Solution;
Given length l ,cross section area a,charge density n and drift velocity Vd;
Total charge is given by;
Q=neAl
e is the charge of a single carrier.
Time taken by charge carriers to move accros the length;
t="\\frac{l}{V_d}"
Current flowing in the conductor is,
I="\\frac Qt=\\frac{neAl}{\\frac{l}{V_d}}=neAV_d"
Voltage across the conductor;
V=IR
Resistance of the conductor is;
R="\\frac VI=\\frac{V}{neAV_d}"
But resistivity of a conductor is given by;
"\\rho=\\frac{RA}{l}=\\frac{V}{neAV_d}\u00d7\\frac Al=\\frac{V}{neV_dl}" Conductivity of the conductor,"\\sigma" is;
"\\sigma=\\frac{1}{\\rho}=\\frac{neV_dl}{V}"
n=charge density
e=electron charge
Vd=drift velocity
l=length of conductor
V=voltage across the conductor.
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