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Magnetic Fields and Two circular loops of wire

  • Thread starter PhilCam
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  • #1
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Homework Statement



Two circular loops of wire, each containing a single turn, have the same radius of 3.6 cm and a common center. The planes of the loops are perpendicular. Each carries a current of 2.0 A. What is the magnitude of the net magnetic field at the common center?

I have been using this equation: B = uI/2(3.14)r

Where u = 4(3.14) x 10^-7, I= 2.0 Amperes, and r = .036 meters.

When I work through the equation this way, I end up with:

B = 1.111 x 10^-5

It tells me that this answer is wrong and I only have one more attempt at this problem.

thanks.
 

Answers and Replies

  • #2
tiny-tim
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Hi PhilCam! :smile:

(have a pi : π and a mu: µ and try using the X2 tag just above the Reply box :wink:)
Two circular loops of wire, each containing a single turn, have the same radius of 3.6 cm and a common center. The planes of the loops are perpendicular. Each carries a current of 2.0 A. What is the magnitude of the net magnetic field at the common center?

I have been using this equation: B = uI/2(3.14)r

Where u = 4(3.14) x 10^-7, I= 2.0 Amperes, and r = .036 meters.

When I work through the equation this way, I end up with:

B = 1.111 x 10^-5

It tells me that this answer is wrong and I only have one more attempt at this problem.

thanks.
Is that the magnitude for just one loop?

You need to add the vectors for the two fields, and then take the magnitude of the resultant vector.
 
  • #3
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Thanks for the reply!

Would the magnitude for the loops be the same?

I assume so, so would my final answer be:

B Final = Square root of B2 + B2..?

Thanks
 
  • #4
tiny-tim
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Thanks for the reply!

Would the magnitude for the loops be the same?

I assume so, so would my final answer be:

B Final = Square root of B2 + B2..?

Thanks
Looks good! :smile:
 
  • #5
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By doing the math, I finished with an answer of 1.571 x 10-5, but it says this is incorrect.

I thought maybe I had done the calculations incorrectly, but double-checked and got the same thing.

Is there something wrong with my formula?
 
  • #6
tiny-tim
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By doing the math, I finished with an answer of 1.571 x 10-5, but it says this is incorrect.
ah, I think you may have had the wrong number of πs in your original formula …

isn't it µ0 (which contains a 4π), divided by 4π, and then multiplied by the length of the wire, which also contains a 2π?

(see http://en.wikipedia.org/wiki/Biot–Savart_law" [Broken])
 
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  • #7
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Thanks, I'll check that out when I have a few more minutes.

Thanks again.
 
  • #8
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I'm having a difficult time applying the law as it is on the wiki page.

It says that the r should be squared. And times by four pi. That four pi would cancel out with the four pi from µ.

So I would just end up with r squared on the denominator?
 
  • #9
tiny-tim
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I'm having a difficult time applying the law as it is on the wiki page.

It says that the r should be squared. And times by four pi. That four pi would cancel out with the four pi from µ.
Hi PhilCam! :smile:

Yes (I think that's something to do with why µ0 is defined with that extra 4π).
So I would just end up with r squared on the denominator?
Yes, but it's partly balanced by a single r in the ∫dl on top (which comes out as the length of the wire, 2πr). :wink:
 
  • #10
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I'm not sure what my final equation would look like. The original looks like

B= uI/2(pi)r

The pi would cancel out from the constant in the "u." I used this and still had the wrong answer. Any other ideas?
 

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