A copper wire is made of two pieces joined end to end, as the figure shows…

A copper wire is made of two pieces joined end to end, as the figure shows. The left piece has diameter $2.0\ \text{mm}$ and the right piece has diameter $1.0\ \text{mm}$ . A steady current of $3.0\ \text{A}$ flows through the wire. The temperature is the same everywhere. For copper, take the number of free electrons per unit volume as $8.5 \times 10^{28}\ \text{m}^{-3}$ and the resistivity as $1.7 \times 10^{-8}\ \Omega\,\text{m}$ . The charge of an electron is $1.6 \times 10^{-19}\ \text{C}$ and its mass is $9.1 \times 10^{-31}\ \text{kg}$ .

A straight copper wire made of two pieces joined end to end. The left piece is thick, with a diameter of 2.0 mm, and the right piece is thinner, with a diameter of 1.0 mm. An arrow on the far left shows a steady current of 3.0 A entering the wire, and a second arrow on the far right shows it leaving. A length of 0.50 m is marked along the thin piece.

Think about what stays the same from one piece to the other, and what does not. Remember that the drift speed is the average speed with which the free electrons move along the wire because of the electric field, on top of their random motion. The relaxation time is the average time between two collisions of an electron.

What is the drift speed in the thick piece?
What is the drift speed in the thin piece?
What is the electric field in the thin piece?
What is the potential difference between two points $0.50\ \text{m}$ apart in the thin piece?
What is the relaxation time of the free electrons? Give it in femtoseconds, where $1\ \text{fs} = 10^{-15}\ \text{s}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $I = 3.0\ \text{A}$ in both pieces. Areas: thick piece $A_1 = \pi (1.0 \times 10^{-3}\ \text{m})^2 = 3.14\ \text{mm}^2$ , thin piece $A_2 = \tfrac{1}{4} A_1$ .

(a) Thick piece. From $I = n A e v_d$ , $v_d = \dfrac{I}{n A e}$ . First, $n A e = 8.5 \times 10^{28} \times 3.142 \times 10^{-6} \times 1.6 \times 10^{-19} = 4.27 \times 10^{4}\ \text{C/m}$ . Then $v_d = \dfrac{3.0\ \text{A}}{4.27 \times 10^{4}\ \text{C/m}} = 0.0702\ \text{mm/s}$ .

(b) Thin piece. The area is four times smaller, so the drift speed is four times larger: $4 \times 0.0702\ \text{mm/s} = 0.2809\ \text{mm/s}$ .

(c) The field. The current density in the thin piece is $J = \dfrac{I}{A_2} = 3.820 \times 10^{6}\ \text{A/m}^2$ . Then $E = \rho J = (1.7 \times 10^{-8}\ \Omega\,\text{m})(3.820 \times 10^{6}\ \text{A/m}^2) = 0.0649\ \text{V/m}$ , that is $64.9\ \text{mV/m}$ .

(d) The potential difference. The field is uniform in the thin piece, so $\Delta V = E L = 0.065\ \text{V/m} \times 0.50\ \text{m} = 0.0325\ \text{V}$ , that is $32.5\ \text{mV}$ .

(e) Relaxation time. In a metal, $\rho = \dfrac{m}{n e^2 \tau}$ , so $\tau = \dfrac{m}{n e^2 \rho}$ . First, $n e^2 \rho = 8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 1.7 \times 10^{-8} = 3.70 \times 10^{-17}$ in SI units. Then $\tau = \dfrac{9.1 \times 10^{-31}}{3.70 \times 10^{-17}}\ \text{s} = 24.6 \times 10^{-15}\ \text{s}$ , which is $24.6\ \text{fs}$ .

Answer: (a) $0.0702\ \text{mm/s}$ . (b) $0.2809\ \text{mm/s}$ . (c) $64.9\ \text{mV/m}$ . (d) $32.5\ \text{mV}$ . (e) $24.6\ \text{fs}$ .

Check. The mobility is $\mu = \dfrac{e\tau}{m} = 4.33 \times 10^{-3}\ \text{m}^2\,\text{V}^{-1}\,\text{s}^{-1}$ . Then $v_d = \mu E$ gives $4.33 \times 10^{-3} \times 0.065\ \text{V/m} = 2.809 \times 10^{-4}\ \text{m/s}$ in the thin piece, the same as (b). The potential drop also equals $I \times \rho L / A_2 = 32.5\ \text{mV}$ .

Watch out. The drift speed is not the same in the two pieces, although the current is. And the drift speed is tiny, a fraction of a millimetre per second, even though the current is $3.0\ \text{A}$ . The relaxation time is a property of the metal, so it is the same in both pieces.

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