Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 31, Problem 47AP

(a)

To determine

The magnetic field at the surface of the inner conductor.

(a)

Expert Solution
Check Mark

Answer to Problem 47AP

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)

To determine

The magnetic field at the inner surface of the outer conductor.

(b)

Expert Solution
Check Mark

Answer to Problem 47AP

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)

To determine

The energy that would be stored in the magnetic field in the space between the conductors.

(c)

Expert Solution
Check Mark

Answer to Problem 47AP

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)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 47AP

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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