Chapter 27, Problem 49P

(a)

To determine

The magnitude of the electric field in the wire.

Answer to Problem 49P

The magnitude of the electric field in the wire is $3.98\text{V}/\text{m}$.

Explanation of Solution

Write the expression for resistance of a wire.

    $R=\rho \frac{l}{A}$                                                                                                                    (I)

Here, $l$ is the length of the wire, $A$ is the area of cross-section, $R$ is the resistance of the wire and $\rho$ is the resistivity of material.

Write the expression for the area of the wire.

    $A=\pi r^2$                                                                                                                    (II)

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

Write the expression for potential difference.

    $V=IR$                                                                                                                     (III)

Here, $I$ is the current and $V$ is the potential difference.

Write the expression for electric field along the wire.

    $E=\frac{V}{l}$                                                                                                                     (IV)

Here, $E$ is the magnitude of electric field.

Conclusion:

Substitute $0.200\text{mm}$ for $r$ in equation (II) to find $A$.

    $\begin{array}{c}A=\pi {\left[\left(0.200\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right]}^2\\ =\pi {\left(0.200\times {10}^{-3}\text{m}\right)}^2\\ =1.256\times {10}^{-7}{\text{m}}^2\end{array}$

Substitute $1.256\times {10}^{-7}{\text{m}}^2$ for $A$, $1\times {10}^{-6}\Omega \cdot \text{m}$ for $\rho$ and $25.0\text{m}$ for $l$ in equation (I) to find $R$.

    $\begin{array}{c}R=\left(1\times {10}^{-6}\Omega \cdot \text{m}\right)\left(\frac{25.0\text{m}}{1.256\times {10}^{-7}{\text{m}}^2}\right)\\ =198.944\Omega \end{array}$

Substitute $198.944\Omega$ for $R$ and $0.500\text{A}$ for $I$ in equation (III) to find $V$.

    $\begin{array}{c}V=\left(0.500\text{A}\right)\left(198.944\Omega \right)\\ =99.47\text{V}\end{array}$

Substitute $25.0\text{m}$ for $l$ and $99.47\text{V}$ for $V$ in equation (IV) to find $E$.

    $\begin{array}{c}E=\frac{99.47\text{V}}{25.0\text{m}}\\ =3.98\text{V}/\text{m}\end{array}$

Therefore, the magnitude of the electric field in the wire is $3.98\text{V}/\text{m}$.

(b)

To determine

The power delivered to the wire.

Answer to Problem 49P

The power delivered to the wire is $49.7\text{W}$.

Explanation of Solution

Write the expression for the power delivered to the wire.

    $P=VI$                                                                                                                      (V)

Here, $P$ is the power delivered to the wire.

Conclusion:

Substitute $99.47\text{V}$ for $V$ and $0.500\text{A}$ for $I$ in equation (V) to find $P$.

    $\begin{array}{c}P=\left(0.500\text{A}\right)\left(99.47\text{A}\right)\\ =49.7\text{W}\end{array}$

Therefore, the power delivered is $49.7\text{W}$.

(c)

To determine

The power delivered to the wire if the temperature is increased to $340°\text{C}$ and the potential difference across the wire remains constant.

Answer to Problem 49P

The power delivered to the wire after temperature change is $44.1\text{W}$.

Explanation of Solution

Write the expression for the resistance of the wire after change in temperature.

    $R_0=R\left(1+\alpha \left(T-T_0\right)\right)$                                                                                            (VI)

Here, $\alpha$ is the temperature coefficient, $R_0$ is the resistance of the wire after temperature change, $T_0$ is the initial temperature of the wire and $T$ is the temperature of wire after temperature change.

Write the expression for power delivered to the wire after temperature change.

    $P^{\prime }=\frac{V^2}{R_0}$                                                                                                                 (VII)

Here, $P^{\prime }$ is the power delivered to the wire after temperature change.

Conclusion:

Substitute $0.4\times {10}^{-3}$ for $\alpha$, $20.0°\text{C}$ for $T_0$, $340°\text{C}$ for $T$ and $198.944\Omega$ for $R$ in equation (VI) to find $R_0$.

    $\begin{array}{c}{R}_0=\left(198.944\Omega \right)\left(1+\left(0.4\times {10}^{-3}°{\text{C}}^{-1}\right)\left(340°\text{C}-20.0°\text{C}\right)\right)\\ =224.4\Omega \end{array}$

Substitute $224.4\Omega$ for $R_0$ and $99.47\text{V}$ for $V$ in equation (VII) to find $P^{\prime }$.

    $\begin{array}{c}{P}^{\prime }=\frac{{\left(99.47\text{V}\right)}^2}{224.4\Omega }\\ =44.1\text{W}\end{array}$

Therefore, the power delivered to the wire after temperature change is $44.1\text{W}$.