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Strong magnetic field and critical current

Short answer

A strong magnetic field destroys superconductivity; the critical field depends on temperature as Hc(T)=H0[1−(T/Tc)²]. For a wire of radius r, Ampere's law gives H=I/(2πr). Setting H=Hc yields critical current Ic=2πrHc and current density Jc=2Hc/r. With an external field Ha, Ic=2πr(Hc−Ha).

How strong magnetic fields effect the superconductor? Derive the relation of critical current in superconductor with necessary diagram.

A superconductor stays superconducting only while the magnetic field at every point is below the critical field HcH_c. A strong field destroys superconductivity, and the critical field depends on temperature:

Hc(T)=H0[1−(TTc)2]H_c(T) = H_0\left[1 - \left(\frac{T}{T_c}\right)^2\right]

where H0H_0 is the critical field at 0 K.

Effect of a strong magnetic field

  • For H<HcH < H_c: flux is expelled (Meissner effect), resistance is zero.
  • For H≥HcH \ge H_c (type-I): flux penetrates completely and the material becomes normal, with normal resistance.
  • For type-II: flux starts entering at Hc1H_{c1} (mixed state), and superconductivity is fully lost at Hc2H_{c2}.
  • A stronger field means a lower temperature is needed to stay superconducting, so TcT_c falls as HH rises.
  H
 H0 |*
    |   *
    |  Super-   *      Normal
    |  conducting  *
    |                *
    +-----------------*---- T
    0                 Tc

Critical current (Silsbee's rule)

A current in the wire itself produces a magnetic field. If this self-field at the surface reaches HcH_c, superconductivity is destroyed. The current at which this happens is the critical current IcI_c.

Consider a long straight superconducting wire of radius rr carrying current II:

        H (circles around wire)
      .-~~~~~-.
     /  .---.  \
    |  ( I ⊙ )  |   radius r
     \  '---'  /
      '-~~~~~-'

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

∮H dl=I  ⇒  H⋅2πr=I  ⇒  H=I2πr\oint H\,dl = I \;\Rightarrow\; H \cdot 2\pi r = I \;\Rightarrow\; H = \frac{I}{2\pi r}

Superconductivity is lost when H=HcH = H_c, so

Ic=2πrHc=2πrH0[1−(TTc)2]\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

Jc=Icπr2=2HcrJ_c = \frac{I_c}{\pi r^2} = \frac{2H_c}{r}

If an external field HaH_a is also applied (along the same direction as the self-field at the surface), the total must stay below HcH_c:

Ha+I2πr≤Hc  ⇒  Ic=2πr (Hc−Ha)H_a + \frac{I}{2\pi r} \le H_c \;\Rightarrow\; I_c = 2\pi r\,(H_c - H_a)

So IcI_c falls linearly as the applied field increases, and falls as temperature rises. This sets the maximum current a superconducting cable or magnet winding can carry.

✓ Answer checked by Claude Opus 5.5 (high effort)

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