Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Mutual Inductance Between Two Long Co-axial Solenoids

4.3.4

Mutual Inductance Between Two Long Co-axial Solenoids

Working through the specific geometry of two long, co-axial solenoids of the same length l, with cross-sectional areas $A_1>A_2$ (solenoid 1 the larger, outer one; solenoid 2 the smaller, inner one) and turn densities $n_1$ and $n_2$ , gives an explicit formula for their mutual inductance and confirms $M_{12}=M_{21}$ for this concrete case. Current $i_1$ in solenoid 1 produces field $B_1=\mu_0 n_1 i_1$ ; because solenoid 2 sits INSIDE solenoid 1, the full field $B_1$ threads solenoid 2's own (smaller) cross-sectional area $A_2$ , giving flux per turn $\Phi_{21}=B_1A_2=\mu_0n_1i_1A_2$ and total flux linkage $N_2\Phi_{21}=(n_2l)(\mu_0n_1i_1A_2)=\mu_0n_1n_2A_2l\,i_1$ . Comparing with $N_2\Phi_{21}=M_{21}i_1$ gives

M_{21} = \mu_0 n_1 n_2 A_2 l \qquad (4.13)

Working the reverse direction: current $i_2$ in solenoid 2 produces field $B_2=\mu_0n_2i_2$ , uniform inside solenoid 2 but essentially zero outside it, so for solenoid 1 the EFFECTIVE overlap area is still the smaller area $A_2$ (not $A_1$ ), even though solenoid 1's own physical cross-section is larger. This gives flux per turn of solenoid 1, $\Phi_{12}=B_2A_2=\mu_0n_2i_2A_2$ , and total flux linkage $N_1\Phi_{12}=(n_1l)(\mu_0n_2i_2A_2)=\mu_0n_1n_2A_2l\,i_2$ , giving

M_{12} = \mu_0 n_1 n_2 A_2 l \qquad (4.14)

Comparing equations (4.13) and (4.14) directly confirms

M_{12}=M_{21}=M \qquad (4.15)

so the general mutual inductance of two long coaxial solenoids is

$M = \mu_0 n_1 n_2 A_2 l \qquad (4.16)$ …

Figure 4.22Mutual inductance of two long co-axial solenoids

What this figure shows. Two solenoids of the same length l share a common axis, with 'Solenoid 1 of $N_1$ turns' shown as the larger-diameter outer winding (cross-sectional area $A_1$ ) and 'Solenoid 2 of $N_2$ turns' shown as the smaller-diameter inner winding (cross-sectional area $A_2 < A_1$ ), nested coaxially one inside the other. The figure sets up the key geometric fact the derivation relies on: because solenoid 2's field is essentially confined to its own (smaller) cross-section, only the smaller area $A_2$ is the EFFECTIVE overlap area for flux linkage between the two solenoids in either direction, regardless of how much larger $A_1$ is -- this is exactly why the final mutual inductance formula …

Misc Example 4.13Mutual inductance of two coplanar coaxial circular coils

Worked out. Two coplanar, coaxial circular coils A (radius 20 cm, 200 turns) and B (radius 2 cm, 1000 turns) are placed with a common axis, and the mutual inductance, along with the induced emf in B and the rate of change of flux through B, are required when the current in A changes from 2 A to 6 A in 0.04 s. Because coil B is much smaller and sits at the centre of coil A, the field of A at that centre, $B_A=\mu_0 N_A i_A/(2r_A)$ , can be treated as uniform over B's small area; the flux linkage of B due to A works out to $N_B\Phi_{BA}\approx7.89\times10^{-4}\,i_A$ Wb-turns, giving mutual inductance $M\approx7.89\times10^{-4}$ H. With $di_A=4$ A over $dt=0.04$ s, the induced emf is $\varepsilon_B = M\,di_A/dt = (7.89\times10^{-4})(4/0.04) \approx 78.9$ mV, and since $\varepsilon_B$ IS $d(N_B\Phi_B)/dt$ by definition, the rate of change of flux linkage through B is likewise 78.9 mWb/s. The problem shows the same c …

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