Chapter 28, Problem 47PQ

(a)

To determine

The magnitude and direction of the electric field in the wire.

Answer to Problem 47PQ

The magnitude and direction of the electric field in the wire are $4.27\text{V/m}$ and from the left.

Explanation of Solution

Write the relation for the electric field in the wire.

   $\begin{array}{c}{E}_x=-\frac{dV}{dx}\\ =-\frac{\Delta V}{\Delta x}\end{array}$                                                                                                    (1)

Here, $E_x$ is electric field in the wire, $\Delta V/\Delta x$ is the slope of the potential which is  from high to low potential points.

Conclusion:

Substitute $\left(0-3.20\text{V}\right)$ for $\Delta V$ and $0.750\text{m}$ for $\Delta x$ to find $E_x$.

  $\begin{array}{c}{E}_x=-\frac{\left(0-3.20\text{V}\right)}{0.750\text{m}}\\ =4.27\text{V/m}\text{from}\text{left}\text{to}\text{right}\end{array}$

Therefore, the magnitude and direction of the electric field in the wire are $4.27\text{V/m}$ and from the left.

(b)

To determine

The resistance of the wire.

Answer to Problem 47PQ

The resistance of the wire is $0.168\text{Ω}$.

Explanation of Solution

Write the relation for the resistance of the cylindrical silver wire.

   $\begin{array}{c}R=\rho \frac{l}{A}\\ =\rho \frac{l}{\pi r^2}\end{array}$                                                                                                     (2)

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

Conclusion:

Substitute $1.586\times {10}^{-8}\text{Ω}\cdot \text{m}$ for $\rho$, $0.750\text{m}$ for $l$, and $0.150\times {10}^{-3}\text{m}$ for $r$ to find $R$ in above equation.

   $\begin{array}{c}R=\left(1.586\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\frac{0.750\text{m}}{\pi {\left(0.150\times {10}^{-3}\text{m}\right)}^2}\\ =0.168\text{Ω}\end{array}$

The resistance of the cylindrical silver wire is $0.168\text{Ω}$.

(c)

To determine

The magnitude and direction of the current in the wire.

Answer to Problem 47PQ

The magnitude and direction of the current in the wire are $19.0\text{A from}\text{left}\text{to}\text{right}$.

Explanation of Solution

Write the relation for the current in the cylindrical silver wire.

   $I=\frac{\Delta V}{R}$                                                                                                    (2)

Here, $I$ is the current in the wire and $\Delta V$ is potential difference.

Conclusion:

Substitute $3.20\text{V}$ for $\Delta V$ and $0.168\Omega$ for $R$ to find $I$.

   $\begin{array}{c}I=\frac{3.20\text{V}}{0.168\Omega }\\ =19.0\text{A}\end{array}$

The current is in positive sign represents that it flows from left to right.

The magnitude and direction of the current in the wire are $19.0\text{A from}\text{left}\text{to}\text{right}$.

(d)

To determine

The current density in the wire.

Answer to Problem 47PQ

The current density in the wire is $2.69\times {10}^8{\text{A/m}}^{\text{3}}$.

Explanation of Solution

Write the relation for the current density in the wire.

   $\begin{array}{c}J=\frac{I}{A}\\ =\frac{I}{\pi r^2}\end{array}$                                                                                                        (3)

Here, $J$ is current density.

Conclusion:

Substitute $19.0\text{A}$ for $I$ and $0.150\times {10}^{-3}\text{m}$ for $r$ in equation (3) to find $J$.

   $\begin{array}{c}J=\frac{19.0\text{A}}{0.150\times {10}^{-3}\text{m}}\\ =2.69\times {10}^8{\text{A/m}}^{\text{2}}\end{array}$

Therefore, the current density in the wire is $2.69\times {10}^8{\text{A/m}}^{\text{3}}$.