Magnetization in a two-state dipole model A material contains N magnetic dipoles per m^3. If Np dipoles per m^3 are aligned parallel to the applied field and Na dipoles per m^3 are aligned anti-parallel, what is the magnetization M (A·m per m^3) of the material?

Introduction / Context:
Magnetization M is the magnetic dipole moment per unit volume. In simple paramagnetic or diamagnetic models with two preferred orientations (parallel and anti-parallel to an applied field), M can be computed directly from the net alignment of microscopic dipoles. This question reinforces that direct counting idea.

Given Data / Assumptions:

• Total dipole number density: N dipoles per m^3.
• Np aligned parallel to the field; Na aligned anti-parallel.
• Each dipole has the same magnetic moment magnitude m0 (A·m^2).
• Only the z-component along the field matters for M (sign handled by direction).

Concept / Approach:

By definition, magnetization M = (vector sum of dipole moments) / volume. If Np dipoles contribute +m0 each and Na contribute −m0 each along the field, the net dipole moment per unit volume is (Np − Na) * m0. Hence M equals that difference times the single-dipole moment magnitude. The remaining fraction (N − Np − Na) is either zero or oriented randomly giving negligible average along the field in this simplified two-state picture.

Step-by-Step Solution:

Verification / Alternative check:

When Np = Na, M = 0 (no net magnetization), which matches physical intuition. When all dipoles are parallel (Na = 0), M = N * m0, the saturation value in this idealized model.

Why Other Options Are Wrong:

(b) ignores opposing alignment; (c) gives saturation regardless of actual alignment; (d) is merely the negative of the correct expression and would reverse the direction convention.

Common Pitfalls:

Forgetting to include the sign of anti-parallel dipoles or assuming an average projection of m0 when the model explicitly counts only two orientations.

Final Answer:

M = (Np - Na) * m0