## GR0877 Problem 92

**PROBLEM STATEMENT**: *This problem is still being typed*.

**SOLUTION**: (**E**) According to Ampere’s law , where is the total current enclosed by the integration path: , is the current density. For the enclosed current is zero, so . For the enclosed current is and from Ampere’s law , so which shows a linear character with respect to on . For the enclosed current is and or . The only graph that reflects such a behavior of is (E).

*Found a typo? Comment!*

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I see we cannot post equations using latex.

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Trying again.

Thanks for the solutions. I think you have a mistake in problem 92. I found the magnetic field for region to be

I got this by setting up since we’re dealing with uniform current density.

The mistake is in your denominator of , but I think it should be bec. we’re comparing the enclosed current to the entire current over the wire.

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The setting up equation was

Thanks for the reply, jon!

However, I see no mistake here. The current flows between the radii and and the cross-section is

P.S. I edited your replies. To insert equations please write “latex [your code]” between two dollar signs $

If you have any questions, please feel free to ask.

Thank you for replying. So you’re saying that is only a fraction of the current found bet. & ?

Yes, is the current

enclosed by the integration path. To find for one must find (since its uniform). . Note also the limiting cases for this formula: for one has and for the enclosed current equals the total (in this case the integration path encloses the whole cylindrical wire).P.S. For latex support information read http://en.support.wordpress.com/latex/