A current of 10A exists in a wire of cross-sectional area of 5 mm² with a drift velocity of 2 × 10⁻³ ms⁻¹. The number of free electrons in each cubic meter of the wire is ___________.

Solution

Physics sir · 4 steps

Use I = neAv_d and solve for n, after turning the area into square metres.

1. The link. Current is the charge that drifts past a cross-section each second. With n free electrons per cubic metre, each carrying charge e, that gives the relation below.

   I = neAv_d ⇒ n = I/eAv_d

2. Convert the area. 1 mm² = 10⁻⁶ m², so A = 5 × 10⁻⁶ m².
3. Put in numbers. The denominator is eAv_d = 1.6 × 10⁻¹⁹ × 5 × 10⁻⁶ × 2 × 10⁻³ = 1.6 × 10⁻²⁷. So n = 10/(1.6 × 10⁻²⁷) = 6.25 × 10²⁷ m⁻³.
4. So. 6.25 × 10²⁷ = 625 × 10²⁵ free electrons per cubic metre, which is option (a).

The idea to keep. Whenever current, area and drift speed appear together, reach for I = neAv_d. Put every quantity in SI units first, especially areas given in mm².

Answer from the NTA final answer key. Solution by PrepWiser.