Answer to Question #103240 in Electric Circuits for Ariel

Question

The current density in a copper wire of diameter 1.02 mm is 1.75x10 to the power of 5 A/m². The number of free electrons per cubic meter of copper is 8.5x10 to the power of 28. Find the current in the wire and magnitude of drift velocity of electrons in the wire.

Answer 1

Solution. According to the condition of the problem d=1.02 mm is diameter of the wire; j=1.75x10^5A/m² is the current density; n=8.5x10^28 m^-3 is the number of free electrons per cubic meter of copper.

Find the current using the formula

$$I=jA=\frac{j\pi d^2}{4}$$

where A is cross section of area of wire. Therefore

$$I=\frac{1.75\times10^5\pi (1.02 \times 10^{-3})^2}{4} \approx 0.143A$$

Find drift velocity of electrons in the wire using the formula

$$v=\frac {I}{qnA}=\frac{j}{qn}$$

where q=1.6×10^-19C is electron charge. As result get

$$v=\frac{1.75\times10^5}{1.6\times10^{-19} \times 8.5\times10^{28}} \approx 1.29\times10^{-5}\frac {m}{s}$$

Answer. I=0.143A

$$I= 0.143A$$

$$v=1.29\times10^{-5}\frac {m}{s}$$