phys 2205 - homework - chapter 25\n25.21 • a current - carrying gold wire has diameter 0.84 mm. the electric…

phys 2205 - homework - chapter 25\n25.21 • a current - carrying gold wire has diameter 0.84 mm. the electric field in the wire is 0.49 v/m. what are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4 m length of this wire?\ntable 25.1 resistivities at room temperature (20°c)\n| substance | $\rho(omegacdot m)$ | substance | $\rho(omegacdot m)$ |\n|--|--|--|--|--|\n| conductors | | | |\n| metals | silver | $1.47\times 10^{-8}$ | semiconductors | |\n| | copper | $1.72\times 10^{-8}$ | pure carbon (graphite) | $3.5\times 10^{-5}$ |\n| | gold | $2.44\times 10^{-8}$ | pure germanium | 0.60 |\n| | aluminum | $2.75\times 10^{-8}$ | pure silicon | 2300 |\n| | tungsten | $5.25\times 10^{-8}$ | insulators | |\n| | steel | $20\times 10^{-8}$ | amber | $5\times 10^{14}$ |\n| | lead | $22\times 10^{-8}$ | glass | $10^{10}-10^{14}$ |\n| | mercury | $95\times 10^{-8}$ | lucite | $>10^{13}$ |\n| alloys | manganin (cu 84%, mn 12%, ni 4%) | $44\times 10^{-8}$ | mica | $10^{11}-10^{15}$ |\n| | constantan (cu 60%, ni 40%) | $49\times 10^{-8}$ | quartz (fused) | $75\times 10^{16}$ |\n| | nichrome | $100\times 10^{-8}$ | sulfur | $10^{15}$ |\n| | | | teflon | $>10^{13}$ |\n| | | | wood | $10^{8}-10^{11}$ |

phys 2205 - homework - chapter 25\n25.21 • a current - carrying gold wire has diameter 0.84 mm. the electric field in the wire is 0.49 v/m. what are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4 m length of this wire?\ntable 25.1 resistivities at room temperature (20°c)\n| substance | $\rho(omegacdot m)$ | substance | $\rho(omegacdot m)$ |\n|--|--|--|--|--|\n| conductors | | | |\n| metals | silver | $1.47\times 10^{-8}$ | semiconductors | |\n| | copper | $1.72\times 10^{-8}$ | pure carbon (graphite) | $3.5\times 10^{-5}$ |\n| | gold | $2.44\times 10^{-8}$ | pure germanium | 0.60 |\n| | aluminum | $2.75\times 10^{-8}$ | pure silicon | 2300 |\n| | tungsten | $5.25\times 10^{-8}$ | insulators | |\n| | steel | $20\times 10^{-8}$ | amber | $5\times 10^{14}$ |\n| | lead | $22\times 10^{-8}$ | glass | $10^{10}-10^{14}$ |\n| | mercury | $95\times 10^{-8}$ | lucite | $>10^{13}$ |\n| alloys | manganin (cu 84%, mn 12%, ni 4%) | $44\times 10^{-8}$ | mica | $10^{11}-10^{15}$ |\n| | constantan (cu 60%, ni 40%) | $49\times 10^{-8}$ | quartz (fused) | $75\times 10^{16}$ |\n| | nichrome | $100\times 10^{-8}$ | sulfur | $10^{15}$ |\n| | | | teflon | $>10^{13}$ |\n| | | | wood | $10^{8}-10^{11}$ |

Answer

Explanation:

Step1: Find the cross - sectional area of the wire

The diameter $d = 0.84\ mm=0.84\times10^{- 3}\ m$, and the cross - sectional area $A=\pi(\frac{d}{2})^2=\pi(\frac{0.84\times 10^{-3}}{2})^2\ m^2$.

Step2: Calculate the current in the wire

The relationship between electric field $E$, conductivity $\sigma$, and current density $J$ is $J = \sigma E$. For gold, from the table $\rho=2.44\times 10^{-8}\ \Omega\cdot m$, and $\sigma=\frac{1}{\rho}$. Also, $J=\frac{I}{A}$, so $I = JA=\frac{\sigma E A}{1}=\frac{E A}{\rho}$. Substituting the values: $\rho = 2.44\times10^{-8}\ \Omega\cdot m$, $E = 0.49\ V/m$, and $A=\pi(\frac{0.84\times 10^{-3}}{2})^2\ m^2$, we get $I=\frac{0.49\times\pi(\frac{0.84\times 10^{-3}}{2})^2}{2.44\times 10^{-8}}\ A\approx2.2\ A$.

Step3: Calculate the potential difference

The potential difference $V$ between two points separated by a distance $L$ in a uniform electric field $E$ is given by $V = EL$. Given $E = 0.49\ V/m$ and $L = 6.4\ m$, so $V=0.49\times6.4\ V = 3.136\ V$.

Step4: Calculate the resistance

We can use Ohm's law $V = IR$. We know $V = 3.136\ V$ and $I\approx2.2\ A$. So $R=\frac{V}{I}=\frac{3.136}{2.2}\ \Omega\approx1.425\ \Omega$.

Answer:

(a) $2.2\ A$ (b) $3.136\ V$ (c) $1.425\ \Omega$