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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

Parallel wires, a distance \(D\) apart, carry a current, \(i\), in opposite directions as shown in the figure. A circular loop, of radius \(R=D / 2\), has the same current flowing in a counterclockwise direction. Determine the magnitude and the direction of the magnetic field from the loop and the parallel wires at the center of the loop as a function of \(i\) and \(R\).

Can an ideal solenoid, one with no magnetic field outside the solenoid, exist? If not, does that render the derivation of the magnetic field inside the solenoid (Section 28.4) void?

The current density in a cylindrical conductor of radius \(R\), varies as \(J(r)=J_{0} r / R\) (in the region from zero to \(R\) ). Express the magnitude of the magnetic field in the regions \(rR .\) Produce a sketch of the radial dependence, \(B(r)\)

A horizontally oriented coil of wire of radius \(5.00 \mathrm{~cm}\) and carrying a current, \(i\), is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle \(\theta\) of \(45.0^{\circ}\) with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is \(B=0.0100 \mathrm{~T}\), the number of turns in the coil is \(N=10.0\), and the total coil mass is \(10.0 \mathrm{~g}\).

Two long, straight parallel wires are separated by a distance of \(20.0 \mathrm{~cm}\). Each wire carries a current of \(10.0 \mathrm{~A}\) in the same direction. What is the magnitude of the resulting magnetic field at a point that is \(12.0 \mathrm{~cm}\) from each wire?
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