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Exam level5 past questions

Self and Mutual Inductance

Self-inductance is the property of a coil to oppose changes in its own current, while mutual inductance is the ability of one coil to induce EMF in another nearby coil.

Why this shows up in the exam

NEET often asks about inductance calculations and their effects in circuits and transformers.

How NEET tests this

Application · 2 QsNumerical · 2 QsDirect recall

Learn the idea

Self‑inductance is the coil’s own opposition to a change in its current (L = NΦ/I). Mutual inductance is the linkage between two coils (M = N₂Φ₁₂/I₁ = N₁Φ₂₁/I₂); the key insight is that M is found by the flux produced by one coil through the area of the other.

🧠 Memory hook: Big loop blows a gentle magnetic breeze; the tiny loop catches it – so M grows with the tiny area (R₂²) and shrinks with the big radius (R₁).

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • L = NΦ/I, emf = ‑L dI/dt, unit Henry (H)
  • M = N₂Φ₁₂/I₁ = N₁Φ₂₁/I₂, emf in each coil = ‑M dI/dt
  • For a single circular loop, B at its centre = μ₀I/(2R)
  • Flux through a second loop = B·area = μ₀IπR₂²/(2R₁) when R₁≫R₂
  • M = μ₀πR₂²/(2R₁) for concentric loops with R₁≫R₂
  • Coupling coefficient k = M/√(L₁L₂), 0 ≤ k ≤ 1

How to approach it

  1. 1Identify which coil’s field links the other (use the larger loop’s field at the centre)
  2. 2Write the magnetic field at that point using B = μ₀I/(2R) for a circular loop
  3. 3Multiply B by the area of the linked loop to obtain the linked flux Φ
  4. 4Relate Φ to the current of the source coil to get M = Φ/I

Worked example — watch it click

Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ > > R₂ the mutual inductance M between them will be directly proportional to:

  • ✅R₂²/R₁
  • B)R₁/R₂
  • C)R₂/R₁
  • D)R₁²/R₂

The concept behind this problem

The example forces you to use the centre‑field of the large loop and the area of the small loop, directly leading to the proportionality M ∝ R₂²/R₁.

Step by step

  1. 1For R₁ >> R₂, field at center due to outer loop is B ≈ μ₀I/(2R₁).
  2. 2Flux through inner loop is Φ ≈ B·πR₂² = μ₀πR₂²I/(2R₁), so M ∝ R₂²/R₁.

Watch out

Students often invert the ratio and pick R₁/R₂ or R₂/R₁ instead of the correct R₂²/R₁.

Common slip-ups that cost marks

  • •Mixing up self‑inductance with mutual inductance – they have different symbols and formulas
  • •Dropping the π factor or the square on the small radius when writing M
  • •Assuming both loops contribute equally; for R₁≫R₂ only the outer loop’s field matters

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5NEET 2006

Two coils of self inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is: (a) 10 mH (b) 6 mH (c) 4 mH (d) 16 mH

Push further

More challenging

3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 3

An AC circuit consists of a capacitor, a resistor, and an inductor connected in series. A soft iron core is inserted into the inductor. How does this action affect the resonant frequency of the circuit?

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