Physics · Magnetism And Matter · NEET
No. H is the field YOU apply from outside (from a solenoid current), and it does not care what material is inside. B is the TOTAL field actually present inside the material, which includes the extra field the material itself creates. In vacuum there is no material to add anything, so B = μ₀H. Inside iron the material adds a huge amount, so B can be hundreds of times larger than μ₀H.
H and M are both 'how many amperes of current-effect per metre'. H comes from the real coil current, M comes from the atomic (bound) currents inside the material. Since they are the same kind of quantity, they add directly: (H + M). Multiplying by μ₀ (unit T·m/A) converts that sum into the actual magnetic field B in tesla. So B = μ₀(H + M): amperes-per-metre times T·m/A gives tesla.
H stays the same as long as the coil current is the same, because H = nI for a solenoid depends only on turns and current, not the core. What changes is M (the iron gets strongly magnetised) and therefore B jumps up. This is exactly why NEET writes B = μ₀(H + M): same H, big M, big B.
Only in vacuum or air (where M ≈ 0). Inside any material you must use the full form B = μ₀(H + M). Using M = χH, this becomes B = μ₀(1 + χ)H = μ₀μ_r H = μH. So B = μH is the general shortcut; B = μ₀H is just the vacuum special case where μ = μ₀.
M is the net magnetic dipole moment per unit volume of the material: M = m_net / V. When atomic dipoles line up with the applied field, the material becomes a tiny magnet itself; M measures how strong that induced magnetism is per cubic metre. For diamagnets M points opposite to H (so χ negative); for para/ferro it points along H (χ positive).
An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A/m. The permeability of the material of the rod is (μ₀ = 4π × 10⁻⁷ T·m/A):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B = μ₀(H + M), where B is total field (tesla), H is applied magnetic intensity (A/m), M is magnetisation (A/m), and μ₀ = 4π × 10⁻⁷ T·m/A.
M depends strongly on the material (and B follows), while H depends only on the free current in the coil, not on the core material.
Since M = χH, B = μ₀(1 + χ)H = μ₀μ_r H = μH. This connects B and H directly through permeability.
B is in tesla (T), while both H and M are in ampere per metre (A/m).
In vacuum M = 0, so B = μ₀H. This is why vacuum permeability μ₀ appears — it is the B-to-H ratio when no material is present.
