Finding Self-Inductance from Flux per Turn and Current (PYQ)

Physics · Electromagnetic Induction · NEET

Self-inductance L equals total flux linkage divided by current: L = NΦ/I, where N is the number of turns, Φ is the flux through ONE turn, and I is the current. The unit is the henry (H). Memory hook: "N times one turn's flux, over I" — the flux you are given is per turn, so you must multiply by N first.
Self-Inductance of a Solenoid: L = NΦ / Iflux per turn ΦIN turns (each links Φ)L = N·Φ / Iflux linkage = N × Φ= 1000 × 4×10⁻³ = 4 WbL = 4 / 4 = 1 H
Self-inductance uses the TOTAL flux linkage NΦ (not the per-turn flux Φ) divided by current I. For the NEET 2016 solenoid: N=1000, Φ=4×10⁻³ Wb, I=4 A, so L = (1000×4×10⁻³)/4 = 1 H.

Your doubts, answered

Is the Φ in L = NΦ/I the flux through one turn or the total flux?

The Φ is the flux through ONE single turn. The total flux linked with the whole coil is N × Φ, called the flux linkage. So the correct form is L = (NΦ)/I. In NEET problems the value given as 'flux linked with each turn' is per turn, so you must multiply by N before dividing by I. Skipping the N is the number one mistake.

Why do we multiply the flux by N?

Each of the N turns sees the same flux Φ. Because the turns are in series, their contributions add up, so the coil links a total flux of NΦ. Self-inductance measures total flux linkage per unit current, so we use NΦ, not just Φ.

What is the difference between flux per turn and flux linkage?

Flux per turn (Φ) is what passes through one loop, measured in weber (Wb). Flux linkage (NΦ) is the total for all N turns, measured in weber-turns. L = flux linkage / current = NΦ/I. Always identify which one the question gives you.

Does self-inductance depend on the current?

No. Even though the formula has I in it, L is a fixed property of the coil. It depends only on geometry (number of turns, area, length) and the core material. If you double I, the flux Φ also doubles, so NΦ/I stays the same. L is constant for a given coil.

What are the units in L = NΦ/I?

L is in henry (H), Φ is in weber (Wb), I is in ampere (A). So 1 H = 1 Wb/A = 1 Wb·turn per ampere. If you get an odd number, check that flux is in Wb and current in A (convert mA and mWb first).

⚠️ The NEET trap
Using L = Φ/I with only the flux per turn (forgetting the 1000 turns), giving L = 4×10⁻³/4 = 1×10⁻³ H = 1 mH.
Multiply the per-turn flux by N first: L = NΦ/I = (1000 × 4×10⁻³)/4 = 4/4 = 1 H.
🧠 The question says 'flux linked with EACH turn' — that word 'each' means per turn, so N must still be multiplied in. Answer is 1 H, not 1 mH.

Real NEET questions

NEET 2016

A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is:

A · 4 H
B · 3 H
C · 2 H
D · 1 H
Solution: Self-inductance links TOTAL flux to current: L = NΦ/I, where NΦ is the flux linkage. Step 1: read the data — N = 1000 turns, flux per turn Φ = 4 × 10⁻³ Wb, current I = 4 A. Step 2: find total flux linkage NΦ = 1000 × 4 × 10⁻³ = 4 Wb·turns. Step 3: divide by current, L = NΦ/I = 4 / 4 = 1 H. The trap is forgetting to multiply by N; 'each turn' means the 4 × 10⁻³ Wb is per turn. Correct answer: D (1 H).

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Frequently asked

What is the formula to find self-inductance from flux?

L = NΦ/I, where N is the number of turns, Φ is the magnetic flux through one turn, and I is the current. The product NΦ is the flux linkage.

What is the SI unit of self-inductance?

The henry (H). One henry equals one weber-turn per ampere: 1 H = 1 Wb/A.

Why is the answer 1 H and not 1 mH in the NEET 2016 solenoid problem?

Because you must multiply the per-turn flux (4×10⁻³ Wb) by N = 1000 turns before dividing by I = 4 A. That gives 4/4 = 1 H. Dropping the 1000 turns wrongly gives 1 mH.

Can I use L = NΦ/I for any coil, not just a solenoid?

Yes. L = NΦ/I is the general definition of self-inductance for any coil. The solenoid formula L = μ₀n²Al is just this general formula worked out for a long solenoid's geometry.

Why does L not change when the current changes?

When I increases, the flux Φ increases in exact proportion (Φ ∝ I). So the ratio NΦ/I stays fixed. L depends only on the coil's shape, turns, and core, not on how much current flows.