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The superconducting transition temperature of lead is 7.26K. If the initial field at 0 K is 64 × 103 amp/m, then the critical field at 5 K is:
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- 0.3364 × 103 amp/m
- 0.364 × 103 amp/m
- 33.640 × 103 amp/m
- 336.400 × 103 amp/m
Correct Option: C
Field intensity at temperature T is given by relation
H (T) = H0 | ![]() | 1 – | ![]() | ||
TC2 |
where, H (T) = Field intensity at temperature T
TC = Critical temperature/transition temperature
H0 = Field intensity at zero degree kelvin
Given TC = 7·26° K
T=5° K
H0 = 64 × 103 amp/m
H (T) =?
Now, H (T) = 64 × 103 | ![]() | 1 – | ![]() | ![]() | 2 | ![]() | |
7.26 |
= 33.64 × 103 amp/m