a_r V[/itex]

I'd just like someone to verify this please, and also I'm not sure about which end would be the anode. I think it'd be the end closest the origin, since the magnetic field is in the outward radial direction, [itex]a_r[/itex], the emf induced is in the [itex]-a_r[/itex] direction, and thus current would be flowing towards the origin, but I'm not sure if I went about handling the direction vectors the right way.
 

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Answers and Replies

  • #2
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Anyone? The exam's tomorrow, I'd just like a quick verification....

Thanks
 
  • #3
Redbelly98
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It sounds like an inadequate description of the motion. 30 cm from the origin in what direction? And, what is the axis of rotation?

My hunch (without knowing the above information) is that the electric and magnetic forces on a test charge Q in the wire would have to be equal in magnitude, opposite in direction.
 

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