Superconductors — Question 5

Question 5

The magnetic field made by a wire can impose its own critical-current limit

For a long round wire of radius a, use the supplied surface-field relation $B_s=\mu_0I/(2\pi a)$. In a simplified type-I model, superconductivity requires $B_s<B_c(T)$, where $B_c(T)=B_{c0}[1-(T/T_c)^2]$. No applied field is present.

1. Find the critical-current boundary as a function of temperature.

2. If radius doubles at fixed temperature, compare the boundary current and its area-averaged density.

Question 5 - Solution

The magnetic field made by a wire can impose its own critical-current limit

(a) Method and calculation

$$\mu_0I_c/(2\pi a)=B_c(T).$$$$I_c(T)=\frac{2\pi aB_{c0}}{\mu_0}[1-(T/T_c)^2].$$

(b) Interpretation and further calculation

$$I_c(2a)/I_c(a)=2,\quad\overline J_c=I_c/(\pi a^2)=2B_c/(\mu_0a).$$$$\overline J_c(2a)/\overline J_c(a)=1/2.$$

Physical check

The self-field constraint scales with radius, not area; real critical-current behavior can involve additional mechanisms.