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Finding current from current density of wire 
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#1
Nov2410, 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. 


#2
Nov2410, 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.



#3
Nov2410, 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.



#4
Nov2410, 05:40 PM

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Finding current from current density of wire



#5
Nov2410, 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/photo31.jpg 


#6
Nov2410, 06:21 PM

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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, φπ=(φπ), and for the other, φπ=φπ.



#7
Nov2410, 07:15 PM

P: 261



#8
Nov2410, 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.



#9
Nov2410, 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?



#10
Nov2410, 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 π^{2}, but you have π, π^{2}, and π^{3}.
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=φπ to do the angular integrals. 


#11
Nov2510, 12:23 PM

P: 261

I seem to have gotten R^{3} Pi^{2}where R=3 mm. Am I close? And of course, the answer is in Amps. Thanks.



#12
Nov2510, 01:01 PM

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PF Gold
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Yup, that matches what I got.



#13
Nov2610, 09:55 AM

P: 261

Ok thanks.



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