A wire of radius $0.15 \mathrm{~mm}$ is made of an unknown metal and carries a current of $8.4 \mathrm{~mA}$. The drift velocity for electrons in the wire is measured to be $36 \times 10^{-6} \mathrm{~m} / \mathrm{s}$. What is the charge carrier density of the metal from which the wire is made?
A) $2.1 \times 10^{28} \mathrm{~m}^{-3}$
B) $4.2 \times 10^{30} \mathrm{~m}^{-3}$
C) $1.3 \times 10^{29} \mathrm{~m}^{-3}$
D) need more information to do this calculation.

Question

Quizplus · Accepted Answer

The charge carrier density ($n$) can be calculated using the formula $I = nAve$, where $I$ is the current, $A$ is the cross-sectional area of the wire, $v$ is the drift velocity, and $e$ is the charge of an electron ($1.6 \times 10^{-19} \mathrm{C}$). Given $I = 8.4 \times 10^{-3} \mathrm{A}$, $v = 36 \times 10^{-6} \mathrm{m/s}$, and the radius $r = 0.15 \times 10^{-3} \mathrm{m}$, the cross-sectional area $A = \pi r^2$. Substituting these values into the formula allows for solving $n$, yielding $n \approx 2.1 \times 10^{28} \mathrm{m}^{-3}$.

A Wire of Radius 0.15 mm0.15 \mathrm{~mm}0.15 mm Is Made of an Unknown Metal and Carries a Current

A Wire of Radius $0.15 \mathrm{~mm}$ Is Made of an Unknown Metal and Carries a Current