Lead has a critical temperature of 7. 2 K and a critical field at 4. 2 K of 430

Question

Lead has a critical temperature of 7. 2 K and a critical field at 4. 2 K of 430 A/cm. (a) What is the critical field at T = 1 K? (b) At 4. 2 K, how much current can a 1-mm-diameter lead wire carry without resistance? (c) If a field of 200 A/cm is applied to a sample of lead at 4. 2 K, what will be the sample magnetization? The relative permeability? The susceptibility? The average magnetic field inside the lead? (d) What will be the values of those four quantities if the applied field is instead. 500 A/cm?

Explanation / Answer

using above formula

Bc=430[1-(4.2/7.2)2]

Bc=283.6805 A/cm

by using above

I=708841.7

B=4*10^-7 M

M=B/4*10^-7

=2.2574*10^8

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