Critical current decreases linearly with field

The critical current $I_c$ is the maximum current a superconductor can carry while remaining superconducting. Beyond it, the magnetic field produced by the current (plus any applied field) exceeds the critical field $H_c$ and the material becomes normal (Silsbee's rule).

Field produced by the current

Take a long superconducting wire of radius $r$ carrying current $I$.

         H_a (applied)
     --------------->
       .-~~~~~~-.
      /  .----.  \   H_I = I/(2πr)
     |  (  I ⊙ ) |   circles round
      \  '----'  /
       '-~~~~~~-'
     --------------->

By Ampere's circuital law, the self-field at the surface is

$$\oint H\,dl = I \;\Rightarrow\; H_I = \frac{I}{2\pi r}$$

Without an applied field, superconductivity is lost when $H_I = H_c$:

$$I_{c0} = 2\pi r H_c$$

With an applied field

Let a uniform external field $H_a$ act on the wire. At the point of the surface where the self-field and applied field point the same way, the total field is the sum:

$$H_{total} = H_a + \frac{I}{2\pi r}$$

The wire stays superconducting only while $H_{total} \le H_c$. The critical condition is

$$\begin{aligned} H_a + \frac{I_c}{2\pi r} &= H_c \\ I_c &= 2\pi r\,(H_c - H_a) \\ I_c &= I_{c0} - 2\pi r\,H_a \end{aligned}$$

This is a straight line in $H_a$ with negative slope $-2\pi r$. So the critical current decreases linearly with the applied field:

• at $H_a = 0$: $I_c = 2\pi r H_c$ (maximum),
• at $H_a = H_c$: $I_c = 0$.

  Ic
 2πrHc|\
      |  \
      |    \
      |      \
      +--------\---- Ha
      0        Hc

Since $H_c = H_0[1 - (T/T_c)^2]$, the whole line moves down as temperature rises. Critical current density is $J_c = I_c/\pi r^2 = 2(H_c - H_a)/r$.