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 × 10 3 MW (the output of a large power plant) at 200 kV, DC, over a distance of 1.00 × 10 3 km without loss. An inner wire of radius a = 2.00 cm, made from the superconductor Nb 3 Sn, 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 × 10 3 km superconducting line? (d) What is the pressure exerted on the outer conductor due to the current in the inner conductor? Figure P31.47

Question

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Answer 1

Textbook Question

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

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

$50.0 mT$ .

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

(b)

Expert Solution

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

$20.0 mT$ .

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

(c)

Expert Solution

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

$2.29 MJ$ .

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

(d)

Expert Solution

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

$318 Pa$ .

Explanation of Solution

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

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 $20.0 mT$ .

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

$iout=(5×103 A)×wl2π(5.00 cm×10−2 m1 cm)=(105 A/m)wl2π$

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

$P=((105 A/m)wl2π)(20.0 mT×10−3 T1 mT)wl=318 Pa$

Conclusion:

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

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Answer 2

Textbook Question

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

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

$50.0 mT$ .

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

(b)

Expert Solution

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

$20.0 mT$ .

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

(c)

Expert Solution

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

$2.29 MJ$ .

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

(d)

Expert Solution

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

$318 Pa$ .

Explanation of Solution

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

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 $20.0 mT$ .

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

$iout=(5×103 A)×wl2π(5.00 cm×10−2 m1 cm)=(105 A/m)wl2π$

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

$P=((105 A/m)wl2π)(20.0 mT×10−3 T1 mT)wl=318 Pa$

Conclusion:

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

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Answer 3

Textbook Question

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

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

$50.0 mT$ .

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

(b)

Expert Solution

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

$20.0 mT$ .

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

(c)

Expert Solution

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

$2.29 MJ$ .

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

(d)

Expert Solution

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

$318 Pa$ .

Explanation of Solution

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

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 $20.0 mT$ .

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

$iout=(5×103 A)×wl2π(5.00 cm×10−2 m1 cm)=(105 A/m)wl2π$

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

$P=((105 A/m)wl2π)(20.0 mT×10−3 T1 mT)wl=318 Pa$

Conclusion:

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

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Answer 4

Textbook Question

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

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

$50.0 mT$ .

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

(b)

Expert Solution

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

$20.0 mT$ .

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

(c)

Expert Solution

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

$2.29 MJ$ .

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

(d)

Expert Solution

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

$318 Pa$ .

Explanation of Solution

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

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 $20.0 mT$ .

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

$iout=(5×103 A)×wl2π(5.00 cm×10−2 m1 cm)=(105 A/m)wl2π$

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

$P=((105 A/m)wl2π)(20.0 mT×10−3 T1 mT)wl=318 Pa$

Conclusion:

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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

$50.0 mT$ .

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

(b)

Expert Solution

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

$20.0 mT$ .

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

(c)

Expert Solution

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

$2.29 MJ$ .

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

(d)

Expert Solution

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

$318 Pa$ .

Explanation of Solution

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

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 $20.0 mT$ .

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

$iout=(5×103 A)×wl2π(5.00 cm×10−2 m1 cm)=(105 A/m)wl2π$

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

$P=((105 A/m)wl2π)(20.0 mT×10−3 T1 mT)wl=318 Pa$

Conclusion:

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

Answer 8

(a)

Expert Solution

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

$50.0 mT$ .

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The magnetic field at the surface of the inner conductor.

Answer 13

Answer to Problem 32.75AP

The magnetic field at the surface of the inner conductor is

$50.0 mT$ .

Answer 14

Explanation of Solution

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

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

$i=1.00×103 MW×106 W1 MW200 kV×103 V1 kV=5×103 A$

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $2.00 cm$ for $a$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×2.00 cm×10−2 m1 cm=0.050 T×103 mT1 T=50.0 mT$

Conclusion:

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

Answer 15

(b)

Expert Solution

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

$20.0 mT$ .

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

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

Answer 20

Answer to Problem 32.75AP

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

$20.0 mT$ .

Answer 21

Explanation of Solution

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

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 $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , and $5.00 cm$ for $b$ to find the $Bin$ .

$Bin=4π×10−7 T⋅m/A×5×103 A2π×5.00 cm×10−2 m1 cm=0.020 T×103 mT1 T=20.0 mT$

Conclusion:

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

Answer 22

(c)

Expert Solution

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

$2.29 MJ$ .

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

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

Answer 27

Answer to Problem 32.75AP

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

$2.29 MJ$ .

Answer 28

Explanation of Solution

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

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).

$U=∫ab((μ0i2πr)22μ0)(2πrldr)=∫ab(μ0i28π2r2)(2πrldr)=μ0i2l4π∫ab(drr)$

Integrate the above equation within limits.

$μ0i2l4πln[ba]$

Substitute $4π×10−7 T⋅m/A$ for $μ0$ , $5×103 A$ for $i$ , $2.00 cm$ for $a$ , $5.00 cm$ for $b$ , and $1.00×103 km$ for $l$ .

$U=(4π×10−7 T⋅m/A)(5×103 A)2(1.00×103 km×103 m1 km)4πln[5.00 cm2.00 cm]=(2.5×106)ln(2.5) J=2.29×106 J×10−6 MJ1 J=2.29 MJ$

Conclusion:

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

Answer 29

(d)

Expert Solution

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

$318 Pa$ .

Explanation of Solution

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

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 $20.0 mT$ .

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

$iout=(5×103 A)×wl2π(5.00 cm×10−2 m1 cm)=(105 A/m)wl2π$

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

$P=((105 A/m)wl2π)(20.0 mT×10−3 T1 mT)wl=318 Pa$

Conclusion:

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

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

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

Answer 34

Answer to Problem 32.75AP

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

$318 Pa$ .

Answer 35

Explanation of Solution

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

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 $20.0 mT$ .

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,

$iout=i×ApCl$