Induced Current

  • Thread starter nahya
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  • #1
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This problem is difficult to describe, so I'll post a picture.
http://img71.imageshack.us/my.php?image=pic1ik.gif

The figure above shows a rod of length L caused to move at a constant speed v along horizontal conducting rails. The magnetic field B (the magnitude and direction of which are qualitatively shown by the figure) is not constant, but is supplied by a long wire parallel to the conducting rails. This wire is a distance a from the rail and has a current i.

L=3.13 cm, v=3.11 m/s, a=15.6 mm, and i=11 A.

What is the induced emf (e) in the rod?

---
B = (u_0 I)/(2pi y), and it is not uniform, so I integrated over y=a...L
I got (u_0 I)/(2pi)*ln(L/a).

emf = vBL = v * (u_0 I)/(2pi)*ln(L/a) * L = 3.11 * 1.544597242E-6 * 0.0313 = 1.503557293E-7 V

That is not the right answer, however.
I double-checked that my calculations are correct. So I'm guessing that my steps are incorrect. Can anyone point me to where I'm going wrong?

Thanks.
 

Answers and Replies

  • #2
Hootenanny
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I would try this;

[tex]B = \frac{\mu_{0} I}{2\pi y}[/tex]

a is a constant; [itex]y = a + L[/itex]

[tex]\frac{dB}{dL} = \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL[/tex]

[tex]B = \int^{0.0313}_{0} \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL[/tex]

See if that works.

~H
 
Last edited:
  • #3
Päällikkö
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Shouldn't the integration be done over y=a..a+L?

EDIT: I'm always too slow :smile:.
 
  • #4
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B was approximately 3.326E-4, and emf was 3.23762818E-5 V, which was still incorrect.
Am I using the right L for emf = vBL? I'm using 3.13 cm (0.0313 m).
 
  • #5
Hootenanny
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I think you may have integrated in correctly you should obtain;

[tex]B = \int^{0.0313}_{0} \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL[/tex]

[tex]B = \left[ \frac{1}{2}\mu_{0}I \log (a + L) \right]^{0.0313}_{0}[/tex]

Also ensure that you are converting correctly, note that a is given in mm. You are using the correct equation here;

nahya said:
Am I using the right L for emf = vBL? I'm using 3.13 cm (0.0313 m)
~H
 
Last edited:
  • #6
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Thanks, that worked.
Weird, though, because I had used my calculator to graph the equation and integrate graphically.
 
  • #7
Hootenanny
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nahya said:
Thanks, that worked.
Weird, though, because I had used my calculator to graph the equation and integrate graphically.
No problem. I prefer pen and paper :wink:

~H
 

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