A lead wire has a critical magnetic field of 6.5103 A/m at 0K. The critical temperature is 7.18K .At what temperature the critical field will drop to 4.5103 A/m?*1 point2.0 K3.98K3.45K2.6K
Question
A lead wire has a critical magnetic field of 6.5103 A/m at 0K. The critical temperature is 7.18K .At what temperature the critical field will drop to 4.5103 A/m?*1 point2.0 K3.98K3.45K2.6K
Solution
The critical magnetic field Hc of a superconductor can be described by the equation:
Hc(T) = Hc(0)[1 - (T/Tc)^2]
where: Hc(T) is the critical magnetic field at temperature T, Hc(0) is the critical magnetic field at 0K, T is the temperature, and Tc is the critical temperature.
We can rearrange this equation to solve for T:
T = Tc * sqrt[1 - (Hc(T)/Hc(0))]
Given that Hc(0) = 6.5103 A/m, Tc = 7.18K, and we want to find T when Hc(T) = 4.5103 A/m, we can substitute these values into the equation:
T = 7.18K * sqrt[1 - (4.5103 A/m / 6.5103 A/m)] T = 7.18K * sqrt[1 - 0.6927] T = 7.18K * sqrt[0.3073] T = 7.18K * 0.5543 T = 3.98K
So, the critical field will drop to 4.5103 A/m at a temperature of approximately 3.98K.
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