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Sagot :
Answer:
Explanation:
Given:
diameter of the wire, [tex]d=3.9~mm[/tex]
current in the wire, [tex]i=1.3~mA[/tex]
a)
Current density:
[tex]\o=\frac{i}{\pi.d^2/4}[/tex]
[tex]\o=\frac{1.3}{\pi\times 3.9^2/4}[/tex]
[tex]\o=0.109~mA/mm^2[/tex]
b)
Given drift velocity, [tex]v_d=1\times 10^{-2} ~cm/s[/tex]
From the formula:
[tex]v_d=\frac{i}{n.e.A}[/tex]
where:
n = charge density (here the charge carriers are electron)
q = quantity of charge on a carrier
A = cross-sectional area of the conductor
[tex]n=\frac{i}{v_d.q.\pi d^2/4}[/tex]
[tex]n=\frac{1.3\times 10^{-3}}{1\times10^{-2}\times(1.6\times 10^{-19})\times\pi\times0.39^2/4 }[/tex]
[tex]n=6.80\times 10^{18}~cm^{-3}[/tex]
c)
Given conductivity of wire, [tex]G=5.5\times 10^{6}~\Omega^{-1}.m^{-1}[/tex]
Using formula of average time between collision:
[tex]\tau=\frac{m.G}{q^2n}[/tex]
here:
m = mass of the a carrier
[tex]\tau=\frac{9.11\times 10^{-31}\times 5.5\times 10^6}{(1.6\times 10^{-19})^2\times 6.80\times 10^{12}}[/tex]
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