What Is Mutual Inductance
Take two coils placed close to one another. Send a current $I_2$ through the second coil. The magnetic field it produces threads the first coil and sets up a flux there. NCERT denotes the flux through the first coil by $\Phi_1$, so that the total flux linkage with the first coil — flux per turn times the number of turns $N_1$ — is proportional to the current that produced it:
$$N_1 \Phi_1 = M_{12}\, I_2$$
The constant $M_{12}$ is the mutual inductance of coil 1 with respect to coil 2, also called the coefficient of mutual induction. It depends only on how the two coils are built and placed, never on the size of the current. The figure below shows the coupling: current in the outer (driving) coil, flux spilling through the inner (pickup) coil.
Current $I_2$ in coil 2 generates a field whose flux $\Phi_1$ threads coil 1. The flux linkage $N_1\Phi_1$ is proportional to $I_2$, with proportionality constant $M_{12}$.
The emf Induced Through Coupling
Mutual inductance becomes physically visible only when the current changes. NCERT recalls Experiment 6.3, in which an emf appeared in coil C1 whenever the current in coil C2 changed. Starting from $N_1\Phi_1 = M I_2$ and differentiating with respect to time, then applying Faraday's law $\varepsilon_1 = -\dfrac{d(N_1\Phi_1)}{dt}$, gives the working equation:
$$\varepsilon_1 = -\,M\,\frac{dI_2}{dt}$$
A varying current in one coil induces an emf in the neighbouring coil. The magnitude of that emf depends on two things only — the rate of change of current and the mutual inductance. The negative sign is the signature of Lenz's law: the induced emf opposes the change of flux that caused it. A steady current ($dI/dt = 0$) induces nothing.
A pair of adjacent coils has a mutual inductance of $1.5\ \text{H}$. If the current in one coil changes from $0$ to $20\ \text{A}$ in $0.5\ \text{s}$, what is the change of flux linkage with the other coil? (NCERT)
The flux linkage with the second coil is $N_2\Phi_2 = M\,I_1$. The change in linkage is therefore
$$\Delta(N_2\Phi_2) = M\,\Delta I_1 = 1.5 \times (20 - 0) = 30\ \text{Wb}.$$
The flux linkage changes by $30\ \text{Wb}$ (weber-turns). Dividing by the time, the average induced emf would be $30 / 0.5 = 60\ \text{V}$, though the question asks only for the change in linkage.
M of Two Coaxial Solenoids
The standard model NCERT solves is two long coaxial solenoids of equal length $l$. The inner solenoid S1 has radius $r_1$ and $n_1$ turns per unit length; the outer solenoid S2 has radius $r_2$ and $n_2$ turns per unit length. Pass a current $I_2$ through S2. Inside a long solenoid the field is uniform, $B = \mu_0 n_2 I_2$, so the flux linking the $n_1 l$ turns of S1 is $N_1\Phi_1 = (n_1 l)(\mu_0 n_2 I_2)(\pi r_1^2)$. Comparing with $N_1\Phi_1 = M_{12}I_2$:
$$M = M_{12} = \mu_0\, n_1 n_2\, \pi r_1^{\,2}\, l = \mu_0\, n_1 n_2\, A\, l$$
where $A = \pi r_1^2$ is the cross-sectional area of the inner solenoid — the field outside it is taken as negligible, so the area that matters is the inner one. The derivation neglects edge effects, valid because the solenoids are long ($l \gg r_2$).
Two long coaxial solenoids of common length $l$. The mutual inductance uses the inner area $A = \pi r_1^2$, giving $M = \mu_0 n_1 n_2 \pi r_1^2 l$.
If the space between the windings is filled with a magnetic medium of relative permeability $\mu_r$ instead of air, every flux value scales up by $\mu_r$, so the mutual inductance becomes $M = \mu_r\,\mu_0\, n_1 n_2\, \pi r_1^2\, l$. This is exactly why transformers use iron cores — a large $\mu_r$ multiplies the coupling between primary and secondary.
Mutual inductance is the two-coil cousin of a single coil's self-inductance — same henry, same opposition to change, but flux from the coil's own current. Read that first if $L$ is unfamiliar.
Reciprocity: M12 = M21
Now reverse the roles. Pass a current $I_1$ through the inner solenoid S1 and ask for the flux linkage it produces in S2, defined by $N_2\Phi_2 = M_{21}I_1$. Because the flux of a long solenoid is confined essentially inside it, the same algebra runs again and yields $M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l$ — an identical expression. Hence:
$$M_{12} = M_{21} = M$$
NCERT stresses that this reciprocity is far more general than the solenoid example used to demonstrate it: for any pair of coils there is a single mutual inductance, independent of which one carries the current. The practical value is computational. If the inner solenoid is much shorter than the outer one, $M_{12}$ is easy to find (the short coil sits in a uniform field) while $M_{21}$ would be extremely hard (the inner coil's field varies across the long outer coil). The equality lets you compute the easy one and assert the other.
$M_{12} = M_{21}$, but $M$ still depends on geometry and coupling
Students misread reciprocity two ways. First, they think $M_{12} = M_{21}$ means the two coils are "interchangeable" — they are not; only the single number $M$ is shared. Second, they assume $M$ is a fixed material constant. It is not: $M$ changes with the number of turns, the cross-sectional area, the length, the medium ($\mu_r$), and crucially the separation and relative orientation of the coils. Slide the coils apart or tilt one, and $M$ falls even though $M_{12} = M_{21}$ stays true.
$M_{12} = M_{21}$ is always true. $M$ itself is set by geometry, medium and coupling — never by the current.
What M Depends On
Collecting the dependencies from NCERT into one place clarifies which knob moves $M$ in a problem. The current is conspicuously absent — $M$ is a structural property of the coil pair.
| Factor | Effect on M | Source relation |
|---|---|---|
| Turns per unit length $n_1, n_2$ | $M \propto n_1 n_2$ | M = μ₀ n₁ n₂ π r₁² l |
| Inner area $A = \pi r_1^2$ | $M \propto A$ | Flux uses inner area |
| Common length $l$ | $M \propto l$ | More linked turns |
| Magnetic medium $\mu_r$ | $M \propto \mu_r$ | M = μ_r μ₀ n₁ n₂ π r₁² l |
| Separation & orientation | $M$ falls as coupling weakens | NCERT: depends on separation and relative orientation |
| The current $I$ | No effect | $M$ is structural |
Two concentric circular coils, one of small radius $r_1$ and the other of large radius $r_2$, with $r_1 \ll r_2$, are placed coaxially with centres coinciding. Find the mutual inductance. (NCERT Example 6.8)
Let current $I_2$ flow in the large coil. The field at its centre is $B_2 = \dfrac{\mu_0 I_2}{2 r_2}$. Since $r_1 \ll r_2$, this field is nearly uniform over the small coil, so the flux through it is
$$\Phi_1 = \pi r_1^2 B_2 = \frac{\mu_0 \pi r_1^2}{2 r_2}\, I_2 = M_{12}\, I_2.$$
Therefore $M = \dfrac{\mu_0 \pi r_1^2}{2 r_2}$, and by reciprocity $M_{12} = M_{21}$. This single-turn result is the one the NEET 2021 PYQ below tests.
The Henry
Inductance is a scalar with dimensions $[\text{M L}^2\text{T}^{-2}\text{A}^{-2}]$ — flux divided by current. Its SI unit is the henry (H), named for Joseph Henry, who discovered electromagnetic induction in the USA independently of Faraday in England. From $N\Phi = MI$ one henry equals one weber per ampere; equivalently, from $\varepsilon = -M\,dI/dt$, a coil pair has $M = 1\ \text{H}$ if a current changing at $1\ \text{A s}^{-1}$ induces $1\ \text{V}$.
Mutual Inductance in One Screen
- Definition: $N_1\Phi_1 = M_{12} I_2$ — flux linkage in one coil per unit current in the other.
- Induced emf: $\varepsilon = -M\,\dfrac{dI}{dt}$; magnitude set by $M$ and the rate of change of current.
- Coaxial solenoids: $M = \mu_0 n_1 n_2 A l$, using the inner area $A = \pi r_1^2$; with a core, multiply by $\mu_r$.
- Reciprocity: $M_{12} = M_{21} = M$ for any coil pair.
- Concentric loops: $M = \dfrac{\mu_0 \pi r_1^2}{2 r_2}$, so $M \propto r_1^2 / r_2$.
- Depends on: turns, area, length, medium, separation, orientation — never the current. Unit: henry.