Chapter 21, Problem 9P

(a)

To determine

The resistivity of the aluminum at a temperature of $50°\text{C}$.

Answer to Problem 9P

The resistivity of the aluminum wire at a temperature of $50°\text{C}$ is $\underset{\_}{31.5\text{n}\Omega \cdot \text{m}}$.

Explanation of Solution

Write the expression for the resistivity.

    $\rho ={\rho }_0\left[1+\alpha \left(T-T_0\right)\right]$        (I)

Here, $\rho$ is the resistivity of the wire at temperature $T$, ${\rho }_0$ is the resistivity of the wire at reference temperature, $\alpha$ is the temperature coefficient, $T_0$ is the reference temperature, $T$ is the temperature.

Conclusion:

Substitute $2.82\times {10}^{-8}\Omega \cdot \text{m}$ for ${\rho }_0$, $3.9\times {10}^{-3}°{\text{C}}^{-1}$ for $\alpha$, $50.0°\text{C}$ for $T$, $20.0°\text{C}$ for $T_0$ in equation (I) to find $\rho$.

    $\begin{array}{c}\rho =\left(2.82\times {10}^{-8}\Omega \cdot \text{m}\right)\left[1+\left(3.9\times {10}^{-3}°{\text{C}}^{-1}\right)\left(50.0°\text{C}-20.0°\text{C}\right)\right]\\ =3.15\times {10}^{-8}\Omega \cdot \text{m}\\ \text{=31}\text{.5}\text{n}\Omega \cdot \text{m}\end{array}$

Therefore, the resistivity of the aluminum at a temperature of $50°\text{C}$ is $\underset{\_}{31.5\text{n}\Omega \cdot \text{m}}$.

(b)

To determine

The current density in the wire.

Answer to Problem 9P

The current density in the wire is $\underset{\_}{6.35\text{MA}/{\text{m}}^2}$.

Explanation of Solution

Write the expression for the current density.

    $J=\frac{I}{A}$        (II)

Here, $J$ is the current density, $I$ is the current, $A$ is the area of the wire.

Write the expression for the current.

    $I=\frac{E}{\rho }A$        (III)

Here, $E$ is the electric field.

Use equation (III) in (II) to solve for $J$.

    $\begin{array}{c}J=\frac{EA/\rho }{A}\\ =\frac{E}{\rho }\end{array}$        (IV)

Conclusion:

Substitute $0.200\text{V}/\text{m}$ for $E$, $31.5\text{n}\Omega \cdot \text{m}$ for $\rho$ in equation (IV) to find $J$.

    $\begin{array}{c}J=\frac{\left(0.200\text{V}/\text{m}\right)}{\left(31.5\text{n}\Omega \cdot \text{m}\times \frac{{10}^{-8}\Omega }{1\text{n}\Omega }\right)}\\ =6.35\times {10}^6\text{A}/{\text{m}}^2\\ =6.35\text{MA}/{\text{m}}^2\end{array}$

Therefore, the current density in the wire is $\underset{\_}{6.35\text{MA}/{\text{m}}^2}$.

(c)

To determine

The total current in the wire.

Answer to Problem 9P

The total current in the wire is $\underset{\_}{49.9\text{mA}}$.

Explanation of Solution

Use equation (II) to solve for $I$.

    $I=JA$        (V)

Write the expression for the $A$.

    $A=\pi {\left(\frac{d}{2}\right)}^2$        (VI)

Here, $d$ is the diameter of the wire.

Use equation (VI) in (V) to solve for $I$.

    $I=J\left(\frac{\pi d^2}{4}\right)$        (VII)

Conclusion:

Substitute $0.100\text{mm}$ for $d$, $6.35\text{MA}/{\text{m}}^2$ for $J$ in equation (VII) to find $I$.

    $\begin{array}{c}I=\left(6.35\text{MA}/{\text{m}}^2\times \frac{{10}^6\text{A}}{1\text{MA}}\right)\left(\frac{\pi {\left(0.100\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}^2}{4}\right)\\ =49.9\times {10}^{-3}\text{A}\\ =49.9\text{mA}\end{array}$

Therefore, the total current in the wire is $\underset{\_}{49.9\text{mA}}$.

(d)

To determine

The drift speed of the conduction electrons.

Answer to Problem 9P

The drift speed of the conduction electrons is $\underset{\_}{659\mu \text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the charge density.

    $n=\frac{N_{\text{A}}{\rho }_e}{M}$        (VIII)

Here, $N_{\text{A}}$ is the Avogadro’s number, ${\rho }_e$ is the density of aluminum, $M$ is the molar mass.

Write the expression for $J$.

    $J=ne{v}_d$        (IX)

Here, $e$ is the electronic charge, $v_d$ is the drift velocity.

Use equation (IX) to solve for $v_d$.

    $v_d=\frac{J}{ne}$        (X)

Conclusion:

Substitute $6.02\times {10}^{23}\text{electrons}$ for $N_{\text{A}}$, $2.70\times {10}^6\text{g}/{\text{m}}^3$ for ${\rho }_e$, $26.98\text{g}$ for $M$ in equation (VIII) to find $n$.

    $\begin{array}{c}n=\frac{\left(6.02\times {10}^{23}\text{electrons}\right)\left(2.70\times {10}^6\text{g}/{\text{m}}^3\right)}{\left(26.98\text{g}\right)}\\ =6.02\times {10}^{28}\text{electrons}/{\text{m}}^3\end{array}$

Substitute $6.02\times {10}^{28}\text{electrons}/{\text{m}}^3$ for $n$, $1.60\times {10}^{-19}\text{C}$ for $e$, $6.35\text{MA}/{\text{m}}^2$ for $J$ in equation (X) to find $J$.

    $\begin{array}{c}{v}_d=\frac{\left(6.35\text{MA}/{\text{m}}^2\times \frac{{10}^6\text{A}}{1\text{MA}}\right)}{\left(6.02\times {10}^{28}\text{electrons}/{\text{m}}^3\right)\left(1.60\times {10}^{-19}\text{C}\right)}\\ =659\times {10}^{-6}\text{m}/\text{s}\\ =659\mu \text{m}/\text{s}\end{array}$

Therefore, the drift speed of the conduction electrons is $\underset{\_}{659\mu \text{m}/\text{s}}$.

(e)

To determine

The potential difference exists between the ends of a $2.00\text{m}$ length of the wire to produce the stated electric field.

Answer to Problem 9P

The potential difference exists between the ends of a $2.00\text{m}$ length of the wire to produce the stated electric field is $\underset{\_}{0.400\text{V}}$.

Explanation of Solution

Write the expression for the potential difference.

    $\Delta V=El$        (XI)

Here, $\Delta V$ is the potential difference, $l$ is the length of the wire.

Conclusion:

Substitute $0.200\text{V}/\text{m}$ for $E$, $2.00\text{m}$ for $l$ in equation (XI) to find $\Delta V$.

    $\begin{array}{c}\Delta V=\left(0.200\text{V}/\text{m}\right)\left(2.00\text{m}\right)\\ =0.400\text{V}\end{array}$

Therefore, the potential difference exists between the ends of a $2.00\text{m}$ length of the wire to produce the stated electric field is $\underset{\_}{0.400\text{V}}$.