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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 . A strong field destroys superconductivity, and the critical field depends on temperature:
where is the critical field at 0 K.
Effect of a strong magnetic field
- For : flux is expelled (Meissner effect), resistance is zero.
- For (type-I): flux penetrates completely and the material becomes normal, with normal resistance.
- For type-II: flux starts entering at (mixed state), and superconductivity is fully lost at .
- A stronger field means a lower temperature is needed to stay superconducting, so falls as 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 , superconductivity is destroyed. The current at which this happens is the critical current .
Consider a long straight superconducting wire of radius carrying current :
H (circles around wire)
.-~~~~~-.
/ .---. \
| ( I ⊙ ) | radius r
\ '---' /
'-~~~~~-'
By Ampere's circuital law, the field at the surface is
Superconductivity is lost when , so
The critical current density is
If an external field is also applied (along the same direction as the self-field at the surface), the total must stay below :
So 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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