Chapter 23, Problem 73P

Review. The use of superconductors has been proposed for power transmission lines. A single coaxial cable (Fig. P23.73) 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 be 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. P23.73

To determine

The magnetic field at the inner surface of the conductor

Answer to Problem 73P

The magnetic field at the inner surface of the conductor has a value of $50\text{mT}$.

Explanation of Solution

Write the equation for the ampere’s law.

    $\begin{array}{c}B(2\pi r)={\mu }_0{I}_{enclosed}\\ B=\frac{{\mu }_0{I}_{enclosed}}{2\pi r}\end{array}$        (I)

Here, $B$ is the magnetic field at the inner surface, ${\mu }_0$ is the permeability in free space, $I_{enclosed}$ is the current enclosed by the inner surface and $r$ is the radius of the inner wire.

Write the equation for the power on a single cable.

    $\begin{array}{c}P=I\Delta V\\ I=\frac{P}{\Delta V}\end{array}$        (II)

Here, $I$ is the current enclosed and $\Delta V$ is the voltage on the single cable. Substitute $1.00\times {10}^9\text{W}$ for $P$ and $200\times {10}^3\text{V}$ for $\Delta V$

    $\begin{array}{c}I=\frac{1.00\times {10}^9\text{W}}{200\times {10}^3\text{V}}\\ =5.00\times {10}^3\text{A}\end{array}$        (III)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_0$, $5.00\times {10}^{-3}\text{A}$ for $I_{enclosed}$ and $0.020\text{m}$ for $r$ in equation (I).

    $\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m/A}\right)\left(5.00\times {10}^3\text{A}\right)}{2\pi \left(0.020\text{m}\right)}\\ =\frac{20\times {10}^{-4}}{4\times {10}^{-2}}\text{T}\\ =0.050\text{T}\\ =50.0\text{mT}\end{array}$

Therefore, the magnetic field at the inner surface of the conductor has a value of $50\text{mT}$.

To determine

The magnetic field at the inner surface of the outer conductor

Answer to Problem 73P

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

Explanation of Solution

Write the equation for the ampere’s law from equation (I).

    $\begin{array}{c}B(2\pi r)={\mu }_0{I}_{enclosed}\\ B=\frac{{\mu }_0{I}_{enclosed}}{2\pi r}\end{array}$

Here, $B$ is the magnetic field at the inner surface, ${\mu }_0$ is the permeability in free space, $I_{enclosed}$ is the current enclosed by the inner surface and $r$ is the radius of the inner wire.

Write the equation for the power on a single cable. From equation (II).

    $\begin{array}{c}P=I\Delta V\\ I=\frac{P}{\Delta V}\end{array}$

Here, $I$ is the current enclosed and $\Delta V$ is the voltage on the single cable. Substituting $1.00\times {10}^9\text{W}$ for $P$ and $200\times {10}^3\text{V}$ for $\Delta V$, the value of $I$ is found to be $5.00\times {10}^3\text{A}$ as in equation (III).

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_0$, $5.00\times {10}^{-3}\text{A}$ for $I_{enclosed}$ and $0.050\text{m}$ for $r$ in equation (I).

    $\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m/A}\right)\left(5.00\times {10}^3\text{A}\right)}{2\pi \left(0.050\text{m}\right)}\\ =\frac{20\times {10}^{-4}}{1\times {10}^{-2}}\text{T}\\ =0.020\text{T}\\ =20.0\text{mT}\end{array}$

Therefore, the magnetic field at the inner surface of the outer conductor has a value of $20.0\text{mT}$.

To determine

The energy stored in the magnetic field

Answer to Problem 73P

The energy stored in the magnetic field in the space between the conductors is $2.29\times {10}^6\text{J}$.

Explanation of Solution

Write the equation for the energy density in the magnetic field.

    $u=\frac{B^2}{2{\mu }_0}$        (IV)

Here, $B$ is the magnetic field and ${\mu }_0$ is the permeability in free space.

Write the equation for the energy stored in the magnetic field.

    $U={\displaystyle \int udV}$        (V)

Here, $u$ is the energy density and $V$ is the volume of the conductor.

Write the equation for the change in volume of the conductor.

    $dV=2\pi rldr$        (VI)

Here, $r$ is the radius which varies from $a$ to $b$ and $l$ is the length of the conductor.

Substitute equation (I) and equation (III) in equation (II).

    $U={\displaystyle \underset{a}{\overset{b}{\int }}\frac{{\left[B\left(r\right)\right]}^2\left(2\pi rldr\right)}{2{\mu }_0}}$        (VII)

Substitute equation (I) in equation (VII).

    $\begin{array}{c}U=\frac{{\mu }_0{I}^{2}l}{4\pi }{\displaystyle \underset{a}{\overset{b}{\int }}\frac{dr}{r}}\\ =\frac{{\mu }_0{I}^{2}l}{4\pi }\ln \left(\frac{b}{a}\right)\end{array}$

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_0$, $5.00\times {10}^{-3}\text{A}$ for $I$, $1000\times {10}^3\text{m}$ for $l$, $5.00\text{cm}$ for $b$ and $2.00\text{cm}$ for $a$.

    $\begin{array}{c}U=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m/A}\right){\left(5.00\times {10}^{-3}\text{A}\right)}^2\left(1000\times {10}^3\text{m}\right)}{4\pi }\ln \left(\frac{5.00\text{cm}}{2.00\text{cm}}\right)\\ =2.29\times {10}^6\text{J}\end{array}$

Therefore, the energy stored in the magnetic field in the space between the conductors is $2.29\times {10}^6\text{J}$.

To determine

The pressure exerted on the outer conductor

Answer to Problem 73P

The pressure exerted on the outer conductor is $318\text{Pa}$.

Explanation of Solution

Consider a small rectangular segment with length $l$ and width $w$ on the outer cylinder. Write the equation for the current carried by the rectangular segment.

    $I=\left(5.00\times {10}^{-3}\text{A}\right)\left(\frac{w}{2\pi \left(0.0050\text{m}\right)}\right)$        (VIII)

Write the equation for the outward force experienced by the rectangular segment.

    $F=IlB\sin \theta$        (IX)

Here, $I$ is the current carried by the rectangular segment, $l$ is the length of the rectangular segment and $B$ is the magnetic field at the inner surface of the outer conductor.

Substitute $\left(5.00\times {10}^{-3}\text{A}\right)\left(w/2\pi \left(0.0050\text{m}\right)\right)$ for $I$, $20\times {10}^{-3}\text{T}$ for $B$ and $90.0°$ for $\theta$.

    $F=\left(5.00\times {10}^{-3}\text{A}\right)\left(\frac{w}{2\pi \left(0.0050\text{m}\right)}\right)l\left(20\times {10}^{-3}\text{T}\right)\sin \left(90.0°\right)$        (X)

Write the equation for the pressure exerted on the outer conductor.

    $\begin{array}{c}P=\frac{F}{A}\\ =\frac{F}{wl}\end{array}$        (XI)

Here, $F$ is the force acting on the rectangular segment and $A$ is the area which is width time the length, $wl$.

Conclusion:

Substitute equation (X) in equation (XI).

$\begin{array}{c}P=\left(5.00\times {10}^{-3}\text{A}\right)\left(\frac{w}{2\pi \left(0.0050\text{m}\right)}\right)l\left(20\times {10}^{-3}\text{T}\right)\sin \left(90.0°\right)\left(\frac{1}{wl}\right)\\ =\frac{\left(5.00\times {10}^{-3}\text{A}\right)\left(20\times {10}^{-3}\text{T}\right)}{2\pi \left(0.0050\text{m}\right)}\\ =318\text{Pa}\end{array}$

Therefore, the pressure exerted on the outer conductor is $318\text{Pa}$.