Difference Between Self-Inductance and Mutual Inductance
Physics · Electromagnetic Induction · NEET
Self-inductance (L) is when ONE coil opposes a change in its OWN current, because its own changing flux induces an emf in itself. Mutual inductance (M) is when a changing current in ONE coil induces an emf in a SECOND, nearby coil. Memory hook: SELF = "self-talk" (one coil talks to itself); MUTUAL = "mutual friends" (two coils, one affects the other).
Self-inductance: one coil's changing current induces a back emf in itself (emf = -L dI/dt). Mutual inductance: a changing current in coil 1 induces an emf in a separate coil 2 (emf2 = -M dI1/dt). Both are measured in henry.
Your doubts, answered
Do I need two coils for self-inductance? (This confuses most students.)
No. Self-inductance needs only ONE coil. When the current in that single coil changes, its own magnetic flux changes, and by Faraday's law this induces a 'back emf' in the same coil that opposes the change. The formula is emf = -L (dI/dt), where L depends only on the coil's shape and turns (for a solenoid, L = mu0 n^2 A l). Mutual inductance is the case that needs TWO coils.
How is mutual inductance different from self-inductance in one line?
Self-inductance: a coil reacts to a change in ITS OWN current (emf = -L dI1/dt in coil 1 itself). Mutual inductance: coil 2 reacts to a change in coil 1's current (emf2 = -M dI1/dt). Same idea (changing flux makes emf), but self = same coil, mutual = a different coil.
Are the formulas and units the same for L and M?
Both L and M have the SAME SI unit: the henry (H). Both come from flux linkage / current. Self: L = N*Phi_self / I (flux the coil makes through itself). Mutual: M = N2*Phi_21 / I1 (flux coil 1 makes through coil 2). The unit is identical, so an exam can trick you by mixing them up. Always check: is the flux linked to the SAME coil (L) or a DIFFERENT coil (M)?
Is mutual inductance symmetric? Does M12 = M21?
Yes. Mutual inductance is reciprocal: the emf induced in coil 2 by coil 1 uses the same M as the emf induced in coil 1 by coil 2, so M12 = M21 = M. This is a favourite NEET fact. Self-inductance is a property of one coil alone, so there is no such 'pairing' for L.
Which depends on more coils' turns, L or M?
Self-inductance L depends on N^2 (turns of that one coil squared, e.g. L = mu0 n^2 A l). Mutual inductance M depends on the product of the two coils' turns (M = mu0 n1 n2 A l for two coaxial solenoids). If you double turns of one coil, L quadruples but M only doubles.
⚠️ The NEET trap ✗ Using the formula L = N*Phi/I but plugging in Phi as the flux per turn AND then forgetting that the 'N*Phi' is total flux linkage, OR treating a two-coil problem as self-inductance. ✓ For self-inductance: L = (total flux linkage)/I = N*Phi/I, where N*Phi is turns times flux-per-turn. For the 2016 PYQ: L = (1000 x 4e-3)/4 = 1 H. For mutual inductance you must use the SECOND coil's flux linkage from the FIRST coil's current: M = N2*Phi_21/I1. 🧠 Ask first: 'Is the induced emf in the SAME coil (self, use L) or a DIFFERENT coil (mutual, use M)?' Choosing the wrong one is the #1 NTA trap here.
Real NEET questions
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 x 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 the coil's own TOTAL flux to its own current: L = N*Phi / I. This is a SELF-inductance problem because the flux and the current belong to the SAME coil. Given N = 1000, flux per turn Phi = 4 x 10^-3 Wb, current I = 4 A. Total flux linkage = N*Phi = 1000 x 4 x 10^-3 = 4 Wb. So L = 4 / 4 = 1 H. Answer: D (1 H).
2021
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:
A · R1^2 / R2^2
B · R2^2 / R1 ✓
C · R1 / R2
D · R2 / R1
Solution: This is a MUTUAL-inductance problem (two coils, current in one links flux to the other). Treat the big loop (R1) as the source. Its field at the common centre is B = mu0*I / (2 R1), which is nearly uniform over the tiny inner loop. Flux through the small loop: Phi = B x (area of small loop) = [mu0*I/(2R1)] x pi*R2^2. Then M = Phi / I = mu0*pi*R2^2 / (2 R1). So M is proportional to R2^2 / R1. Answer: B.
Solved Electromagnetic Induction NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the main difference between self-inductance and mutual inductance?
Self-inductance involves ONE coil opposing a change in its own current (its own flux induces its own back emf). Mutual inductance involves TWO coils, where a changing current in the first coil induces an emf in the second coil.
What are the units of self-inductance and mutual inductance?
Both have the same SI unit, the henry (H). 1 henry = 1 weber per ampere = 1 volt-second per ampere.
What are the formulas for L and M?
Self-inductance: emf = -L (dI/dt) and L = N*Phi/I (same coil). Mutual inductance: emf2 = -M (dI1/dt) and M = N2*Phi_21/I1 (second coil's flux from first coil's current). For two coaxial solenoids, M = mu0 n1 n2 A l.
Why is mutual inductance the same both ways (M12 = M21)?
By reciprocity, the flux-linkage-per-current is identical whichever coil is the source. So the coupling constant M is a single number for the pair, and M12 = M21.
Does self-inductance depend on the current?
No. Both L and M depend only on geometry (size, shape, number of turns, and the core material), not on the current. The current only decides how big the induced emf is via dI/dt.
