Given,
d = 1.6 mm
I = 130 mA
n = 5.8 * 10^28 m^-3
d = 1.0 cm = 0.01 m
Area of the wire,
A = (1/4)*π*d^2
A = (1/4)*π*(1.6 * 10^-3)^2
A = 2.01 * 10^-6
Ww know,
I = nqA(Vd)
130 * 10^-3 = 5.8 * 10^28 * 1.6 * 10^-19 * 2.01 * 10^-6 * Vd
Vd = 6.97 * 10^-6 m/s
t = d/Vd
t = 0.01/(6.97 * 10^-6)
t = 1434.72 s
t = 23.9 min
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