The number density of free electrons in copper is nearly 8 × 10²⁸ m⁻³. A copper wire has its area of cross section =2 × 10⁻⁶ m² and is carrying a current of 3.2 A. The drift speed of the electrons is ___________ × 10⁻⁶ ms⁻¹

Solution

Physics sir · 4 steps

Current is just charge density times area times drift speed, so turn I = neAv_d around and solve for v_d.

1. Link current and drift. In time t, the electrons in a length v_d t of wire pass a cross section. That is nAv_d t electrons carrying charge neAv_d t, so the current is I = neAv_d.

   v_d = I/neA

2. Find the denominator. neA = (8 × 10²⁸) × (1.6 × 10⁻¹⁹) × (2 × 10⁻⁶) = 25.6 × 10³ C m⁻¹.
3. Divide. v_d = 3.2/(25.6 × 10³) = 0.125 × 10⁻³ m s⁻¹ = 1.25 × 10⁻⁴ m s⁻¹.
4. So. Written in the form the blank asks for, v_d = 125 × 10⁻⁶ m s⁻¹, so the answer is 125.

The common slip: 1.25. Stopping at 1.25 × 10⁻⁴ and entering the coefficient gives 1.25, not 125. Convert to × 10⁻⁶ before you fill the blank.

The idea to keep. Drift speed in a metal is tiny, about a tenth of a millimetre per second. Use v_d = I/(neA), then rewrite your answer in exactly the power of ten the blank asks for.

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