InterviewSolution
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InterviewSolution
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(a) Define the term conductivity of a metallic wire. Write its SI unit.(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity ofa wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E. |
Answer» Solution : On APPLYING a potential difference V across the ends of a CONDUCTOR of length l, the electric field `E = V/l` Force experienced by an electron F = e E` = (eE)/(l)`in a DIRECTION opposite to that of E. ` therefore ` Acceleration of electron in a direction opposite to that of E will be ` a= F/m = e/m. V/l` If the AVERAGE time between two successive collisions suffered by an electron be `tau` , the drift velocity of electron will be `v_d = a tau= e/m. V/l.tau ` But in terms of drift velocity magnitude of electric current is given by `I = nAev_d = nAe.e/m.V/l.tau = (nAe^2)/(m) tau.V/l` ` therefore ` Conductance `G = I/V = (nAe^2)/(ML) tau` and conductivity`sigma = (Gl)/(A) = ("ne"^2)/(m) tau` ` therefore ` current density `J = I/A = ("ne"^2 tau V)/(ml) = ("ne"^2 tau)/(m) . E = sigma E` In term of vectors , we have`vecJ = sigma vecE` |
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