brThe diagram below show the inner section of a circular conductor of inner part and outside radius bthis transfer current.magnetic field at radial=?a) r< ab) ab

According to Ampere's Law, ∮B→.dl→ = μ0Ienclosed note- considering the conductor carries a current…

Answered: b r The diagram below show the inner…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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b
r
The diagram below show the inner section of a circular conductor of inner part and outside radius b
this transfer current.
magnetic field at radial=?
a) r< a
b) a<r<b
c) r>b

Expert Solution

Step 1

According to Ampere's Law,

$\oint \vec{B} . \vec{d l} = μ_{0} I_{e n c l o s e d}$

note-

considering the conductor carries a current I

a) for r<a

The total charged enclosed in the region r<a is zero,

$I_{e n c l o s e d} = 0$

using the ampere's law,

$\oint \vec{B} . \vec{d l} = μ_{0} I_{e n c l o s e d} \oint \vec{B} . \vec{d l} = 0 \vec{B} = 0$

so the magnetic field inside the region r<a is 0.

Step 2

b) for a<r<b

The current enclosed in a cylindrical shell of inner radius 'a' and outer radius 'r' is

$I_{e n c l o s e d} = \frac{I}{π (b^{2} - a^{2})} \times π (r^{2} - a^{2}) I_{enclosed} = \frac{I}{(b^{2} - a^{2})} (r^{2} - a^{2})$

From ampere's law,

$\oint \vec{B} . \vec{d l} = μ_{0} I_{e n c l o s e d} \vec{B} . \oint \vec{d l} = μ_{0} \frac{I (r^{2} - a^{2})}{b^{2} - a^{2}} \vec{B} (2 πr) = μ_{0} \frac{I (r^{2} - a^{2})}{b^{2} - a^{2}} \vec{B} = \frac{μ_{0}}{2 πr} \frac{I (r^{2} - a^{2})}{b^{2} - a^{2}}$

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