A 1 m long copper wire carries a current of 1 A. If the cross section of the wire is 2.0 mm² and the resistivity of copper is 1.7 × 10⁻⁸ Ω m, the force experienced by moving electron in the wire is ____________ × 10⁻²³ N.
(charge on electorn =1.6 × 10⁻¹⁹ C)

Solution

Physics sir · 3 steps

Find the electric field inside the wire from E = ρ J, then the force on each electron is just eE.

1. Field from current. The field that pushes the electrons is set by the current density: E = ρ J = ρ I/A. This comes from J = σ E with σ = 1/ρ.
2. Put in numbers. Convert the area first: A = 2.0 mm² = 2.0 × 10⁻⁶ m². Then E = (1.7 × 10⁻⁸ × 1)/(2.0 × 10⁻⁶) = 8.5 × 10⁻³ V m⁻¹.

   F = eE = 1.6 × 10⁻¹⁹ × 8.5 × 10⁻³ = 1.36 × 10⁻²¹ N

3. So. F = 136 × 10⁻²³ N, so the answer is 136.

The common slip: 0.136. Writing 2.0 mm² as 2.0 × 10⁻³ m² instead of 2.0 × 10⁻⁶ m² makes the field 1000 times too small and gives 0.136, not 136.

The idea to keep. Force on a drifting electron always means F = eE. Get E from E = ρ I/A, or from V/L if you find the voltage first. Both give the same field.

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