Consider a co-axial cable which consists of an inner wire of radius \(a\) surrounded by an outer shell of inner and outer radii \(b\) and \(c\) respectively. The inner wire carries a current \(I\) and outer shell carries an equal and opposite current. The magnetic field at a distance \(x\) from the axis where \(b<x<c\) is

\((a)\) \(\frac{\mu_{0} I(c^2 - b^2)}{2 \pi x(c^2 - a^2)}\)

\((b)\) \(\frac{\mu_{0} I(c^2 - x^2)}{2 \pi x(c^2 - a^2)}\)

\((c)\) \(\frac{\mu_{0} I(c^2 - x^2)}{2 \pi x(c^2 - b^2)}\)

\((d)\) \(0\)

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## Comments

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TopNewestThis question is based on the application of Ampere's circuital law. Give it a try at least. :)

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thnx buddy bt i had already done it... it was the question that i felt was good to be posted because i got stuck in this particular question!!!!

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with your help... :)

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hey... Is your AGE = 14 ...................?? Please Reply

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the answer will be C

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8.02x, huh?

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magnetic field due to inner wire:

mu I / (2 pi x)current between b and x:

- I (x^2 - b^2) / (c^2 - b^2)magnetic field due to current between b and x:

- [mu I / (2 pi x)] [(x^2 - b^2) / (c^2 - b^2)]magnetic field at x:

[mu I / (2 pi x)] [(c^2 - x^2) / (c^2 - b^2)]which isCLog in to reply

Ah Ramon, I was waiting for Advitiya to post his attempt. You should not have given the solution. It would be a great experience for Advitiya if he solved this by himself. :)

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i rarely give away full solutions. usually i just give hints. i just feel like giving the solution this time.

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Sorry, option b)!!!

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C

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I think it will be option c)!!!

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Comment deleted May 21, 2013

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hmm.. :)

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