A copper wire of uniform circular cross-section carries a current of 2.50 A. The wire has diameter 0.60 mm. Copper has n = 8.5×10^28 free (conduction) electrons per cubic metre, and the charge on a single electron has magnitude q = 1.60×10^-19 C. (a) State the equation relating…

Answer: ((a)) I = Anvq; drift velocity is the (slow) average velocity with which charge carriers move along the conductor as a result of the applied electric field. ((b)) A = 2.83×10^-7 m^2 ((c)) v = 6.50×10^-4 m s^-1 ((d)) v doubles to 1.30×10^-3 m s^-1, since v I when A, n and q are fixed ((e)) The electric field driving the current is set up throughout the wire almost instantaneously (at a speed close to the speed of light), so all the free electrons throughout the circuit start drifting together at once; the lamp does not have to wait for an individual electron to tr…. Part (a): The equation and meaning of drift velocity The current in a conductor is related to the motion of its charge carriers by: I = Anvq where A is the cross-sectional area, n is the number density (number per unit volume) of charge ca…

AS Level Physics 9702: Drift Velocity of Electrons Using I = Anvq

A copper wire of uniform circular cross-section carries a current of $2.50\text{ A}$ . The wire has diameter $0.60\text{ mm}$ . Copper has $n = 8.5\times10^{28}$ free (conduction) electrons per cubic metre, and the charge on a single electron has magnitude $q = 1.60\times10^{-19}\text{ C}$ .

(a) State the equation relating current $I$ , cross-sectional area $A$ , number density of charge carriers $n$ , drift velocity $v$ and charge on each carrier $q$ , and identify what is meant by "drift velocity". [2]

(b) Calculate the cross-sectional area $A$ of the wire, in $\text{m}^2$ . [2]

(d) The current in the wire is increased to $5.00\text{ A}$ , with $A$ , $n$ and $q$ unchanged. State and calculate the new drift velocity. [2]

(e) The drift velocity found in (c) is extremely small, yet a lamp connected to this wire lights up almost instantly when the circuit is switched on. Explain why. [1]

Part (a): The equation and meaning of drift velocity

The current in a conductor is related to the motion of its charge carriers by: $I = Anvq$

where $A$ is the cross-sectional area, $n$ is the number density (number per unit volume) of charge carriers, $v$ is their drift velocity and $q$ is the charge on each carrier. The drift velocity is the (small) average velocity with which the charge carriers move along the conductor, superimposed on their much faster random thermal motion, as a result of the electric field set up by the source.

Part (b): Cross-sectional area

The radius is half the diameter, converted to metres: $r = \frac{0.60\times10^{-3}}{2} = 3.00\times10^{-4}\text{ m}$

The cross-section is a circle, so: $A = \pi r^2 = \pi \times (3.00\times10^{-4})^2 = \pi \times 9.00\times10^{-8}$ $A = 2.83\times10^{-7}\text{ m}^2 \ (3\text{ s.f.})$

Part (c): Drift velocity

Rearranging $I = Anvq$ for $v$ : $v = \frac{I}{Anq}$

Substituting $I = 2.50\text{ A}$ , $A = 2.8274\times10^{-7}\text{ m}^2$ (unrounded), $n = 8.5\times10^{28}\text{ m}^{-3}$ and $q = 1.60\times10^{-19}\text{ C}$ :

First find the denominator: $Anq = 2.8274\times10^{-7} \times 8.5\times10^{28} \times 1.60\times10^{-19} = 3845\ (3\text{ s.f.})$

Then: $v = \frac{2.50}{3845} = 6.50\times10^{-4}\text{ m s}^{-1} \ (3\text{ s.f.})$

Check: $A n v q = 2.8274\times10^{-7} \times 8.5\times10^{28} \times 6.50\times10^{-4} \times 1.60\times10^{-19} \approx 2.50\text{ A}$ , which matches $I$ . $\checkmark$

Part (d): Drift velocity at a larger current

Since $A$ , $n$ and $q$ are unchanged, $v = I/(Anq)$ shows that $v$ is directly proportional to $I$ . Doubling the current to $5.00\text{ A}$ therefore doubles the drift velocity: $v_{\text{new}} = \frac{5.00}{3845} = 1.30\times10^{-3}\text{ m s}^{-1} \ (3\text{ s.f.})$

This is exactly double the value found in (c), as expected.

Part (e): Why the lamp lights up almost instantly

Although each individual electron drifts extremely slowly (of the order of $10^{-4}\text{ m s}^{-1}$ ), switching on the circuit sets up an electric field along the entire length of the wire almost instantaneously (this field propagates at a speed close to the speed of light). This field acts on the free electrons everywhere in the circuit at once, so electrons already present in the filament of the lamp begin drifting (and colliding, transferring energy) as soon as the switch closes. No single electron needs to travel from the switch to the lamp for the lamp to light.

Final answers

(a) $I = Anvq$ ; drift velocity $=$ average velocity of charge carriers along the conductor due to the applied field.
(b) Cross-sectional area $=$ $2.83\times10^{-7}\text{ m}^2$
(c) Drift velocity $=$ $6.50\times10^{-4}\text{ m s}^{-1}$
(d) New drift velocity $=$ $1.30\times10^{-3}\text{ m s}^{-1}$ (double, since $v \propto I$ )
(e) The field driving the electrons is established almost instantly throughout the circuit, so the lamp does not wait for individual electrons to travel from the switch.