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Physics · Ch 12 — Electromagnetic Induction

Mutual Inductance (M)

12.14

Mutual Inductance (M)

Now consider two separate coils placed near each other, coil 1 and coil 2. A steady current I1I_1 flowing in coil 1 sets up a magnetic field B⃗1(x,y,z)\vec{B}_1(x,y,z) throughout the surrounding region, including at coil 2; the resulting flux linked with coil 2's own area s2s_2 is Φ21=∫s2B⃗1⋅da⃗\Phi_{21}=\int_{s_2}\vec{B}_1\cdot d\vec{a}. Since B⃗1\vec{B}_1 (and hence Φ21\Phi_{21}) is proportional to I1I_1 as long as the coils' relative position is fixed, Φ21=M21I1\Phi_{21}=M_{21}I_1, where the proportionality constant M21M_{21} is the mutual inductance of coil 2 WITH RESPECT TO coil 1. If I1I_1 varies (slowly enough that the quasi-static field relation still holds), Φ21\Phi_{21} varies in step, inducing an emf in coil 2: e21=−dΦ21dt=−M21dI1dte_{21}=-\frac{d\Phi_{21}}{dt}=-M_{21}\frac{dI_1}{dt}.

By an identical argument with the roles reversed -- a current I2I_2 in coil 2 producing flux Φ12=M12I2\Phi_{12}=M_{12}I_2 through coil 1, and hence an emf e12=−M12dI2dte_{12}=-M_{12}\frac{dI_2}{dt} in coil 1 -- a remarkable symmetry emerges: M12=M21=MM_{12}=M_{21}=M, regardless of the two coils' shapes, sizes or relative geometry. This lets M be defined equally validly as the flux linked with one circuit per unit current in the OTHER: M=Φ21I1=Φ12I2M=\frac{\Phi_{21}}{I_1}=\frac{\Phi_{12}}{I_2}, or (more commonly, calling the current-carrying coil the PRIMARY and the other the SECONDARY) as the flux Φs\Phi_s linked with the secondary per unit current IpI_p in the primary, giving the standard relation es=MdIpdte_s = M\frac{dI_p}{dt}. The unit of M, like L, is the henry: a mutual inductance of 1 H means a current changing at 1 A/s in the primary induces exactly 1 V in the secondary. (An equivalent, less commonly used, definition treats M as the mutual potential energy W=MI1I2W=MI_1I_2 of the two circuits per unit current flowing in each.)

The extent of coupling between the two coils is captured by the coefficient of coupling K (always ≤1\le 1), related to M and the two self-inductances by M=KL1L2M=K\sqrt{L_1L_2}. K depends on the fraction of coil 1's flux that actually reaches coil 2: coils wound on a common iron core couple almost perfectly (K≈1K\approx1); air-core coils in typical use are tightly coupled if K > 0.5 and loosely coupled if K < 0.5 (K for radio-frequency coils, for instance, is typically only 0.001-0.05); and coupling is minimised (K small) by orienting the coils' axes at right angles and placing them as far apart as possible -- a trick deliberately used in appliances such as clothes dryers, where the heating coils are counter-wound specifically to cancel their combined field and keep any induced emf on the metal casing safely low, since (unlike in a transformer, where a LARGE M is exactly what's wanted) a large stray M is a hazard there. …

Figure 12.13Fig. 12.13: Mutual inductance of two coils
Fig. 12.13 — Fig. 12.13: Mutual inductance of two coils

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Shows two separate coils, coil 1 and coil 2, positioned near each other (e.g. coaxially, one behind the other, or side by side), with coil 1 carrying a current I1I_1 from an external source and coil 2 connected only to a galvanometer or left open. Field lines from coil 1's current are drawn spreading outward, with a portion of them shown passing through (linking) coil 2's own turns, representing the flux Φ21\Phi_{21}. The figure establishes the two-coil geometry that defines mutual inductance: how much of one …

Misc Ex.12.9Example 12.9: Mutual inductance of a wireless (inductive) battery-charging system

Worked out. Models a wireless charger's base unit as a solenoid (coil B) of length l, NBN_B turns and cross-sectional area A carrying current iBi_B, with the handle's coil (coil H, NHN_H turns) fitting coaxially and completely around the base solenoid when docked. The base solenoid's interior field is Bsolenoid=μ0NBliBB_{solenoid}=\mu_0\frac{N_B}{l}i_B, so the flux through each turn of the surrounding handle coil is ΦH=BsolenoidA\Phi_H=B_{solenoid}A, giving flux linkage NHΦHN_H\Phi_H and hence mutual inductance M=μ0NBNHlAM=\mu_0\frac{N_BN_H}{l}A -- the general (symbolic) result this worked example derives, with no numeric values …

Misc Ex.12.10Example 12.10: Mutual inductance of two coupled coils, from their self-inductances and coupling coefficient

Worked out. Two coils with self-inductances L1L_1=75 mH and L2L_2=55 mH are coupled with coefficient of coupling K=0.75. Using M=KL1L2M=K\sqrt{L_1L_2}, the mutual inductance evaluates to M=0.75×75×55=0.75×4125≈0.75×64.24≈48.18M=0.75\times\sqrt{75\times55}=0.75\times\sqrt{4125}\approx0.75\times64.24\approx48.18 mH, matching the example's own printed result. …

Misc Ex.12.11Example 12.11: Coefficient of coupling of two coils, from their mutual and self-inductances

Worked out. Two coils have self-inductances L1L_1=5 H and L2L_2=4 H, and mutual inductance M=1.5 H. Using K=ML1L2K=\frac{M}{\sqrt{L_1L_2}}, the coefficient of coupling evaluates to K=1.55×4=1.520≈1.54.472≈0.335K=\frac{1.5}{\sqrt{5\times4}}=\frac{1.5}{\sqrt{20}}\approx\frac{1.5}{4.472}\approx0.335, i.e. about 33.5%, matching the example's own printed result. …