# 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

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

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

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

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

5 years, 4 months ago

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

- 5 years, 4 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!!!!

- 5 years, 4 months ago

- 5 years, 4 months ago

- 5 years, 4 months ago

- 5 years, 4 months ago

8.02x, huh?

- 5 years, 4 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

- 5 years, 4 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. :)

- 5 years, 4 months ago

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

- 5 years, 4 months ago

hints...this doesn't even require any hint!!

- 5 years, 4 months ago

Sorry, option b)!!!

- 5 years, 4 months ago

C

- 5 years, 4 months ago

I think it will be option c)!!!

- 5 years, 4 months ago

Comment deleted May 21, 2013

hmm.. :)

- 5 years, 4 months ago