Mutual Inductance

Inductance: The Geometry Constant

Whether flux through a coil comes from a neighbouring coil's current or its own, one proportionality always holds (for fixed geometry):

$$\Phi_B \propto I$$

For a closely wound coil of $N$ turns, the same flux links every turn, so we use the flux linkage $N\Phi_B$, and write

$$N\Phi_B \propto I$$

The constant of proportionality is called inductance. Like capacitance, it depends only on geometry (and intrinsic material properties such as $\mu_r$) - not on the current.

• Inductance is a scalar.
• Dimensions: $[ML^2T^{-2}A^{-2}]$ (flux linkage over current).
• SI unit: henry (H), honouring Joseph Henry, who discovered induction in the USA independently of Faraday.

Key Point: 1 henry = 1 weber per ampere = 1 volt-second per ampere. Equivalently, a coil has $L = 1\,\text{H}$ if a current changing at 1 A/s induces 1 V in it.

Two flavours follow: mutual inductance (this section) — one coil's changing current induces emf in another; and self-inductance (next section) — a coil induces emf in itself.

Coaxial Solenoids: Calculating M

Take two long coaxial solenoids, each of length $l$: inner solenoid S$_1$ (radius $r_1$, $n_1$ turns/length) and outer S$_2$ (radius $r_2$, $n_2$ turns/length).

Current in the outer, flux in the inner. Pass $I_2$ through S$_2$. Its interior field is $B_2 = \mu_0 n_2 I_2$, and the flux linkage of S$_1$ (total turns $N_1 = n_1 l$, each of area $\pi r_1^2$) is

$$N_1\Phi_1 = (n_1 l)(\pi r_1^2)(\mu_0 n_2 I_2) = MI_2$$

$$\boxed{M = \mu_0 n_1 n_2 \pi r_1^2 l}$$

Current in the inner, flux in the outer. Now pass $I_1$ through S$_1$. Its field is confined inside S$_1$ (long solenoid), threading area $\pi r_1^2$ of each of S$_2$'s $n_2 l$ turns:

$$N_2\Phi_2 = (n_2 l)(\pi r_1^2)(\mu_0 n_1 I_1) = MI_1$$

$$\boxed{M_{12} = M_{21} = M}$$

Note two things worth stressing: (i) it is the smaller radius $r_1$ that appears - only the inner area carries shared flux; (ii) the equality $M_{12} = M_{21}$ (reciprocity) is completely general, and is a free gift in hard problems: if one direction is difficult to compute (flux of a short inner coil through a long outer one), compute the easy direction instead!

The Working Law and the Concentric-Coils Classic

From $N_1\Phi_1 = MI_2$, differentiating when $I_2$ varies with time:

$$\varepsilon_1 = -\frac{d(N_1\Phi_1)}{dt} = -M\frac{dI_2}{dt}$$

A varying current in one coil induces an emf in a neighbouring coil, proportional to the rate of change of current. This is exactly Experiment 3 of Section 1 - the tapping key - now quantified. It is also the heart of the transformer (next chapter).

The concentric-coils result: a small coil of radius $r_1$ sits at the centre of a large coplanar, concentric coil of radius $r_2$ ($r_1 \ll r_2$). Current $I_2$ in the big coil makes a nearly uniform field $B_2 = \frac{\mu_0 I_2}{2r_2}$ over the small coil's area, so

$$\Phi_1 = \pi r_1^2 B_2 = \frac{\mu_0 \pi r_1^2}{2 r_2} I_2 \quad\Rightarrow\quad \boxed{M = \frac{\mu_0 \pi r_1^2}{2 r_2}}$$

[JEE Tip] Notice the reciprocity trick in action: computing the small coil's (non-uniform!) flux through the big coil directly would be very hard - but $M_{21} = M_{12}$ lets us use the easy direction. This exact reasoning (and formula) is a JEE regular.

[NEET Important] $M$ depends on geometry (sizes, turns, separation, orientation) and the medium ($\mu_r$ multiplies it if a magnetic material fills the space) — never on the currents themselves.

Solved Examples

Example 1: Concentric coils

Two concentric, coplanar circular coils have radii $r_1$ and $r_2$ with $r_1 \ll r_2$. Obtain the mutual inductance.

Solution:

1. Let current $I_2$ flow in the outer coil. At its centre, $B_2 = \frac{\mu_0 I_2}{2r_2}$ — effectively uniform over the tiny inner coil.
2. Flux through the inner coil: $\Phi_1 = \pi r_1^2 B_2 = \frac{\mu_0 \pi r_1^2}{2r_2} I_2$.
3. Comparing with $\Phi_1 = M_{12}I_2$: $M = \frac{\mu_0\pi r_1^2}{2r_2}$ (and by reciprocity $M_{21}$ is the same).
4. The approximation is good precisely because $r_1 \ll r_2$.

Example 2: Coaxial solenoids numerical

Two coaxial solenoids of length 1.0 m each have $n_1 = n_2 = 1000$ turns/m; the inner one has radius 2.0 cm. Find $M$.

Solution:

1. Formula: $M = \mu_0 n_1 n_2 \pi r_1^2 l$.
2. $\pi r_1^2 = \pi (0.02)^2 = 1.26 \times 10^{-3}$ m$^2$.
3. $M = 4\pi \times 10^{-7} \times 10^6 \times 1.26 \times 10^{-3} \times 1.0$.
4. Answer: $M \approx 1.6 \times 10^{-3}$ H $= 1.6$ mH.

Example 3: EMF from a changing current

Two coils have mutual inductance 1.5 H. The current in one changes from 0 to 20 A in 0.5 s. Find (a) the change of flux linkage with the other coil, (b) the average induced emf in it.

Solution:

1. (a) $\Delta(N\Phi) = M\,\Delta I = 1.5 \times 20 = 30$ Wb.
2. (b) $\varepsilon = M\frac{\Delta I}{\Delta t} = 1.5 \times \frac{20}{0.5} = 60$ V.
3. The faster the switch-on, the larger the induced emf — the tapping-key effect, quantified.

Example 4: Concentric coils numerical

A small coil of radius 1.0 cm lies at the centre of a large coil of radius 20 cm. Find $M$.

Solution:

1. $M = \frac{\mu_0\pi r_1^2}{2r_2} = \frac{4\pi \times 10^{-7} \times \pi \times (0.01)^2}{2 \times 0.2}$.
2. Numerator: $4\pi \times 10^{-7} \times 3.14 \times 10^{-4} \approx 3.95 \times 10^{-10}$.
3. Answer: $M \approx 9.9 \times 10^{-10}$ H — about 1 nH. Small coils far apart couple weakly.

Example 5: Why reciprocity is a gift

A short coil is placed well inside a very long solenoid. Computing the flux of the short coil's (complicated) field through the long solenoid is hard. How do you find $M$?

Solution:

1. Use $M_{12} = M_{21}$: compute the easy direction — the long solenoid's uniform field $\mu_0 n I$ through the short coil's turns.
2. If the short coil has $N$ turns of area $A$: $M = \mu_0 n N A$.
3. Reciprocity converts an impossible integral into one line — this is exactly the useful trick.

Example 6: Flux linkage from $M$

Two coils have $M = 5$ mH. What flux linkage does a 4 A current in one produce through the other?

Solution:

1. $N_1\Phi_1 = MI_2 = 5 \times 10^{-3} \times 4$.
2. Answer: $2 \times 10^{-2}$ Wb of flux linkage — no emf yet; emf needs this to change.

Example 7: Defining the henry

Define 1 henry of mutual inductance in two equivalent ways.

Solution:

1. Via flux: $M = 1$ H if 1 A in one coil produces a flux linkage of 1 Wb in the other ($M = N\Phi/I$).
2. Via emf: $M = 1$ H if a current changing at 1 A/s in one coil induces 1 V in the other ($\varepsilon = M\,dI/dt$).
3. Both follow from the same equation — Wb/A = V s/A = H.

Example 8: Coil inside a long solenoid [JEE Numerical]

A short 100-turn coil of area $2 \times 10^{-4}$ m$^2$ sits well inside a long solenoid of 1500 turns/m. Find $M$, and the emf in the coil when the solenoid's current changes at 50 A/s.

Solution:

1. Easy direction (reciprocity): the solenoid's field $\mu_0 n I$ threads the coil's $N$ turns: $M = \mu_0 n N A$.
2. $M = 4\pi \times 10^{-7} \times 1500 \times 100 \times 2 \times 10^{-4} \approx 3.8 \times 10^{-5}$ H.
3. $|\varepsilon| = M\frac{dI}{dt} = 3.77 \times 10^{-5} \times 50 \approx 1.9 \times 10^{-3}$ V $\approx 1.9$ mV.

Example 9: Scaling the solenoids

Both turn densities $n_1$ and $n_2$ of a coaxial-solenoid pair are doubled. By what factor does $M$ change?

Solution:

1. $M = \mu_0 n_1 n_2 \pi r_1^2 l \propto n_1 n_2$.
2. Doubling both: $M \to 2 \times 2 \times M = 4M$.
3. Answer: four times. (Doubling only the inner radius would also give 4x — the $r_1^2$ dependence.)

Example 10: The tapping key, quantified

In Experiment 3 (Section 1), the key press raises $I_2$ from 0 to 5 A in 0.01 s; the coil pair has $M = 20$ mH. Find the average emf induced in C$_1$ at make, and explain the zero deflection while the key stays pressed.

Solution:

1. At make: $\varepsilon = M\frac{\Delta I}{\Delta t} = 20 \times 10^{-3} \times \frac{5}{0.01} = 10$ V.
2. Key held: $dI_2/dt = 0$, so $\varepsilon = 0$ — steady current, no induction.
3. At break, the same magnitude appears with opposite polarity (flux collapsing). The 190-year-old experiment is one line of algebra now.