Chapter 31, Problem 47AP

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

To determine

The magnetic field at the surface of the inner conductor.

Answer to Problem 47AP

The magnetic field at the surface of the inner conductor is $50.0\text{mT}$.

Explanation of Solution

Given info: The power carry by the coaxial cable is $1.00\times {10}^3\text{MW}$, potential difference is $200\text{kV}$, total distance is $1.00\times {10}^3\text{km}$ inner radius is $2.00\text{cm}$ and surrounding cylinder radius is $5.00\text{cm}$.

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

$i=\frac{P}{\Delta V}$

Here,

$i$ is the current flow in the coaxial cable.

$P$ is the power deliver.

$\Delta V$ is the potential difference.

Substitute $1.00\times {10}^3\text{MW}$ for $P$ and $200\text{kV}$ for $\Delta V$ to find $i$.

$\begin{array}{c}i=\frac{1.00\times {10}^3\text{MW}\times \frac{{10}^6\text{W}}{1\text{MW}}}{200\text{kV}\times \frac{{10}^3\text{V}}{1\text{kV}}}\\ =5\times {10}^3\text{A}\end{array}$

Thus, the current flow in the coaxial cable is $5\times {10}^3\text{A}$.

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

$B_{\text{in}}=\frac{{\mu }_{0}i}{2\pi a}$

Here,

$B_{\text{in}}$ is the magnetic field at inner conductor.

${\mu }_0$ is the absolute permeability of free space.

$a$ is the radius of inner cable.

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $5\times {10}^3\text{A}$ for $i$, and $2.00\text{cm}$ for $a$ to find the $B_{\text{in}}$.

$\begin{array}{c}{B}_{\text{in}}=\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\times 5\times {10}^3\text{A}}{2\pi \times 2.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\\ =0.050\text{T}\times \frac{{10}^3\text{mT}}{1\text{T}}\\ =50.0\text{mT}\end{array}$

Conclusion:

Therefore, the magnetic field at the surface of the inner conductor is $50.0\text{mT}$.

To determine

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

Answer to Problem 47AP

The magnetic field at the inner surface of the outer conductor is $20.0\text{mT}$.

Explanation of Solution

Given info: The power carry by the coaxial cable is $1.00\times {10}^3\text{MW}$, potential difference is $200\text{kV}$, total distance is $1.00\times {10}^3\text{km}$ inner radius is $2.00\text{cm}$ and surrounding cylinder radius is $5.00\text{cm}$.

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

$B_{\text{out}}=\frac{{\mu }_{0}i}{2\pi b}$

Here,

$B_{\text{out}}$ is the magnetic field at inner surface of outer conductor.

$b$ is the radius of outer conductor.

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $5\times {10}^3\text{A}$ for $i$, and $5.00\text{cm}$ for $b$ to find the $B_{\text{in}}$.

$\begin{array}{c}{B}_{\text{in}}=\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\times 5\times {10}^3\text{A}}{2\pi \times 5.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\\ =0.020\text{T}\times \frac{{10}^3\text{mT}}{1\text{T}}\\ =20.0\text{mT}\end{array}$

Conclusion:

Therefore, the magnetic field at the surface of the inner conductor is $20.0\text{mT}$.

To determine

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

Answer to Problem 47AP

The energy that stored in the magnetic field in the space between the conductors is $2.29\text{MJ}$.

Explanation of Solution

Given info: The power carry by the coaxial cable is $1.00\times {10}^3\text{MW}$, potential difference is $200\text{kV}$, total distance is $1.00\times {10}^3\text{km}$ inner radius is $2.00\text{cm}$ and surrounding cylinder radius is $5.00\text{cm}$.

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

$u_{\text{B}}=\frac{B^2}{2{\mu }_0}$

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

$U={\displaystyle {\int }_a^b{u}_{\text{B}}dV}$ (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.

$dV=2\pi rldr$

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,

$B=\frac{{\mu }_{0}i}{2\pi r}$

Substitute $\frac{{\mu }_{0}i}{2\pi r}$ for $B$ and $2\pi rldr$ for $dV$ in equation (1).

$\begin{array}{c}U={\displaystyle {\int }_a^b\left(\frac{{\left(\frac{{\mu }_{0}i}{2\pi r}\right)}^2}{2{\mu }_0}\right)\left(2\pi rldr\right)}\\ ={\displaystyle {\int }_a^b\left(\frac{{\mu }_0{i}^2}{8{\pi }^2{r}^2}\right)\left(2\pi rldr\right)}\\ =\frac{{\mu }_0{i}^{2}l}{4\pi }{\displaystyle {\int }_a^b\left(\frac{dr}{r}\right)}\end{array}$

Integrate the above equation within limits.

$\frac{{\mu }_0{i}^{2}l}{4\pi }\ln \left[\frac{b}{a}\right]$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $5\times {10}^3\text{A}$ for $i$, $2.00\text{cm}$ for $a$, $5.00\text{cm}$ for $b$, and $1.00\times {10}^3\text{km}$ for $l$.

$\begin{array}{c}U=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right){\left(5\times {10}^3\text{A}\right)}^2\left(1.00\times {10}^3\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}{4\pi }\ln \left[\frac{5.00\text{cm}}{2.00\text{cm}}\right]\\ =\left(2.5\times {10}^6\right)\ln \left(2.5\right)\text{J}\\ =2.29\times {10}^6\text{J}\times \frac{{10}^{-6}\text{MJ}}{1\text{J}}\\ =2.29\text{MJ}\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 47AP

The pressure exerted on the conductor due to the current in the inner conductor is $318\text{Pa}$.

Explanation of Solution

Given info: The power carry by the coaxial cable is $1.00\times {10}^3\text{MW}$, potential difference is $200\text{kV}$, total distance is $1.00\times {10}^3\text{km}$ inner radius is $2.00\text{cm}$ and surrounding cylinder radius is $5.00\text{cm}$.

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 $20.0\text{mT}$.

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

$A_{\text{p}}=wl$

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

$C_{\text{l}}=2\pi b$

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

$i_{\text{out}}=i\times \frac{A_{\text{p}}}{C_{\text{l}}}$

Here,

$A_{\text{x}}$ is the projection area.

$C_{\text{l}}$ is the circumferential length.

Substitute $wl$ for $A_{\text{p}}$ and $2\pi b$ for $A_{\text{c}}$.

$i_{\text{out}}=i\times \frac{wl}{2\pi b}$

Substitute $5\times {10}^3\text{A}$ for $i$ and $5.00\text{cm}$ for $b$ to find $i_{\text{out}}$.

$\begin{array}{c}{i}_{\text{out}}=\left(5\times {10}^3\text{A}\right)\times \frac{wl}{2\pi \left(5.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =\frac{\left({10}^5\text{A}/\text{m}\right)wl}{2\pi }\end{array}$

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

$F=i_{\text{out}}Bl$

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

$P=\frac{F}{A}$

Substitute $i_{\text{out}}Bl$ for $F$ and $wl$ for $A$ to find $P$.

$P=\frac{i_{\text{out}}Bl}{wl}$

Substitute $\frac{\left({10}^5\text{A}/\text{m}\right)wl}{2\pi }$ for $i_{\text{out}}$, $20.0\text{mT}$ for $B$.

$\begin{array}{c}P=\left(\frac{\left({10}^5\text{A}/\text{m}\right)wl}{2\pi }\right)\frac{\left(20.0\text{mT}\times \frac{{10}^{-3}\text{T}}{1\text{mT}}\right)}{wl}\\ =318\text{Pa}\end{array}$

Conclusion:

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