Most of you may be aware of the fact that the drift speed of electrons in a current carrying conductor is very small compared to the speed with which a current flows in a circuit. The drift speed is typically of the order of millimeter per second while the speed with which a current flows in a circuit is nearly equal to the speed of light. Understand that the electron need not move from one end of the conductor to the other end for a current to flow. What happens is that an electric field is established in each region of a conductor (at nearly the speed of light) and the electrons in the region drift under its influence.

Now, consider the following MCQ:

**A current of 4.4 A is flowing in a copper wire of radius 1 mm. Density of copper is 9****×10 ^{3} kg m^{–3} and its atomic mass is 63.5 u. **

**If every atom of copper contributes one conduction electron then the drift velocity of electrons is nearly**

**(a) 0.1 mm s**^{–1} **(b) 0.5 mm s**^{–1} **(c) 1 mm s**^{–1} ** (d) 1.5 mm s**^{–1} ** (e) 5 mm s**^{–1} ** **** **** **

The current in the wire is given by

I = nAve where ‘n’ is the number of conduction electrons per unit volume, ‘A’ is the area of cross section of the wire, ‘v’ is the drift velocity of the electrons and ‘e’ is the electronic charge.

From this equation, v = I/nAe.

Since the atomic mass of copper is given as 63.5 g, the number of copper atoms in 0.0635 kg is 6.02 ×10^{23}. The density of copper being 9×10^{3} kg m^{–3}, unit volume of copper contains (9×10^{3}/0.0635) ×6.02×10^{23} = 8.5×10^{28} atoms.

Therefore, number of conduction electrons per unit volume, n = 8.5×10^{28} (since each atom contributes one conduction electron).

Therefore, drift velocity, v = I/nAe = 4.4/[8.5×10^{28}×π(1×10^{–3})^{2}×1.6×10^{–19}] = 0.1×10^{–3} ms^{–1}(nearly).

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper:

**When a current I flows through a wire, the drift velocity of the electrons is ‘v’. When current 2I flows through another wire of the same material having double the length and double the area of cross section, the drift velocity of the electrons will be**

**(a) v/8 (b) v/4 (c) v/2 (d) v (e) 2v**

The drift velocity is given by v = I/nAe.

Since the wires are of the same material, the number density (n) of conduction electrons is the same. The electronic charge (e) is a constant. The current and area of cross section are doubled. Therefore, the drift velocity is *unchanged*. ** **

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