Question

A small but measurable current of 1.2 × 10–10 A exists in a copper wire whose
diameter is 2.5 mm. Assuming that the current is uniform, calculate (a) the
current density and (b) the electron drift speed. The density of free electrons is
8.5 × 1028 electrons/m3
( Show your solution)

Answer 1

Answer:

A small but measurable current of 1.2×10−10A. exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers per unit volume is 8.49×1028m−3.

Answer 2

Given info : 1.2 × 10¯¹⁰ A current exists in a copper wire whose diameter is 2.5 mm. The density of free electrons in copper wire is 8.5 × 10^28 /m³.

To find : (a) The current density is...

(b) the drift speed of electron is ...

solution : current , i = 1.2 × 10¯¹⁰ A

d = 2.5 mm = 2.5 × 10¯³ m

current density of copper wire, J = i/A

= (i)/(πd²/4)

= (1.2 × 10¯¹⁰ A)/( 3.14 × 6.25 × 10¯⁶/4) A/m²

= 0.245 × 10¯⁴ A/m²

the drift speed of electron, v = J/ne

= (0.245 × 10¯⁴ A/m²)/(8.5 × 10^28 /m³ × 1.6 × 10^-19 C)

= 0.018 × 10¯¹³ m/s

= 1.8 × 10¯¹⁵ m/s

Therefore the drift speed of electron in the copper wire is 1.8 × 10¯¹⁵ m/s.