A current of 2 A flows through a wire of cross-sectional area 25.0 mm². The number of free electrons in a cubic meter are 2.0 × 10²⁸. The drift velocity of the electrons is __________ × 10⁻⁶ ms⁻¹ (given, charge on electron =1.6 × 10⁻¹⁹ C ).

Solution

Physics sir · 4 steps

Use I = neAv_d, and convert the area from mm² to m² before you put it in.

1. Current and drift. In time t, every free electron in a length v_d t of wire crosses a section. That charge is neAv_d t, so the current is I = neAv_d.

   v_d = I/neA

2. Convert the area. 1 mm² = 10⁻⁶ m², so A = 25 × 10⁻⁶ m².
3. Put in numbers. neA = 2.0 × 10²⁸ × 1.6 × 10⁻¹⁹ × 25 × 10⁻⁶ = 8 × 10⁴ C m⁻¹. Then v_d = 2/(8 × 10⁴) = 2.5 × 10⁻⁵ m s⁻¹.
4. So. v_d = 25 × 10⁻⁶ m s⁻¹, so the blank is 25.

The common slip: 0.025. Treating 1 mm² as 10⁻³ m² makes the area 1000 times too large and gives 0.025, not 25.

The idea to keep. Drift velocity comes straight from I = neAv_d. Put every quantity in SI units first, especially an area given in mm², which is 10⁻⁶ m².

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