Finding current from current density of wire


by maherelharake
Tags: current, density, wire
maherelharake
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#1
Nov24-10, 01:42 PM
P: 261
1. The problem statement, all variables and given/known data

A cylindrical wire of radius 3mm has current density, J=3s |φ-[tex]\pi[/tex]| z_hat. Find the total current in the wire.


2. Relevant equations



3. The attempt at a solution
I believe all I have to do is integrate over the area, but for some reason I can't get it to work. Is the differential area going to be da=s ds dφ? In that case s goes from 0 to 3mm and φ goes from 0 to 2Pi? The 'z' direction is throwing me off a bit. Thanks.
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vela
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#2
Nov24-10, 05:06 PM
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PF Gold
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Yes, that's right. Post your work if you still can't get it to work out so we can see where you're going wrong.
maherelharake
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#3
Nov24-10, 05:27 PM
P: 261
I just tried to work it out on a napkin, because I am not near a scanner at the moment. I ended up with a net result of 0 though. If you don't think this is correct, I can try to rewrite it and take a picture with my phone and upload it. Thanks again.

berkeman
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#4
Nov24-10, 05:40 PM
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Finding current from current density of wire


Quote Quote by maherelharake View Post
I just tried to work it out on a napkin, because I am not near a scanner at the moment. I ended up with a net result of 0 though. If you don't think this is correct, I can try to rewrite it and take a picture with my phone and upload it. Thanks again.
Yes, please upload it. Or you could use the Latex editor in the Advanced Reply window to write out your equations. Click on the [tex]\Sigma[/tex] symbol to the right in the toolbar to see your Latex options.
maherelharake
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#5
Nov24-10, 05:48 PM
P: 261
Alright here you go. If you can't read it, let me know. Thanks.
http://i77.photobucket.com/albums/j7...e/photo-31.jpg
vela
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#6
Nov24-10, 06:21 PM
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PF Gold
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You're not dealing with the absolute value correctly. Break the integral over the angle into two ranges, one from 0 to π and the other from π to 2π. For the first integral, |φ-π|=-(φ-π), and for the other, |φ-π|=φ-π.
maherelharake
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#7
Nov24-10, 07:15 PM
P: 261
Hmm ok. How's this look.

http://i51.tinypic.com/kf4lrd.jpg
vela
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#8
Nov24-10, 09:12 PM
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PF Gold
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Your integrals look fine, but you made a mistake somewhere evaluating them.
maherelharake
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#9
Nov24-10, 09:17 PM
P: 261
Hmm I can't find it. I checked it a few times after I posted it. Did you work it out and get a different result?
vela
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#10
Nov24-10, 09:33 PM
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PF Gold
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I entered it into Mathematica and got a different result. It looks like you messed up the angular integrations in several spot. Every integral should be proportional to π2, but you have π, π2, and π3.

You can simplify the algebra a bit by separating the s integral and φ integral:

[tex]I=\int_0^{R} 3s^2ds \int_0^{2\pi} |\varphi-\pi|\,d\varphi[/tex]

and using the substitution u=φ-π to do the angular integrals.
maherelharake
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#11
Nov25-10, 12:23 PM
P: 261
I seem to have gotten R3 Pi2where R=3 mm. Am I close? And of course, the answer is in Amps. Thanks.
vela
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#12
Nov25-10, 01:01 PM
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PF Gold
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Yup, that matches what I got.
maherelharake
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#13
Nov26-10, 09:55 AM
P: 261
Ok thanks.


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