Magnetic Classification of Substances

μᵣ = 1 + χₘ; for diamagnetic, χₘ < 0 so μᵣ < 1

Step 1: Starting from B = μ₀(H + M) and M = χₘ H, we get B = μ₀(1 + χₘ)H. Step 2: Since B = μH = μᵣμ₀H, comparing gives μᵣ = 1 + χₘ. Step 3: Diamagnetic substances are magnetised opposite to the applied field, so M is antiparallel to H, giving χₘ = M/H < 0. Step 4: Therefore μᵣ = 1 + (negative number) < 1 for diamagnetics. Wrong options: Option B uses a minus sign which is incorrect; Option C gives a completely wrong formula; Option D correctly states the formula but incorrectly assigns χₘ > 0 to diamagnetics.

Energy dissipated as heat per unit volume per cycle of magnetisation

Step 1: During each cycle of magnetisation, the domains in the ferromagnetic material undergo repeated realignment, consuming energy. Step 2: When the external field is removed, domain walls do not return to original positions (hysteresis effect), so the energy input during magnetisation is not fully recovered. Step 3: The unrecovered energy is released as heat — this is called hysteresis loss. Step 4: Mathematically, the area enclosed by the B-H (or M-H) hysteresis loop equals the energy lost per unit volume per complete magnetisation cycle. Wrong options: Total magnetic flux has different un

χ₁ is in ferromagnetic region (undefined by Curie-Weiss); χ₂ = C/(600 − 500) = C/100

Step 1: The Curie-Weiss law χₘ = C/(T − Tᶜ) applies only above the Curie temperature (T > Tᶜ). Step 2: At T = 300 K < Tᶜ = 500 K, the substance is in the ferromagnetic region. Curie-Weiss law does NOT apply here. χ₁ cannot be calculated using this law. Step 3: At T = 600 K > Tᶜ = 500 K, the substance behaves paramagnetically and Curie-Weiss law applies: χ₂ = C/(600 − 500) = C/100. Step 4: Therefore, only χ₂ is calculable; the ratio χ₁/χ₂ is not simply obtainable. This is a conceptually tricky question — the key point is the domain of validity of Curie-Weiss law.

P for transformer core; Q for permanent magnet

Step 1: Transformer core requirements — the material undergoes continuous cyclic magnetisation, so it must have low hysteresis loss (narrow loop), low coercivity (easy to magnetise/demagnetise), and high permeability. Material P (narrow loop, low coercivity) fits perfectly. Step 2: Permanent magnet requirements — the material must retain magnetism strongly (high retentivity) and resist demagnetisation by external fields (high coercivity). Material Q (wide loop, high coercivity) fits perfectly. Step 3: The confusion arises because P has high retentivity, but its low coercivity makes it unsuitab