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Sample 1: GCE Nov '95 Question 3(b)

A copper wire of cross-sectional area 0.50mm² and length 3.0m carries a steady current of 200mA. Each copper atom in the wire contributes one free electron to the conduction process. Find the average time taken for an electron to drift from one end of the wire to the other.

(Relative atomic mass of copper = 64; density of copper = 8.9 x 10³kgm^-3)

Thinking Process:

Step 1:

Which equation should we use? It should be I = nevA.

I is the current
n the carrier density, ie. number of electrons per m³
v the drift velocity
A the cross-sectional area of the wire.

Step 2:

We know I is 200mA (given); A is 0.5mm² (given); e is 1.6 x 10^-19C (from data sheet).

The only unknown is n (number of electrons per m³).

Step 3:

The relative atomic mass of copper is 64. This means that one mole (ie. 6.02 x 10²³) copper atoms has a mass of 64g or 0.064kg.

Next, we know that the density of copper is 8.9 x 10³kgm^-3. This means that 8.9 x 10³kg of copper occupies a volume of 1m³.

Therefore, the number of copper atoms in 1m³ is:

(8.9x10³/0.064) x 6.02 x 10²³ = 8.37x10²⁸

Since one copper atom contributes one electron, the number of electrons per m³ is also 8.37x10²⁸

Step 4:

Substituting all known data into the equation I = nevA, we obtain the drift velocity v of the electrons as 2.99 x 10^-5 ms^-1

Step 5:

Since the length of the copper wire is 3.0m long, the time taken for an electron to drift from one end of the wire to the other is:

3.0 / 2.99 x 10^-5 = 1.0 x 10⁵ s

Remark:

A sizeable current can still flows even though the drift velocity of electrons in a conductor is very low.

This is because the carrier density (ie. number per unit volume) of electrons in copper is very large.

Imagine one whole column of soldiers marching (30 soldiers side-by-side like in German troop parades). Even though the soldiers (electrons) march slowly, the number of soldiers passing the spectator stand per unit time (current) is considerable.