Physics · Electromagnetic Induction · NEET
The magnetic field of a long solenoid is uniform INSIDE it and almost zero outside. So when the outer solenoid S₂ carries current, its field B = μ₀n₂I₂ fills the whole tube. But the inner solenoid S₁ only occupies the area π r₁². Field outside S₁'s turns exists but does not pass THROUGH S₁'s loops. Flux is field times the area the loops actually enclose, so we use π r₁². Rule: mutual flux is limited by the SMALLER coil's area.
n₁ and n₂ are turns per unit length (turns/metre) of the two solenoids. Over the shared length l, solenoid S₁ has N₁ = n₁ l total turns. Each of those N₁ turns catches the same flux Φ₁, so total flux linkage is N₁Φ₁ = (n₁ l)(B₂)(π r₁²). The l comes from counting how many turns of S₁ are linked. That is why M grows with length.
Definition: N₁Φ₁ = M₁₂ I₂. Plug in N₁Φ₁ = (n₁ l)(μ₀ n₂ I₂)(π r₁²). Divide both sides by I₂ and I₂ cancels: M₁₂ = μ₀ n₁ n₂ π r₁² l. Notice current cancels — M depends only on geometry (turns, radius, length), never on the current value.
Drive the inner one instead: its field μ₀n₁I₁ fills only π r₁² (its own area). The outer solenoid's N₂ = n₂l turns each link exactly this flux (field outside S₁ is zero, so extra outer area adds nothing). You get M₂₁ = μ₀ n₁ n₂ π r₁² l — identical. This equality is called reciprocity; NEET expects you to know M₁₂ = M₂₁ = M.
A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per metre. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π² ohm, the total charge flowing through the coil during this time is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
M = μ₀ n₁ n₂ π r₁² l, where μ₀ = 4π×10⁻⁷ T·m/A, n₁ and n₂ are turns per metre, r₁ is the inner solenoid radius, and l is the common length. With a magnetic core of relative permeability μ_r, multiply by μ_r: M = μ₀ μ_r n₁ n₂ π r₁² l.
No. When you derive M, the current I cancels out. M depends only on geometry — number of turns per metre, the inner radius, the length, and the medium. This is why M is a fixed property of the coil pair, measured in henry (H).
They look similar but differ. Self-inductance of a solenoid is L = μ₀ n² π r² l (one solenoid, so one n squared). Mutual inductance uses n₁ n₂ (two different solenoids) and the inner radius r₁. See difference-self-and-mutual-inductance for a full comparison.
For an ideal long solenoid (length ≫ radius), the field is strong and uniform inside and essentially zero outside. This lets us treat B₂ = μ₀ n₂ I₂ as constant over the whole inner solenoid and ignore edge effects, giving a clean flux = B × area calculation.
