Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 32, Problem 32.75AP

Review. The use of superconductors has been proposed for power transmission lines. A single coaxial cable (Fig. P31.47) could carry a power of 1.00 × 103 MW (the output of a large power plant) at 200 kV, DC, over a distance of 1.00 × 103 km without loss. An inner wire of radius a = 2.00 cm, made from the superconductor Nb3Sn, carries the current I in one direction. A surrounding superconducting cylinder of radius b = 5.00 cm would carry the return current I. In such a system, what is the magnetic field (a) at the surface of the inner conductor and (b) at the inner surface of the outer conductor? (c) How much energy would he stored in the magnetic field in the space between the conductors in a 1.00 × 103 km superconducting line? (d) What is the pressure exerted on the outer conductor due to the current in the inner conductor?

Figure P31.47

Chapter 32, Problem 32.75AP, Review. The use of superconductors has been proposed for power transmission lines. A single coaxial

(a)

Expert Solution
Check Mark
To determine
The magnetic field at the surface of the inner conductor.

Answer to Problem 32.75AP

The magnetic field at the surface of the inner conductor is 50.0mT .

Explanation of Solution

Given info: The power carry by the coaxial cable is 1.00×103MW , potential difference is 200kV , total distance is 1.00×103km inner radius is 2.00cm and surrounding cylinder radius is 5.00cm .

Formula to calculate the current flow in the coaxial cable is,

i=PΔV

Here,

i is the current flow in the coaxial cable.

P is the power deliver.

ΔV is the potential difference.

Substitute 1.00×103MW for P and 200kV for ΔV to find i .

i=1.00×103MW×106W1MW200kV×103V1kV=5×103A

Thus, the current flow in the coaxial cable is 5×103A .

Formula to calculate the magnetic field at inner conductor from Ampere’s law is,

Bin=μ0i2πa

Here,

Bin is the magnetic field at inner conductor.

μ0 is the absolute permeability of free space.

a is the radius of inner cable.

Substitute 4π×107Tm/A for μ0 , 5×103A for i , and 2.00cm for a to find the Bin .

Bin=4π×107Tm/A×5×103A2π×2.00cm×102m1cm=0.050T×103mT1T=50.0mT

Conclusion:

Therefore, the magnetic field at the surface of the inner conductor is 50.0mT .

(b)

Expert Solution
Check Mark
To determine
The magnetic field at the inner surface of the outer conductor.

Answer to Problem 32.75AP

The magnetic field at the inner surface of the outer conductor is 20.0mT .

Explanation of Solution

Given info: The power carry by the coaxial cable is 1.00×103MW , potential difference is 200kV , total distance is 1.00×103km inner radius is 2.00cm and surrounding cylinder radius is 5.00cm .

Formula to calculate the magnetic field at inner surface of outer conductor from Ampere’s law is,

Bout=μ0i2πb

Here,

Bout is the magnetic field at inner surface of outer conductor.

b is the radius of outer conductor.

Substitute 4π×107Tm/A for μ0 , 5×103A for i , and 5.00cm for b to find the Bin .

Bin=4π×107Tm/A×5×103A2π×5.00cm×102m1cm=0.020T×103mT1T=20.0mT

Conclusion:

Therefore, the magnetic field at the surface of the inner conductor is 20.0mT .

(c)

Expert Solution
Check Mark
To determine
The energy that would be stored in the magnetic field in the space between the conductors.

Answer to Problem 32.75AP

The energy that stored in the magnetic field in the space between the conductors is 2.29MJ .

Explanation of Solution

Given info: The power carry by the coaxial cable is 1.00×103MW , potential difference is 200kV , total distance is 1.00×103km inner radius is 2.00cm and surrounding cylinder radius is 5.00cm .

Formula to calculate the energy density store in magnetic field is,

uB=B22μ0

Formula to calculate the total energy stored in the magnetic field in the space between the conductors is,

U=abuBdV (1)

Here,

U is the total energy stored in the magnetic field in the space between the conductors.

dV is the small arbitrary volume.

Write the expression for the small arbitrary volume.

dV=2πrldr

Here,

r is the distance from the center of the coaxial cable.

l is the length of the coaxial cable.

Formula to calculate the magnetic field from Ampere’s law is,

B=μ0i2πr

Substitute μ0i2πr for B and 2πrldr for dV in equation (1).

U=ab((μ0i2πr)22μ0)(2πrldr)=ab(μ0i28π2r2)(2πrldr)=μ0i2l4πab(drr)

Integrate the above equation within limits.

μ0i2l4πln[ba]

Substitute 4π×107Tm/A for μ0 , 5×103A for i , 2.00cm for a , 5.00cm for b , and 1.00×103km for l .

U=(4π×107Tm/A)(5×103A)2(1.00×103km×103m1km)4πln[5.00cm2.00cm]=(2.5×106)ln(2.5)J=2.29×106J×106MJ1J=2.29MJ

Conclusion:

Therefore, the energy that stored in the magnetic field in the space between the conductors is 2.29MJ .

(d)

Expert Solution
Check Mark
To determine
The pressure exerted on the conductor due to the current in the inner conductor.

Answer to Problem 32.75AP

The pressure exerted on the conductor due to the current in the inner conductor is 318Pa .

Explanation of Solution

Given info: The power carry by the coaxial cable is 1.00×103MW , potential difference is 200kV , total distance is 1.00×103km inner radius is 2.00cm and surrounding cylinder radius is 5.00cm .

The magnetic field created by the inner conductor exerts a force of repulsion on the current in the outer sheath. The strength of this magnetic field is calculated in part (b) that is 20.0mT .

Write the expression for the projection area of the outer conductor.

Ap=wl

Write the expression for the circumferential area of the outer conductor.

Cl=2πb

Formula to calculate the current flow in the outer cylinder is,

iout=i×ApCl

Here,

Ax is the projection area.

Cl is the circumferential length.

Substitute wl for Ap and 2πb for Ac .

iout=i×wl2πb

Substitute 5×103A for i and 5.00cm for b to find iout .

iout=(5×103A)×wl2π(5.00cm×102m1cm)=(105A/m)wl2π

Formula to calculate the force experience by the outer conductor is,

F=ioutBl

Formula to calculate the pressure exerted on the conductor due to the current is,

P=FA

Substitute ioutBl for F and wl for A to find P .

P=ioutBlwl

Substitute (105A/m)wl2π for iout , 20.0mT for B .

P=((105A/m)wl2π)(20.0mT×103T1mT)wl=318Pa

Conclusion:

Therefore, the pressure exerted on the conductor due to the current in the inner conductor is 318Pa .

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Chapter 32 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 32 - Prob. 32.6OQCh. 32 - Prob. 32.7OQCh. 32 - Prob. 32.1CQCh. 32 - Prob. 32.2CQCh. 32 - A switch controls the current in a circuit that...Ch. 32 - Prob. 32.4CQCh. 32 - Prob. 32.5CQCh. 32 - Prob. 32.6CQCh. 32 - The open switch in Figure CQ32.7 is thrown closed...Ch. 32 - After the switch is dosed in the LC circuit shown...Ch. 32 - Prob. 32.9CQCh. 32 - Discuss the similarities between the energy stored...Ch. 32 - Prob. 32.1PCh. 32 - Prob. 32.2PCh. 32 - Prob. 32.3PCh. 32 - Prob. 32.4PCh. 32 - An emf of 24.0 mV Ls induced in a 500-turn coil...Ch. 32 - Prob. 32.6PCh. 32 - Prob. 32.7PCh. 32 - Prob. 32.8PCh. 32 - Prob. 32.9PCh. 32 - Prob. 32.10PCh. 32 - Prob. 32.11PCh. 32 - A toroid has a major radius R and a minor radius r...Ch. 32 - Prob. 32.13PCh. 32 - Prob. 32.14PCh. 32 - Prob. 32.15PCh. 32 - Prob. 32.16PCh. 32 - Prob. 32.17PCh. 32 - Prob. 32.18PCh. 32 - Prob. 32.19PCh. 32 - When the switch in Figure P32.18 is closed, the...Ch. 32 - Prob. 32.21PCh. 32 - Show that i = Iiet/ is a solution of the...Ch. 32 - Prob. 32.23PCh. 32 - Consider the circuit in Figure P32.18, taking =...Ch. 32 - Prob. 32.25PCh. 32 - The switch in Figure P31.15 is open for t 0 and...Ch. 32 - Prob. 32.27PCh. 32 - Prob. 32.28PCh. 32 - Prob. 32.29PCh. 32 - Two ideal inductors, L1 and L2, have zero internal...Ch. 32 - Prob. 32.31PCh. 32 - Prob. 32.32PCh. 32 - Prob. 32.33PCh. 32 - Prob. 32.34PCh. 32 - Prob. 32.35PCh. 32 - Complete the calculation in Example 31.3 by...Ch. 32 - Prob. 32.37PCh. 32 - A flat coil of wire has an inductance of 40.0 mH...Ch. 32 - Prob. 32.39PCh. 32 - Prob. 32.40PCh. 32 - Prob. 32.41PCh. 32 - Prob. 32.42PCh. 32 - Prob. 32.43PCh. 32 - Prob. 32.44PCh. 32 - Prob. 32.45PCh. 32 - Prob. 32.46PCh. 32 - In the circuit of Figure P31.29, the battery emf...Ch. 32 - A 1.05-H inductor is connected in series with a...Ch. 32 - A 1.00-F capacitor is charged by a 40.0-V power...Ch. 32 - Calculate the inductance of an LC circuit that...Ch. 32 - An LC circuit consists of a 20.0-mH inductor and a...Ch. 32 - Prob. 32.52PCh. 32 - Prob. 32.53PCh. 32 - Prob. 32.54PCh. 32 - An LC circuit like the one in Figure CQ32.8...Ch. 32 - Show that Equation 32.28 in the text Ls Kirchhoffs...Ch. 32 - In Figure 31.15, let R = 7.60 , L = 2.20 mH, and C...Ch. 32 - Consider an LC circuit in which L = 500 mH and C=...Ch. 32 - Electrical oscillations are initiated in a series...Ch. 32 - Review. Consider a capacitor with vacuum between...Ch. 32 - Prob. 32.61APCh. 32 - An inductor having inductance I. and a capacitor...Ch. 32 - A capacitor in a series LC circuit has an initial...Ch. 32 - Prob. 32.64APCh. 32 - When the current in the portion of the circuit...Ch. 32 - At the moment t = 0, a 24.0-V battery is connected...Ch. 32 - Prob. 32.67APCh. 32 - Prob. 32.68APCh. 32 - Prob. 32.69APCh. 32 - At t = 0, the open switch in Figure P31.46 is...Ch. 32 - Prob. 32.71APCh. 32 - Prob. 32.72APCh. 32 - Review. A novel method of storing energy has been...Ch. 32 - Prob. 32.74APCh. 32 - Review. The use of superconductors has been...Ch. 32 - Review. A fundamental property of a type 1...Ch. 32 - Prob. 32.77APCh. 32 - In earlier times when many households received...Ch. 32 - Assume the magnitude of the magnetic field outside...Ch. 32 - Prob. 32.80CPCh. 32 - To prevent damage from arcing in an electric...Ch. 32 - One application of an RL circuit is the generation...Ch. 32 - Prob. 32.83CP
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