Chapter 32, Problem 75AP

(a)

To determine

The magnetic field at the surface of the inner conductor.

Answer to Problem 75AP

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

Explanation of Solution

Write the expression for power.

    $P=IV$

Her, $P$ is the power, $I$ is the current and $V$ is the voltage.

Rewrite the above equation for $I$.

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

Write the expression for magnetic field.

    $B=\frac{{\mu }_{0}I}{2\pi r}$                                                                                                                   (II)

Here, $B$ is the magnetic field, $r$ is the distance from point to electric flow, $I$ is the current flow and ${\mu }_0$ is the permeability of free space.

Substitute $\frac{P}{V}$ for $I$ in Equation (II) to calculate $B$.

    $\begin{array}{c}B=\frac{{\mu }_0\left(\frac{P}{V}\right)}{2\pi r}\\ =\frac{{\mu }_{0}P}{2\pi rV}\end{array}$                                                                                                             (III)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $1.00\times {10}^3\text{MW}$ for $P$, $200\text{kV}$ for $V$ and $2.00\text{cm}$ for $r$ in Equation (III) to calculate $B$.

    $\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(1.00\times {10}^3\text{MW}\left(\frac{1\times {10}^6\text{W}}{1\text{MW}}\right)\right)}{2\pi \left(2.00\text{cm}\left(\frac{1\text{cm}}{1\times {10}^2\text{m}}\right)\right)\left(200\text{kV}\left(\frac{1\times {10}^3\text{V}}{1\text{kV}}\right)\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(1.00\times {10}^9\text{W}\right)}{2\pi \left(2.00\times {10}^{-2}\text{m}\right)\left(200\times {10}^3\text{V}\right)}\\ =\frac{20}{400}\text{T}\\ =\text{0}\text{.05}\text{T}\left(\frac{1\times {10}^3\text{mT}}{1\text{T}}\right)\end{array}$

Further solve the above equation.

    $B=50.0\text{mT}$

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

(b)

To determine

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

Answer to Problem 75AP

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

Explanation of Solution

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $1.00\times {10}^3\text{MW}$ for $P$, $200\text{kV}$ for $V$ and $5.00\text{cm}$ for $r$ in Equation (III) to calculate $B$.

    $\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(1.00\times {10}^3\text{MW}\left(\frac{1\times {10}^6\text{W}}{1\text{MW}}\right)\right)}{2\pi \left(5.00\text{cm}\left(\frac{1\text{cm}}{1\times {10}^2\text{m}}\right)\right)\left(200\text{kV}\left(\frac{1\times {10}^3\text{V}}{1\text{kV}}\right)\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(1.00\times {10}^9\text{W}\right)}{2\pi \left(5.00\times {10}^{-2}\text{m}\right)\left(200\times {10}^3\text{V}\right)}\\ =\frac{20}{1000}\text{T}\\ =\text{0}\text{.02}\text{T}\left(\frac{1\times {10}^3\text{mT}}{1\text{T}}\right)\end{array}$

Further solve the above equation.

    $B=20.0\text{mT}$

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

(c)

To determine

The energy stored in the magnetic field in the space between the conductors in a $1.00\times {10}^3\text{km}$ superconducting line.

Answer to Problem 75AP

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

Explanation of Solution

Write the expression for energy stored in the magnetic field.

    $dU=\frac{1}{2{\mu }_0}{B}^2\left(2\pi rldr\right)$                                                                                           (IV)

Here, $dU$ is the energy stored, $l$ is the length of the cylinder.

Substitute, $\frac{{\mu }_{0}I}{2\pi r}$ for $B$ in Equation (IV) to calculate $U$.

    $\begin{array}{c}dU=\frac{1}{2{\mu }_0}{\left(\frac{{\mu }_{0}I}{2\pi r}\right)}^2\left(2\pi rldr\right)\\ U=\frac{1}{2{\mu }_0}{\displaystyle {\int }_{\text{a}}^{\text{b}}{\left(\frac{{\mu }_{0}I}{2\pi r}\right)}^2\left(2\pi rldr\right)}\\ =\frac{{\mu }_0{I}^{2}l}{4\pi }{\displaystyle {\int }_{\text{a}}^{\text{b}}\frac{dr}{r}}\\ =\frac{{\mu }_0{I}^{2}l}{4\pi }\ln \left(\frac{b}{a}\right)\end{array}$                                                                                 (V)

Conclusion:

Substitute $1.00\times {10}^3\text{MW}$ for $P$, $200\text{kV}$ for $U$ in Equation (I) to calculate $I$.

    $\begin{array}{c}I=\frac{1.00\times {10}^3\text{MW}\left(\frac{1\times {10}^6\text{W}}{1\text{MW}}\right)}{200\text{kV}\left(\frac{1\times {10}^3\text{V}}{1\text{kV}}\right)}\\ =\frac{1.00\times {10}^9\text{W}}{200\times {10}^3\text{V}}\\ =5000\text{A}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $5000\text{A}$ for $I$, $1.00\times {10}^3\text{km}$ for $l$, $2.00\text{cm}$ for $a$ and $5.00\text{cm}$ for $b$ in Equation (V) to calculate $U$

    $\begin{array}{c}U=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right){\left(5000\text{A}\right)}^2\left(1.00\times {10}^3\text{km}\left(\frac{1\times {10}^3\text{m}}{1\text{km}}\right)\right)}{4\pi }\ln \left(\frac{5.00\text{cm}}{2.00\text{cm}}\right)\\ =\left(1\times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right){\left(5000\text{A}\right)}^2\left(1.00\times {10}^6\text{m}\right)\ln \left(2.5\right)\\ =2.29\times {10}^6\text{J}\left(\frac{1\text{MJ}}{1\times {10}^6\text{J}}\right)\\ =2.29\text{MJ}\end{array}$

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

(d)

To determine

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

Answer to Problem 75AP

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

Explanation of Solution

Consider a small rectangular section of the outer cylinder of length $l$ and width $w$. It is shown that current carried by the wire of surface area $2\pi bl$ is $5000\text{A}$, then current carried by that rectangle of area $wl$ is,

    $\begin{array}{c}\frac{5000\text{A}}{I}=\frac{2\pi bl}{wl}\\ I=\frac{\left(5000\text{A}\right)w}{2\pi b}\end{array}$

Write the expression for the force.

    $F=BIl\sin \theta$                                                                                                          (VI)

Here, $F$ is the force.

Substitute $\frac{\left(5000\text{A}\right)w}{2\pi b}$ for $I$ in Equation (I) to calculate $F$.

    $F=B\left(\frac{\left(5000\text{A}\right)w}{2\pi b}\right)l\sin \theta$

Write the expression for pressure.

    $P=\frac{F}{A}$                                                                                                                    (VII)

Here, $P$ is the pressure and $A$ is the area.

Substitute $B\left(\frac{\left(5000\text{A}\right)w}{2\pi b}\right)l\sin \theta$ for $F$ and $wl$ for $A$ in Equation (VII) to calculate $P$.

    $\begin{array}{c}P=\frac{\left(B\left(\frac{\left(5000\text{A}\right)w}{2\pi b}\right)l\sin \theta \right)}{wl}\\ =\left(B\left(\frac{\left(5000\text{A}\right)}{2\pi b}\right)\sin \theta \right)\end{array}$                                                                             (VIII)

Conclusion:

Substitute $20.0\text{mT}$ for $B$, $90°$ for $\theta$ and $5.00\text{cm}$ for $b$ in Equation (VIII) to calculate $P$.

    $\begin{array}{c}P=\left(\left(20.0\text{mT}\right)\left(\frac{1\times {10}^{-3}\text{T}}{1\text{mT}}\right)\left(\frac{\left(5000\text{A}\right)}{2\pi \left(5.00\text{cm}\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\right)\sin 90°\right)\\ =\left(\left(20.0\times {10}^{-3}\text{T}\right)\left(\frac{\left(5000\text{A}\right)}{2\pi \left(0.05\text{m}\right)}\right)\right)\\ =318\text{Pa}\end{array}$

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