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Chapter 28: Problem 41

A current of constant density, \(J_{0}\), flows through a very long cylindrical conducting shell with inner radius \(a\) and outer radius \(b\). What is the magnetic field in the regions \(rb\) ? Does \(B_{ab}\) for \(r=b\) ?

Short Answer

Expert verified
Question: Calculate the magnetic field in the three regions inside (rb) the conducting shell. According to the derived formula, the magnetic field for the three regions are as follows: 1. Inside the shell (rb): \(B_{r>b}=\cfrac{\mu_0J_0 (b^2 - a^2)}{2r}\)

Step by step solution

01

Define Ampere's Law for a Cylindrical Shell

Ampere's Law states that the integral of the magnetic field (\(\vec{B}\)) along a closed path is equal to the encircled current multiplied by the permeability of free space (\(\mu_0\)): $$ \oint_C \vec{B}\cdot\vec{dl}=\mu_0I_{enclosed} $$ For a cylindrical shell with symmetry, we can choose a closed path in the form of a circle centered on the axis of the cylinder.
02

Calculate the Magnetic Field for \(r

For the region inside the conducting shell (r<a), there's no current enclosed within the chosen circle. Therefore, the integral of the magnetic field along a closed path will be zero: $$ \oint_C \vec{B}\cdot\vec{dl}=\mu_0 \times 0 \Rightarrow B_{r<a}=0 $$
03

Calculate the Magnetic Field for \(a

Let's consider a circular path with radius \(r\) such that \(a<r<b\). The enclosed current can be found using the current density, \(J_0\), and the area of the enclosed circle with radius \(r<a\): $$ I_{enclosed}=J_0 \times \pi a^2 $$ Now, we can use Ampere's Law to find the magnetic field: $$ B_{a<r<b}(2\pi r)=\mu_0J_0\pi a^2 \Rightarrow B_{a<r<b}=\cfrac{\mu_0J_0 a^2}{2r} $$
04

Calculate the Magnetic Field for \(r>b\)

For the region outside the conducting shell (r>b), we consider a circular path with radius \(r\). In this case, the enclosed current is the total current passing through the cylindrical shell: $$ I_{enclosed}=J_0 \times \pi (b^2 - a^2) $$ Using Ampere's Law, we can find the magnetic field: $$ B_{r>b}(2\pi r)=\mu_0J_0\pi (b^2 - a^2) \Rightarrow B_{r>b}=\cfrac{\mu_0J_0 (b^2 - a^2)}{2r} $$
05

Check the Magnetic Field Equality for \(r=b\)

At \(r=b\), let's evaluate if \(B_{ab}\): $$ \cfrac{\mu_0J_0 a^2}{2b}=\cfrac{\mu_0J_0 (b^2 - a^2)}{2b} $$ Divide both sides by \(\cfrac{\mu_0 J_0}{2}\): $$ \cfrac{a^2}{b}=\cfrac{(b^2 - a^2)}{b} $$ This equation holds true, so the magnetic field is indeed equal within and outside the shell at \(r=b\). In conclusion, the magnetic field in the three regions are: 1. \(B_{rb}=\cfrac{\mu_0J_0 (b^2 - a^2)}{2r}\)

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Most popular questions from this chapter

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