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CP CALC A Coaxial Cable. A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d \Phi_{B}\) through a narrow strip of length \(I\) parallel to the axis, of width \(d r\), at a distance \(r\) from the axis of the cable and lying in a planc containing the axis. (c) Intcgrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(/\) of the cable is $$ I_{.}=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right) $$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length \(l\) of the cable.

Short Answer

Expert verified
The magnetic field at any point between the conductors is \( B = \frac{\mu_{0} i}{2 \pi r} \). The total magnetic flux produced by a current \(i\) in the central conductor is \( \Phi_{B} =\frac{\mu_{0} i l}{ 2 \pi } \ln \left(\frac{b}{a}\right) \). The inductance of a length \(l\) of the cable is \( I_{.}=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)\) . Lastly, the energy stored in the magnetic field for a length \(l\) of the cable is \( E_{m}= \frac{1}{2} \left[l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)\right]i^{2} \).

Step by step solution

01

Find the Magnetic Field

We will start by using Ampere's law to find the magnetic field between the conductors. As both conductors are carrying the same current \(i\), but in the opposite directions, it produces a net current inside the volume of the tube between radius \(a\) and \(b\). According to Ampere's law, the magnetic field can be calculated using \( B = \frac{\mu_{0} i}{2 \pi r} \), where \( r \) is the radius from the point to the axis within the volume.
02

Write the Expression for the Flux

Now, we can write an expression for the magnetic flux. By using a narrow strip of length \( l \) and width \( dr \), we find that \( d \Phi_{B} = B \cdot da \), where \( da \) is the area of the strip \( ldr \). Replacing \( B \) with our equation from step 1 we get \( d \Phi_{B} = \frac{\mu_{0} i l dr} {2 \pi r} \).
03

Find Total Flux

To find the total magnetic flux between the two conductors, integrate the expression obtained in step 2 from \(a\) to \(b\). The result is \( \Phi_{B} = \int_{a}^{b} \frac{\mu_{0} i l dr} {2 \pi r} \) . This will yield \( Ph_{B} =\frac{\mu_{0} i l}{ 2 \pi } \ln \left(\frac{b}{a}\right) \)
04

Prevent Self-Inductance

The inductance \(I_{.}\) of a length \(l\) of the cable can be calculated using the formula \( I_{.}= \frac{\Phi_{B}}{i} \). Using the expression from Step 3 for \(\Phi_{B}\), we can write \(I_{.}=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)\)
05

Compute Energy Stored in Magnetic Field

Lastly, the energy stored in the magnetic field for a length \(l\) of the cable is given by \(E_{m}=\frac{1}{2}I_{.}i^{2}\). Substituting for the inductance obtained in Step 4, we find that \(E_{m}= \frac{1}{2} \left[l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)\right]i^{2}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Calculation
Understanding how to calculate the magnetic field within a coaxial cable is essential when tackling problems in electromagnetism. Ampere's law, a fundamental principle in electromagnetism, relates the integrated magnetic field around a closed loop to the electric current passing through the loop. For a coaxial cable with currents flowing in opposite directions in the inner and outer conductors, the situation simplifies significantly.

To calculate the magnetic field at any point between the two conductors, we apply Ampere's law in the form \( B = \frac{\mu_{0} i}{2 \pi r} \) where \( \mu_{0} \) is the permeability of free space, \( i \) is the current, and \( r \) is the radial distance from the cable's axis. This formula holds true in the region between the conductors and gives us the magnitude of the magnetic field generated by the current within that region.

This concept is paramount for understanding various phenomena such as the transmission of electrical signals through cables and the design of electrical circuits.
Magnetic Flux Expression
Magnetic flux provides us with a measure of the total magnetic field passing through a given area. When dealing with coaxial cables, it is particularly useful to determine the link between the magnetic field and the geometry of the cable.

The infinitesimal magnetic flux \(d \Phi_{B}\) through a narrow strip inside a coaxial cable can be written as \( d \Phi_{B} = B \cdot da \), where \( da \) is the area of the strip, which can be expressed as \( l dr \) with \( l \) being the length of the cable and \( dr \) a small change in radius. Utilizing the previous calculation for \( B \), the expression becomes \( d \Phi_{B} = \frac{\mu_{0} i l dr} {2 \pi r} \). This integral will reveal the total magnetic flux between the inner and outer conductors which is important for further calculations such as finding the self-inductance of the coaxial cable.
Self-Inductance
Self-inductance is a property of a conductor that quantifies its ability to oppose changes in the electric current passing through it, a concept known as Lenz's law. When current flows through a coaxial cable, a magnetic field is created around it, and a change in this current will induce a changing magnetic field which, in turn, induces an electromotive force (EMF).

The self-inductance \( L \) of a coaxial cable can be directly calculated by the magnetic flux through the cable for a given current. According to the exercise, the inductance per unit length \( l \) can be expressed as \( L_\cdot = l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right) \). The ratio \( \frac{b}{a} \) compares the radii of the coaxial cable’s outer and inner conductors, and \( \mu_{0} \) is again the permeability of free space.

Knowing the self-inductance is critical in electronics to predict how a circuit will respond to changing currents, which affects the performance of devices such as transformers and inductors.
Energy Stored in Magnetic Field
The energy stored in the magnetic field is a result of electrical energy being converted into magnetic energy when a current flows through a conductor. For a coaxial cable, this energy is directly proportional to the self-inductance of the cable as well as the square of the current passing through it.

The expression for the energy \( E_{m} \) stored in the magnetic field for a length \( l \) of the coaxial cable is \( E_{m}=\frac{1}{2}L_\cdot i^{2} \), where \( i \) is the current and \( L_\cdot \) is the inductance per unit length. Substituting the formula for \( L_\cdot \) yields the equation \( E_{m}= \frac{1}{2} \left[l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)\right]i^{2} \). This energy represents the potential to do work and it is what enables inductors in electric circuits to release energy when needed, for instance, to smooth out electrical currents or to provide a surge of power.

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Most popular questions from this chapter

An air-filled toroidal solenoid has a mean radius of \(15.0 \mathrm{~cm}\) and a cross-sectional area of \(5.00 \mathrm{~cm}^{2}\). When the current is \(12.0 \mathrm{~A}\), the encrgy stored is \(0.390 \mathrm{~J}\). How many turns does the winding have?

A long. straight solenoid has 800 turns. When the current in the solenoid is \(2.90 \mathrm{~A}\), the average flux through each turn of the solenoid is \(3.25 \times 10^{-3} \mathrm{~Wb}\). What must be the magnitude of the rate of change of the current in order for the sclf-induced cmf to cqual \(6.20 \mathrm{mV} ?\)

When the currcnt in a toroidal solcnoid is changing at a rate of \(0.0260 \mathrm{~A} / \mathrm{s},\) the magnitude of the induced \(\mathrm{cmf}\) is \(12.6 \mathrm{mV}\). When the current equals \(1.40 \mathrm{~A},\) the average flux through each turn of the solenoid is 0.00285 Wb. Ilow many turns does the solenoid have?

Solar Magnetic Energy. Magnetic fields within a sunspot can be as strong as 0.4 T. (By comparison, the earth's magnetic field is about \(1 / 10,000\) as strong. ) Sunspots can be as large as \(25,000 \mathrm{~km}\) in radius. The material in a sunspot has a density of about \(3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(\mu\) for the sunspot material is \(\mu_{0}\). If \(100 \%\) of the magneticfield energy stored in a sunspot could be used to eject the sunspot's material away from the sun's surface, at what speed would that material be cjected? Compare to the sun's escape speed, which is about \(6 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) (Hint: Calculate the kinetic energy the magnetic field could supply to \(1 \mathrm{~m}^{3}\) of sunspot material.)

\(A 15.0 \mu F\) capacitor is charged by a 150.0 V power supply, then disconnected from the power and connected in scries with a \(0.280 \mathrm{mH}\) inductor. Calculate: (a) the oscillation frequency of the circuit: (b) the energy stored in the capacitor at time \(t=0 \mathrm{~ms}\) (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t=1.30 \mathrm{~ms}\)

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