Mean drift velocity: (a) mean drift velocity of charge carriers

The idea

In a conductor, charge carriers move randomly with high speeds but with no net displacement until an electric field is applied. The field gives each carrier a small average velocity along the conductor, called the mean drift velocity v. The current is I = nAvq, where n is the number density of charge carriers, A is the cross-sectional area, v is the mean drift velocity and q is the charge on each carrier. For copper, n is about 8.5 × 10²⁸ m⁻³, so a current of 1 A in a wire of cross-sectional area 1 × 10⁻⁶ m² gives v = I ÷ (nAq) ≈ 7 × 10⁻⁵ m s⁻¹, a fraction of a millimetre per second. The random thermal speeds are much larger, around 10⁵ m s⁻¹, but average to zero.

What earns the marks

• Mean drift velocity is the average velocity of charge carriers along a conductor due to an applied electric field.
• The relationship is I = nAvq, where n is the number density of charge carriers, A is the cross-sectional area, v is the mean drift velocity and q is the charge on each carrier.
• Carriers also have much larger random thermal speeds, but these average to zero and do not contribute to the current.
• For a given current, a smaller cross-sectional area or a smaller number density gives a larger mean drift velocity.
• Typical values of mean drift velocity in metals are of the order of 10⁻⁵ to 10⁻⁴ m s⁻¹, far smaller than the random thermal speeds.

Common mistakes and how to put them right

• Confusing mean drift velocity with the speed at which the electrical signal travels.

  Put it right: The signal propagates much faster than the carriers drift.

• Using the cross-sectional area in cm² without converting to m².

  Put it right: The equation requires SI units, so convert 1 cm² to 1 × 10⁻⁴ m².

• Forgetting that n is the number density of charge carriers per cubic metre, not the total number of carriers in the wire.

Exam tips

• Rearrange I = nAvq for the quantity required before substituting values, and check that all units are SI.
• Remember that for electrons q = 1.60 × 10⁻¹⁹ C, and use the magnitude when calculating v.
• If a question gives the number of carriers per unit volume, use it directly as n; if it gives the total number, divide by the volume of the conductor.