Perfect diamagnetism in superconductors (Meissner state) For an ideal superconductor in its superconducting state, what is the relationship between magnetisation M and applied magnetic field H (SI units), assuming B = μ0 (H + M) and complete flux expulsion?

Introduction / Context:
In the Meissner state, superconductors expel magnetic flux from the bulk. This perfect diamagnetism is quantified by the constitutive relation linking B, H, and M. Recognizing the correct relation is fundamental in magnet design and cryogenic applications.

Given Data / Assumptions:

• Superconductor is ideal and in the superconducting state (T < Tc, H below critical).
• Bulk region considered far from edges (ignoring small penetration depth).
• SI form: B = μ0 (H + M).

Concept / Approach:
Perfect diamagnetism implies B = 0 in the interior. Setting B = 0 in B = μ0 (H + M) gives H + M = 0 → M = −H. This large negative magnetisation cancels the applied field inside the superconductor, producing flux expulsion.

Step-by-Step Solution:
Write interior condition: B_in = 0.Use relation: 0 = μ0 (H + M).Solve: M = −H.

Verification / Alternative check:
Experimental observations of the Meissner effect (levitation, susceptibility near −1 in SI for bulk) are consistent with M = −H in the ideal limit.

Why Other Options Are Wrong:

• 'Extremely small' or 'zero': contradicts perfect diamagnetism requirement.
• '− I': uses the wrong symbol; I is electric current, not magnetic field intensity.
• Small positive slope: describes weak paramagnetism, not superconductivity.

Common Pitfalls:

• Confusing B = 0 inside with H = 0; it is M that cancels H.
• Neglecting limits (type II mixed state allows flux lines; this question assumes full Meissner state).

Final Answer:
- H