Chapter 27, Problem 27.67AP

(a)

To determine

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

Answer to Problem 27.67AP

The magnitude of the electric field in the wire is $8.00\text{V}/\text{m}$ in the positive $x$ direction.

Explanation of Solution

Given information: Length of the cylindrical wire is $0.500\text{m}$, diameter of the cylindrical wire is $0.200\text{mm}$, resistivity of the wire is $4.00\times {10}^{-8}\text{Ωm}$, potential maintained at left end of the wire at $x=0$ is $4.00\text{V}$, potential maintained at the wire at $x=0.500\text{m}$ is $0\text{V}$.

Explanation:

Formula to calculate the magnitude of the electric field in the wire.

$\overrightarrow{E}=\frac{V}{L}\widehat{\text{i}}$ (1)

Here,

$\overrightarrow{E}$ is the electric field in the wire.

$V$ is the potential maintained at the wire.

$L$ is the length of the cylindrical wire.

Substitute $4.00\text{V}$ for $V$, $0.500\text{m}$ for $L$ in equation (1) to find $\overrightarrow{E}$,

$\begin{array}{c}\overrightarrow{E}=\frac{4.00\text{V}}{0.500\text{m}}\widehat{\text{i}}\\ =8.00\widehat{x}\text{V}/\text{m}\end{array}$

Thus, the magnitude and direction of the electric field in the wire is $8.00\text{V}/\text{m}$ in the positive $x$ direction.

Conclusion:

Therefore, the magnitude and direction of the electric field in the wire is $8.00\text{V}/\text{m}$ in the positive $x$ direction.

(b)

To determine

The resistance of the wire.

Answer to Problem 27.67AP

The resistance of the wire is $0.637\text{Ω}$.

Explanation of Solution

Given information: Length of the cylindrical wire is $0.500\text{m}$, diameter of the cylindrical wire is $0.200\text{mm}$, resistivity of the wire is $4.00\times {10}^{-8}\text{Ωm}$, potential maintained at left end of the wire at $x=0$ is $4.00\text{V}$, potential maintained at the wire at $x=0.500\text{m}$ is $0\text{V}$.

Explanation:

Write the expression for the area of cross section of the cylindrical wire.

$A=\frac{\pi }{4}{d}^2$ (2)

Here,

$A$ is the area of cross section of the cylindrical wire.

$d$ is the diameter of the cylindrical wire.

Substitute $0.200\text{mm}$ for $d$ in equation (2) to find $A$,

$\begin{array}{c}A=\frac{\pi }{4}{\left(0.200\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\\ =3.14\times {10}^{-8}{\text{m}}^2\end{array}$

Thus, the area of cross section of the cylindrical wire is $3.14\times {10}^{-8}{\text{m}}^2$.

Formula to calculate the resistance of the wire.

$R=\rho \frac{L}{A}$ (3)

Here,

$R$ is the resistance of the wire.

$L$ is the the length of the wire.

$\rho$ is the resistivity of the wire.

Substitute $0.500\text{m}$ for $L$, $4.00\times {10}^{-8}\text{Ωm}$ for $\rho$, $3.14\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (3) to find $R$,

$\begin{array}{c}R=\left(4.00\times {10}^{-8}\text{Ωm}\right)\times \frac{0.500\text{m}}{3.14\times {10}^{-8}{\text{m}}^2}\\ =0.6369\text{Ω}\\ \approx 0.637\text{Ω}\end{array}$

Thus, the resistance of the wire is $0.637\text{Ω}$.

Conclusion:

Therefore, the resistance of the wire is $0.637\text{Ω}$.

(c)

To determine

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

Answer to Problem 27.67AP

The magnitude and direction of the electric current in the wire is $6.28\text{A}$ in the positive $x$ direction.

Explanation of Solution

Given information: Length of the cylindrical wire is $0.500\text{m}$, diameter of the cylindrical wire is $0.200\text{mm}$, resistivity of the wire is $4.00\times {10}^{-8}\text{Ωm}$, potential maintained at left end of the wire at $x=0$ is $4.00\text{V}$, potential maintained at the wire at $x=0.500\text{m}$ is $0\text{V}$.

Explanation:

Formula to calculate the magnitude of the electric current in the wire.

$\overrightarrow{I}=\frac{V}{R}\widehat{\text{i}}$ (4)

Here,

$\overrightarrow{I}$ is the electric current in the wire.

$V$ is the potential maintained at the wire.

$L$ is the length of the cylindrical wire.

Substitute $4.00\text{V}$ for $V$, $0.637\text{Ω}$ for $R$ in equation (4) to find $\overrightarrow{I}$,

$\begin{array}{c}\overrightarrow{I}=\frac{4.00\text{V}}{0.637\text{Ω}}\widehat{\text{i}}\\ =6.279\widehat{\text{i}}\text{A}\\ \approx 6.28\widehat{\text{i}}\text{A}\end{array}$

Thus, the magnitude and direction of the electric current in the wire is $6.28\text{A}$ in the positive $x$ direction.

Conclusion:

Therefore, the magnitude and direction of the electric current in the wire is $6.28\text{A}$ in the positive $x$ direction.

(d)

To determine

The current density in the wire.

Answer to Problem 27.67AP

The current density in the wire is $200\text{MA}/{\text{m}}^2$.

Explanation of Solution

Given information: Length of the cylindrical wire is $0.500\text{m}$, diameter of the cylindrical wire is $0.200\text{mm}$, resistivity of the wire is $4.00\times {10}^{-8}\text{Ωm}$, potential maintained at left end of the wire at $x=0$ is $4.00\text{V}$, potential maintained at the wire at $x=0.500\text{m}$ is $0\text{V}$.

Explanation:

Formula to calculate the current density in the wire.

$J=\frac{\overrightarrow{I}}{A}$ (5)

Here,

$J$ is the current density in the wire.

Substitute $6.28\text{A}$ for $\overrightarrow{I}$, $3.14\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (5) to find $J$,

$\begin{array}{c}J=\frac{6.28\text{A}}{3.14\times {10}^{-8}{\text{m}}^2}\\ =200000000\text{A}/{\text{m}}^2\\ =200\text{MA}/{\text{m}}^2\end{array}$

Thus, the current density in the wire is $200\text{MA}/{\text{m}}^2$.

Conclusion:

Therefore, the current density in the wire is $200\text{MA}/{\text{m}}^2$.

(e)

To determine

To show: The expression for electric field in the wire is given by $E=\rho J$

Answer to Problem 27.67AP

The expression for electric field in the wire is given by $E=\rho J$.

Explanation of Solution

Given information: Length of the cylindrical wire is $0.500\text{m}$, diameter of the cylindrical wire is $0.200\text{mm}$, resistivity of the wire is $4.00\times {10}^{-8}\text{Ωm}$, potential maintained at left end of the wire at $x=0$ is $4.00\text{V}$, potential maintained at the wire at $x=0.500\text{m}$ is $0\text{V}$.

Explanation:

From equation (1), write the expression for the electric field in the wire.

$\overrightarrow{E}=\frac{V}{L}\widehat{\text{i}}$

Multiply by $\rho$ in numerator and denominator in above expression.

$\overrightarrow{E}=\rho \left(\frac{V}{\rho L}\right)\widehat{\text{i}}$ (6)

From equation (5), formula to calculate the current density in the wire.

$J=\frac{\overrightarrow{I}}{A}$

From equation (4), formula to calculate the magnitude of the electric current in the wire.

$\overrightarrow{I}=\frac{V}{R}\widehat{\text{i}}$

Substitute $\frac{V}{R}\widehat{\text{i}}$ for $\overrightarrow{I}$ in equation (4) to find $J$,

$\begin{array}{c}J=\frac{\left(\frac{V}{R}\widehat{\text{i}}\right)}{A}\\ =\frac{V}{RA}\widehat{\text{i}}\end{array}$ (7)

From equation (3), formula to calculate the resistance of the wire.

$R=\rho \frac{L}{A}$

Substitute $\rho \frac{L}{A}$ for $R$ in equation (7) to find $J$,

$\begin{array}{c}J=\frac{V}{\left(\rho \frac{L}{A}\right)A}\widehat{\text{i}}\\ =\frac{V}{\rho L}\widehat{\text{i}}\end{array}$

Substitute $J$ for $\frac{V}{\rho L}\widehat{\text{i}}$ in equation (6) to find $\overrightarrow{E}$,

$E=\rho J$

Thus, the expression for electric field in the wire is given by $E=\rho J$.

Conclusion:

Therefore, the expression for electric field in the wire is given by $E=\rho J$.