Question

A potential difference of 1.0 V is applied across a conductor of length 5.0 m and area of cross-section 1.0 mm². When current of 4.25 A is passed through the conductor, calculate

1. the drift speed and
2. relaxation time, of electrons.

(Given number density of electrons in the conductor, n = 8.5 × 10²⁸ m⁻³).

Answer 1

V = 1.0 V, I = 5 m

A = 1.0 mm² = 1 × 10⁻⁶ m²

n = 8.5 × 10²⁸

I = 4.25 A

i. l = V_denA

`V_d = I/(enA)`

= `4.25/(1.6 xx 10^-19 xx 8.5 xx 10^28 xx 1 xx 10^-6)`

= `4.25/13.6 xx 10^-3`

= 0.313 × 10⁻³

= 3.13 × 10⁻³

ii. `V_d = (-eE)/mtau, E = V/l`

`V_d = (-eV)/(ml)tau`

`tau = (V_dml)/(eV)`

= `(3.13 xx 10^-3 xx 9.1 xx 10^-31 xx 5)/(1.6 xx 10^-19 xx 1.0)`

= `(142.415 xx 10^-34)/(1.6 xx 10^-20)`

τ = 89 × 10⁻¹⁴ s

τ = 8.9 × 10⁻¹⁵ s