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Please help with this electromagnetic calculation and design problem!!!

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Rodney

Fri Nov 20 2009, 02:11AM

Registered Member #2474 Joined: Fri Nov 20 2009, 01:22AM
Location:
Posts: 4

I am currrently having problem with this 2 problems please advice thank you.

1:Develop an equation for the resistance of a copper conductor as shown.
Picture in question 1 file attached below.

2:Design a circular shell radius such that the surface current density Js=10r.
Picture in question 2 file attached below.

Please help me on this..... Thank you

rp181

Fri Nov 20 2009, 02:25AM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529

1:
That would be difficult, the best i can think of is thinking of the cone as stacked cylinders.
"resistance (Rt, Omega/cm2) is derived from the following equations (1) Rt = 1/S = rho L/(pi (r/2)2 F) where rho is copper resistivity (1.67 Ã— 10-6 Omega cm), L is wire length (3.6 Ã— 10-3 cm), r is cross-sectional diameter of copper wires â€¦."

wylie

Fri Nov 20 2009, 02:51AM

Registered Member #882 Joined: Sat Jul 07 2007, 04:32AM
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Posts: 103

I can't shake the feeling that that's some sort of homework......

klugesmith

Fri Nov 20 2009, 04:11AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

wylie wrote ...

I can't shake the feeling that that's some sort of homework......

Me too, esp. as first post by a newcomer.
Problem #1 needs a bit of calculus or cleverness, unlike #2.
How 'bout if we all let this rest a week before contributing answers?

[edit] Came back to report an attitude adjustment.
* Would not frown on prompt homework help, if OP says it's homework.
* Giving answers is inappropriate, esp. when OP does not show any effort.
But giving hints (Dr Slack beat me to it) is what teachers, parents, and 4hvers do.

Rich

Dr. Slack

Fri Nov 20 2009, 08:10AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

1) Calculus (hint, the definite integral)
2) arithmetic (hint, what's the circumference of a half sphere)

Any more help than that and you won't meet the learning objective for your homework. Research FTW.

Rodney

Fri Nov 20 2009, 10:23AM

Registered Member #2474 Joined: Fri Nov 20 2009, 01:22AM
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Posts: 4

Its sort of 2 question i found out in a book but without explaination and answer..... thats y i tried to post and ask it..... thanks for all the suggestion. will look into it.

Rodney

Thu Dec 03 2009, 06:44AM

Registered Member #2474 Joined: Fri Nov 20 2009, 01:22AM
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Posts: 4

Hi,

Thanks for all the help I have successgully done question number 1... Can anyone help me check whether i did it correctly or not?

Can anyone help me on question 2 ? I got stuck in that question.....

klugesmith

Thu Dec 03 2009, 08:40AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Rodney wrote ...

Hi,
Thanks for all the help I have successgully done question number 1... Can anyone help me check whether i did it correctly or not?

Hi Rodney.
Looks pretty good, but I got a different answer (unreviewed) by doing just a single integration along the length. Hint: try, early on, replacing r(x) with x * r1/x1. (I am using x instead of lowercase l for typographic clarity).

Then where you get the factor 0.9 after integrating r^-2 from 1 to 10, I integrated x^-2 and got the very similar form
R = (0.9 * x1)/(sigma*pi*r1^2) = (0.1 * L)/(sigma * pi * r1^2 ).
Solving for R = 50 ohms, I got L = 3 * a power of 10 * pi meters.
Want to try matching that? Be careful with centimeter/meter conversion.

If the real conductivity of copper were used, I think the length of 50 ohm tapered bar would be roundly geodetic. Could be wrong!

You did so well, how can problem 2 be a problem after Dr Slack's hint?
[edit]If you can visualize the drawing in 3 dimensions, think about applying the principle of symmetry. Valuable in current spreading problems. Don't be deterred by infinite current density & voltage drop at point sources; they are nicely behaved everywhere else.

Mattski

Thu Dec 03 2009, 08:44AM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

I think for the second problem they are expecting you to assume that the current density will spread out evenly across the sphere as the cross-section expands, in which case it becomes a simple matter of geometry.

But I don't think that that is what would actually happen if you were to actually solve the problem, I suspect that the current will be bunched up more at the center of the half-sphere. I think a full solution would need to solve the differential equation presented by div(J)=0, using image theory where you can replace the (presumably perfect conductor) at the base of the half-sphere with the mirror image, i.e. make it a whole sphere. Then you define dot(normal, J) = 0 on the sphere's surface, as no current flows outward, except at the top and bottom where you have a point source J = 50A. Sounds to me a like a very tricky problem to solve analytically, but probably not too bad numerically.

Rodney

Sun Dec 13 2009, 06:51AM

Registered Member #2474 Joined: Fri Nov 20 2009, 01:22AM
Location:
Posts: 4

wow.... got it... haha it nv occur to me that it is so easy haaa

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