A) The magnitude of the current density in the wire; J = 8.163 × 10⁵ A/m²
B) The electric field in the wire is; E = 0.0082 V/m
C) The time required for an electron to travel the length of the wire is; t = 19.91 hours
A) We are given;
Current; I = 3.6 A
Length of side if square; L = 2.1 mm = 2.1 × 10⁻³ m
density of free electrons; n = 8.5 × 10²⁸ m³
Area of square; A = L² = (2.1 × 10⁻³)²
A = 4.41 × 10⁻⁶ m²
Current density; J = I/A
J = 3.6/(4.41 × 10⁻⁶)
J = 8.163 × 10⁵ A/m²
B) The formula for the electric field here is;
E = J
Where;
is resistivity of copper wire = 1.72 × 10⁸ Ω·m
E = 1.72 × 10⁻⁸ × 8.163 × 10⁵
E = 0.0081872
Thus;
E = 0.0082 V/m
C) Let us first calculate the drift velocity from the formula;
v_d = I/nqA
where;
I is current
q is charge on one electron = 1.6 × 10⁻¹⁹ C
n is density of free electrons
Thus;
v_d = 3.6/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ 4.41 × 10⁻⁶)
v_d = 0.00006 m/s
Formula for the time is;
t = L/v_d
where L is length of wire which in this case = 4.3 m
t = 4.3/0.00006
t = 71666.67 s
Converting to hours gives t = 19.91 hours
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