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# Coaxial Cable!

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

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

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

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

$$(d)$$ $$0$$

3 years, 8 months ago

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This question is based on the application of Ampere's circuital law. Give it a try at least. :) · 3 years, 8 months ago

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!!!! · 3 years, 7 months ago

with your help... :) · 3 years, 7 months ago

the answer will be C · 3 years, 8 months ago

8.02x, huh? · 3 years, 7 months ago

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 · 3 years, 8 months ago

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. :) · 3 years, 7 months ago

i rarely give away full solutions. usually i just give hints. i just feel like giving the solution this time. · 3 years, 7 months ago

hints...this doesn't even require any hint!! · 3 years, 7 months ago

Sorry, option b)!!! · 3 years, 8 months ago

C · 3 years, 8 months ago

I think it will be option c)!!! · 3 years, 8 months ago

Comment deleted May 21, 2013