Magnetic field, current with arced path

  • Thread starter iRaid
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  • #1
iRaid
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Homework Statement


attachment.php?attachmentid=63204&d=1382399172.png



Homework Equations





The Attempt at a Solution


[tex]B=\frac{\mu_{0}I}{4\pi}\int \frac{dl\times \hat{r}}{r^{2}}=\frac{\mu_{0}I}{4\pi r^{2}}\int dl\times \hat{r}[/tex]
So I think since when you cross dl with r, you end up with just dl.
[tex]\frac{\mu_{0} I}{4\pi r^{2}}\int dl[/tex]
l=rθ so dl=rdθ, substituting:
[tex]\frac{\mu_{0} I}{4\pi r^{2}}\int rd\theta=\frac{\mu_{0} I}{4\pi r}\int d\theta=\frac{\mu_{0} I}{4\pi r}\theta[/tex]
Plugging in values I end up with: 2.618x10-7 T

Just need someone to look over work/logic, I think it's correct.
 

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Answers and Replies

  • #2
iRaid
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Just realized I could of just done the integral at ∫dl and plugged in rθ after, but whatever. Same result.
 
  • #3
rude man
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Magnitude is correct.

Direction?
 

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