A Cu wire, 1m long with a cross-sectional area=(1mm) 2 is placed in a circuit.(

Question

A Cu wire, 1m long with a cross-sectional area=(1mm)2 is placed in a circuit.( assume the wire connecting the battery has no resistance). Ohm’s law measurements are done at T1=10ok and T2=273K. At T1, for V1=1.55*10-3 volts, I=1amp. At T2, for V1=7.75*10-6 volts, I=1amp

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Calculate the resistivity of Cu wire at T1 and T2, PT1Cu and PT2Cu.

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Assume that 1e- per cu atom contributes to the conductivity and that nAtomicCu=8.47*1028/m3.

Calculate the Drude relaxation time T at T1 and T2.

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Calculate the electron mobility at T1 and T2.

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Calculate the electric field and drift velocities at T1 and T2.

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Calculate the e- thermal (classical) velocity (3kbT/me)1/2 at T1 and T2.

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Calculate the mean free path lmfp between collisions at T1 and T2.

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We will see that when we include pauli principle that

Nwith paulithermal = 10* Nclassical(no pauli)thermal at t=273K and the value is approximately independent of Te. Recalculate lmfp

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If L1=5 m( instead of 1m), would ohm’s law be observed? Why or why not?

Explanation / Answer

Given data:
length ,l = 1 meter
area ,a = 1 mm^2 = 10^-6 m^2.

T1 = 10 K --> V1 = 1.55*10^-3 volt, I1 = 1 Amp --> R1 = 1.55*10^-3 Ohm

T2 = 273 K --> V2 = 7.75*10^-6 volt, I2 = 1 Amp --> R2 = 7.75*10^-6 Ohm

(1) The resistivity,
= ( Resistor * Area )/ ( length)

The resistivity at T1 = (R1 * a) / l
        = (1.55*10^-3 *10^-6)/ (1)
        = 1.55*10^-9 Ohm meter
       
The resistivity at T2 = (R2 * a) / l
        = (7.75*10^-6 *10^-6)/ (1)
        = 7.55*10^-12 Ohm meter
       

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