Chapter 27, Problem 33P

(a)

To determine

The resistivity of the aluminum at $50.0\text{°C}$ temperature.

Answer to Problem 33P

The resistivity of the aluminum is $31.5\text{nΩ}\cdot \text{m}$ at $50.0\text{°C}$ temperature.

Explanation of Solution

Write the expression to obtain the resistivity of the aluminum at $50.0\text{°C}$.

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

Here, $\rho$ is the resistivity of aluminum at $50.0\text{°C}$, ${\rho }_0$ is the initial resistivity, $\alpha$ is the coefficient of thermal expansion $T_0$ is the initial temperature and $T$ is the final temperature.

Conclusion:

Substitute $2.82\times {10}^{-8}\text{Ω}\cdot \text{m}$ for ${\rho }_0$, $3.9\times {10}^{-3}{\text{°C}}^{-1}$ for $\alpha$, $20\text{°C}$ for $T_0$ and $50\text{°C}$ for $T$ in the above equation to calculate $\rho$.

    $\begin{array}{c}\rho =2.82\times {10}^{-8}\text{Ω}\cdot \text{m}\left(1+3.9\times {10}^{-3}{\text{°C}}^{-1}\left(50\text{°C}-20\text{°C}\right)\right)\\ =31.5\times {10}^{-9}\text{Ω}\cdot \text{m}\\ =31.5\times {10}^{-9}\text{Ω}\cdot \text{m}\left(\frac{1\text{nΩ}\cdot \text{m}}{{10}^{-9}\text{Ω}\cdot \text{m}}\right)\\ =31.5\text{nΩ}\cdot \text{m}\end{array}$

Therefore, the resistivity of the aluminum is $31.5\text{nΩ}\cdot \text{m}$ at $50.0\text{°C}$ temperature.

(b)

To determine

The current density in the wire.

Answer to Problem 33P

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

Explanation of Solution

Write the expression to obtain the current density in the wire.

    $J=\frac{E}{\rho }$

Here, $\rho$ is the resistivity, $E$ is the electric field and $J$ is the current density.

Conclusion:

Substitute $3.15\times {10}^{-8}\text{Ω}\cdot \text{m}$ for $\rho$ and $0.200\text{V}/\text{m}$ for $E$ in the above equation to calculate $J$.

    $\begin{array}{c}J=\frac{0.200\text{V}/\text{m}}{3.15\times {10}^{-8}\text{Ω}\cdot \text{m}}\\ =6.35\times {10}^6\text{A}/{\text{m}}^2\\ =6.35\times {10}^6\text{A}/{\text{m}}^2\left(\frac{1\text{MA}/{\text{m}}^2}{{10}^6\text{A}/{\text{m}}^2}\right)\\ =6.35\text{MA}/{\text{m}}^2\end{array}$

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

(c)

To determine

The current in the wire.

Answer to Problem 33P

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

Explanation of Solution

Write the expression to obtain the current in the wire.

    $I=JA$

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

Substitute $\pi r^2$ for $A$ in the above equation.

    $I=J\left(\pi r^2\right)$

Here, $r$ is the radius of the wire.

Conclusion:

Calculate the radius of the wire.

    $\begin{array}{c}r=\frac{0.100}{2}\text{mm}\\ =0.05\text{mm}\end{array}$

Substitute $0.05\text{mm}$ for $r$ and $6.35\times {10}^6\text{A}/{\text{m}}^2$ for $J$ in the above equation to calculate $I$.

    $\begin{array}{c}I=6.35\times {10}^6\text{A}/{\text{m}}^2\left(\pi {\left(0.05\text{mm}\right)}^2\right)\\ =6.35\times {10}^6\text{A}/{\text{m}}^2\left(\pi {\left(0.50\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\right)\\ =49.9\times {10}^{-3}\text{A}\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ =49.9\text{mA}\end{array}$

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

(d)

To determine

The drift velocity of the conduction electron.

Answer to Problem 33P

The drift velocity of the conduction electron is $658\mu \text{m}/\text{s}$.

Explanation of Solution

Write the expression to obtain the electron density.

    $n=\frac{N_{\text{A}}{\rho }_{\text{Al}}}{M}$                                                                                                                (I)

Here, $n$ is the electron density, $N_{\text{A}}$ is the Avogadro’s number, $M$ is the aluminum molar mass and ${\rho }_{\text{Al}}$ is the aluminum density.

Write the expression to obtain the drift velocity.

    $v_{\text{d}}=\frac{J}{ne}$                                                                                                                    (II)

Here, $v_{\text{d}}$ is the drift velocity, $J$ is the current density, $n$ is the electron density and $e$ is the charge on electron.

Conclusion:

Substitute $6.02\times {10}^{23}\text{atoms}/\text{mol}$ for $N_{\text{A}}$, $2.7\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{Al}}$ and $26.98\text{g}/\text{mol}$ for $M$ in equation (I) to calculate $n$.

    $\begin{array}{c}n=\frac{\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)\left(2.7\times {10}^3\text{kg}/{\text{m}}^3\right)}{26.98\text{g}/\text{mol}}\\ =\frac{\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)\left(2.7\times {10}^3\text{kg}/{\text{m}}^3\right)}{\left(26.98\text{g}/\text{mol}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}\\ =6.02\times {10}^{28}{\text{m}}^{-3}\end{array}$

Substitute $6.35\times {10}^6\text{A}/{\text{m}}^2$ for $J$, $6.02\times {10}^{28}{\text{m}}^{-3}$ for $n$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (II) to calculate $v_{\text{d}}$.

    $\begin{array}{c}{v}_{\text{d}}=\frac{6.35\times {10}^6\text{A}/{\text{m}}^2}{\left(6.02\times {10}^{28}{\text{m}}^{-3}\right)\left(1.6\times {10}^{-19}\text{C}\right)}\\ =658\times {10}^{-6}\text{m}/\text{s}\\ =\left(658\times {10}^{-6}\text{m}/\text{s}\right)\left(\frac{1\mu \text{m}/\text{s}}{{10}^{-6}\text{m}/\text{s}}\right)\\ =658\mu \text{m}/\text{s}\end{array}$

Therefore, the drift velocity of the electron is $658\mu \text{m}/\text{s}$.

(e)

To determine

The applied voltage.

Answer to Problem 33P

The applied voltage is $0.400\text{V}$.

Explanation of Solution

Write the expression to obtain the applied voltage.

    $V=El$

Here, $V$ is the applied voltage, $E$ is the electric field and $l$ is the length of wire.

Conclusion:

Substitute $0.200\text{V}/\text{m}$ for $E$ and $2.00\text{m}$ for $l$ in the above equation to calculate $V$.

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

Therefore, the applied voltage is $0.400\text{V}$.