Critical magnetic field and critical current

Critical magnetic field

The critical magnetic field $H_c$ is the minimum magnetic field which, applied to a superconductor at a temperature below $T_c$, destroys its superconductivity and returns it to the normal state.

• It is largest ($H_0$) at 0 K and falls to zero at $T = T_c$.
• The temperature variation is nearly parabolic:

$$H_c(T) = H_0\left[1 - \left(\frac{T}{T_c}\right)^2\right]$$

  Hc
  H0|*
    |    *
    | supercon-  *     normal
    | ducting      *
    |               *
    +----------------*--- T
    0                Tc

• Points under the curve are superconducting; points above it are normal.
• Type-I materials have a single $H_c$ (about 0.01 to 0.2 T). Type-II have $H_{c1}$ (flux starts entering) and $H_{c2}$ (fully normal), with $H_{c2}$ as high as tens of tesla.
• Example: lead has $H_0 \approx 0.08$ T and $T_c = 7.2$ K.

Critical current

The critical current $I_c$ is the maximum current a superconductor can carry without losing superconductivity. A current produces its own magnetic field; when this field at the surface reaches $H_c$, the material turns normal (Silsbee's rule).

For a long straight wire of radius $r$ carrying current $I$, Ampere's law gives the surface field:

$$H \cdot 2\pi r = I \;\Rightarrow\; H = \frac{I}{2\pi r}$$

Setting $H = H_c$:

$$\begin{aligned} I_c &= 2\pi r H_c \\ &= 2\pi r H_0\left[1 - \left(\frac{T}{T_c}\right)^2\right] \end{aligned}$$

The critical current density is

$$J_c = \frac{I_c}{\pi r^2} = \frac{2 H_c}{r}$$

With an external field $H_a$ adding to the self-field, $I_c = 2\pi r (H_c - H_a)$.

So the critical current depends on wire size, temperature and applied field. It falls as $T$ rises towards $T_c$ and becomes zero at $T_c$. A thicker wire carries a larger $I_c$ (though a smaller $J_c$). Practical superconducting cables use many fine filaments in a copper matrix, mainly for stability against flux jumps and quenching.