×

# 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

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

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

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

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

Note by Advitiya Brijesh
4 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

- 4 years, 7 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!!!!

- 4 years, 6 months ago

with your help... :)

- 4 years, 6 months ago

- 4 years, 7 months ago

the answer will be C

- 4 years, 7 months ago

8.02x, huh?

- 4 years, 6 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

- 4 years, 7 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. :)

- 4 years, 6 months ago

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

- 4 years, 6 months ago

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

- 4 years, 6 months ago

Sorry, option b)!!!

- 4 years, 7 months ago

C

- 4 years, 7 months ago

I think it will be option c)!!!

- 4 years, 7 months ago

Comment deleted May 21, 2013

hmm.. :)

- 4 years, 7 months ago