Chapter 18, Problem 50P

(a)

To determine

The resistance of wire.

Answer to Problem 50P

  $0.029\Omega$

Explanation of Solution

Given info:

Voltage, $V=22.0mV=22.0\times {10}^{-3}V$

Current, $R=750mA=0.75A$

Formula used:

Expression to calculate voltage:

  $V=I\times R$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

Calculation:

Rearrange the expression $V=I\times R$ . So, $R=\frac{V}{I}$ .

Substitute all the given values in the expression, $R=\frac{V}{I}$

  $\begin{array}{c}R=\frac{22.0\times {10}^{-3}V}{0.75A}\\ =0.2933\Omega \approx 0.029\Omega \end{array}$

Conclusion: The resistance is $0.029\Omega$

(b)

To determine

The resistivity of wire.

Answer to Problem 50P

  $1.6\times {10}^{-8}\Omega \cdot m$

Explanation of Solution

Given info:

Voltage, $V=22.0mV=22.0\times {10}^{-3}V$

Current, $R=750mA=0.75A$

Resistance, $R=0.2933\Omega$

Length, $L=5.80m$

Formula used:

Expression to calculate resistance:

  $R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

  $\pi$ is constant and $r$ is a radius.

Calculation:

Rearrange the expression $R=\frac{\rho L}{\pi r^2}$ . So, $\rho =\frac{R\pi r^2}{L}$ .

Substitute all the given values in the expression, $\rho =\frac{R\pi r^2}{L}$

  $\rho =\frac{\left(0.02933\right)\pi {\left(1.0\times {10}^{-3}m\right)}^2}{5.80m}=1.6\times {10}^{-8}\Omega \cdot m$

Conclusion: The resistivity is $1.6\times {10}^{-8}\Omega \cdot m$

(c)

To determine

The number of free electrons per unit volume.

Answer to Problem 50P

The answer is $8.8\times {10}^{28}{e}^{-}/m^3$ .

Explanation of Solution

Given info:

Voltage, $V=22.0mV=22.0\times {10}^{-3}V$

Current, $R=750mA=0.75A$

Resistance, $R=0.2933\Omega$

Length, $L=5.80m$

The resistivity is $1.6\times {10}^{-8}\Omega \cdot m$

Radius, $r=1.0\times {10}^{-3}m$

Drift speed, $v_d=1.7\times {10}^{-5}m/s$

Formula used:

Expression to calculate resistance:

  $I=neA{v}_d$

Where,

  $v_d$ is drift speed.

  $I$ is the current,

  $A$ is the cross-sectional area of the wire.

  $n$ is number of free electrons,

  $e$ is charges,

Calculation:

Rearrange the expression $I=neA{v}_d$ . So, $n=\frac{I}{eA{v}_d}=\frac{I}{e\pi r^2{v}_d}$ .

Substitute all the given values in the expression, $n=\frac{I}{e\pi r^2{v}_d}$

  $n=\frac{\left(0.75A\right)}{\left(1.60\times {10}^{-19}C\right)\pi {\left(1.0\times {10}^{-3}m\right)}^2\left(1.7\times {10}^{-5}m/s\right)}=8.8\times {10}^{28}{e}^{-}/m^3$

Conclusion: The number of free electrons per unit volume is $8.8\times {10}^{28}{e}^{-}/m^3$ .