Physics · Electromagnetic Induction · NEET
Because L is defined as flux linkage per unit current: L = NΦ/I. The flux linkage NΦ is itself proportional to I (since B = μ₀nI grows with I). So when you divide NΦ by I, the current cancels top and bottom. That is the whole point: L is a fixed property of the coil's shape and core, not something that changes when you turn the current up or down. NEET loves options that hide an I in them — reject any answer for L that still contains the current.
It is small n squared, where n = N/l = turns per unit length (unit: per metre). Capital N = total number of turns = n × l. In the derivation you use N = nl in the flux linkage, and B = μ₀nI uses small n. When you combine them you get one nl and one n, which multiply to n²l. So the standard form is L = μ₀n²Al with small n. If a question gives you total turns N instead, use L = μ₀N²A/l (note the divide by l).
For a solenoid wound on a circular tube of radius r, the cross-section area is A = πr². So L = μ₀n²Al simply becomes L = μ₀n²(πr²)l = μ₀πn²r²l. Both are the same equation — one keeps A general, the other plugs in a circle. The ReNEET 2026 answer μ₀πn²r²l is exactly this. Do not confuse r (the coil radius) with l (the coil length); L needs the AREA (∝ r²) times the LENGTH.
No. L depends only on geometry (n, A, l) and the core material (μ₀ for air, or μ₀μᵣ for iron). Changing the current changes the flux and the induced emf, but L stays the same. This is exactly like resistance R staying fixed while V and I change. NEET traps use this: they may vary the current and ask for L — the current is a distractor for finding L.
It is the magnetic field inside a long ideal solenoid, from Ampère's law (Chapter: Moving Charges and Magnetism). Inside a long solenoid the field is uniform and equals μ₀ times turns-per-length times current: B = μ₀nI. You treat this B as constant over the whole cross-section A, so the flux through one turn is Φ = BA. This is why the solenoid must be 'long' — end effects are ignored so B is uniform.
Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ the permeability of free space, the inductance of the solenoid is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
L = μ₀n²Al for an air core, where n is turns per unit length, A is cross-section area, and l is length. With a circular cross-section it becomes L = μ₀πn²r²l, and with total turns N it is L = μ₀N²A/l.
Only on the coil's geometry (turns per length n, area A, length l) and the core material (μ₀ for air, μ₀μᵣ for a magnetic core). It does not depend on the current or the emf.
The henry (H). From L = NΦ/I, 1 H = 1 weber per ampere (Wb/A) = 1 volt-second per ampere (V·s/A).
Replace μ₀ with μ₀μᵣ, where μᵣ is the relative permeability of the core. So L = μ₀μᵣn²Al. Iron (large μᵣ) can raise L by hundreds or thousands of times, which is why inductors and transformers use iron cores.
A long solenoid has a uniform field B = μ₀nI inside and nearly zero outside, so the flux per turn Φ = BA is the same for every turn. For a short solenoid the field is non-uniform at the ends, so this simple formula only approximates.
