Answer to Question #103177 in Electricity and Magnetism for Ajay

Question

A copper wire of diameter 1 mm and length 30 m is connected across a battery of 2V.
Calculate the current density in the wire and drift velocity of the electrons. The resistivity
of copper is 1.72 ×10^–8Ωm and n = 8.0 ×10^28 electrons m^–3.
.

Answer 1

a) Let’s first find the resistance of the copper wire from the formula:

$$R = \rho \dfrac{l}{A},$$

here, $\rho$ is the resistivity of the copper wire, $l$ is the length of the wire, $A = \dfrac{\pi d^2}{4}$ is the cross-sectional area of the wire and $d$ is the diameter of the wire.

Then, we get:

$$R = \rho \dfrac{l}{A} = \rho \dfrac{4l}{\pi d^2},$$ $$R = 1.72 \cdot 10^{-8} \Omega \cdot m \cdot \dfrac{4 \cdot 30m}{\pi \cdot (1.0 \cdot 10^{-3}m)^2} = 0.65 \Omega.$$

Then, from the Ohm’s law we can find the current flowing through the copper wire:

$$I = \dfrac{V}{R} = \dfrac{2.0 V}{0.65 \Omega} = 3.1A.$$

Finally, we can find the current density in the wire:

$$J = \dfrac{I}{A} = \dfrac{4I}{\pi d^2} = \dfrac{4 \cdot 3.1A}{\pi \cdot (1.0 \cdot 10^{-3}m)^2} = 3.95 \cdot 10^6 \dfrac{A}{m^2}.$$

b) We can find the drift velocity of the electrons from the formula:

$$v = \dfrac{I}{nAq},$$

here, $I$ is the current flowing through the wire, $n$ is the number of free electrons per unit volume of the copper wire, $A$ is the cross-sectional area of the wire and $q$ is the charge on each electron. Then, we get:

$$v = \dfrac{I}{nAq} = \dfrac{4I}{n\pi d^2 q},$$

$v = \dfrac{4 \cdot 3.1A}{8.0 \cdot 10^{28} \dfrac{electrons}{m^3} \cdot \pi \cdot (1.0 \cdot 10^{-3}m)^2 \cdot 1.6 \cdot 10^{-19}C} = 3.08 \cdot 10^{-4} \dfrac{m}{s}.$

Answer:

a) $J = 3.95 \cdot 10^6 \dfrac{A}{m^2}.$

b) $v = 3.08 \cdot 10^{-4} \dfrac{m}{s}.$