Physics · Magnetism And Matter · NEET
No. B is the total magnetic field inside the material, and it includes the material's own contribution. H is only the part set by the free current you pass (the coil current). NCERT splits the field: B = μ₀(H + M), where μ₀H is the coil's part and μ₀M is the material's part. So H ≠ B; they even have different units (H in A/m, B in tesla).
H is measured in ampere per metre (A/m), the same unit as magnetisation M. This makes sense from H = nI: n is turns per metre (per metre) and I is current (ampere), so nI has units of A/m. Do not write H in tesla — tesla is only for B.
For a long solenoid (or the core inside it), H = nI, where n is the number of turns per unit length and I is the current. This is the direct NCERT example: 1000 turns/m carrying 2 A gives H = nI = 2000 A/m. Notice H does NOT contain μ_r or χ — H is decided purely by the current setup, not by the material inside the coil.
No, and this is the key idea NEET tests. H = nI depends only on how you wound the coil and how much current flows. If you slide in an iron core, H stays the same for the same current — but B and M shoot up because the material responds. That is why H is called the 'magnetising field': it is the cause, and M is the material's response (χ = M/H).
Start from B = μ₀(H + M). Rearranging gives H = B/μ₀ − M. Also M = χH, so B = μ₀(H + χH) = μ₀(1 + χ)H = μ₀μ_r H. So once you know H and the material's χ (or μ_r), you can get both M and B. H is the input, M is the material's answer, B is the total.
An iron rod of susceptibility 599 is subjected to a magnetising field (magnetic intensity) of 1200 A m⁻¹. The permeability of the material of the rod is (μ₀ = 4π × 10⁻⁷ T m A⁻¹):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Because B mixes two very different sources: your coil current and the material's own magnetisation. H isolates the part you control (the free current). This makes problems easy: you fix H with the coil, then the material adds M = χH, and B = μ₀(H + M). Splitting the field this way is why NCERT introduces H.
Roughly yes. H is the 'magnetising field' set by the free current (the cause). The material responds with magnetisation M = χH, and the total field B = μ₀(H + M) is the result. For NEET, treat H as the input and B, M as outputs.
Since H = nI has units A/m, its dimensions are [A L⁻¹], i.e. current per length. The same as magnetisation M. Both are A/m, which is why H = B/μ₀ − M is dimensionally consistent.
H depends only on the free current arrangement, so H = nI holds whether the core is air or iron. B does depend on the core: B = μ₀μ_r H = μ₀μ_r nI. In vacuum (μ_r = 1), B = μ₀nI, which is the familiar solenoid field.
