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) 2πx(c2−a2)μ0I(c2−b2)
(b) 2πx(c2−a2)μ0I(c2−x2)
(c) 2πx(c2−b2)μ0I(c2−x2)
(d) 0
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This 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!!!!
with your help... :)
hey... Is your AGE = 14 ...................?? Please Reply
the answer will be C
I think it will be option c)!!!
Sorry, option b)!!!
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C
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 is C
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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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8.02x, huh?