Mutual Inductance of Two Coaxial Solenoids: 04-BS-9, May 2017, Q6

Question 6: Mutual Inductance of Two Coaxial Solenoids (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Outer solenoid radius $r_{\text{out}}$, turns $N_{\text{out}}$	$2$mm, $100$ turns
Inner solenoid radius $r_{\text{in}}$, turns $N_{\text{in}}$	$1$mm, $50$ turns
Length $L$ (both, coaxial)	$5$cm $=0.05$m

Find. The mutual inductance $M$.

Two coaxial solenoids of equal length; the inner one's smaller radius sets the area over which flux is shared.

Approach. Treat both as long (ideal) solenoids. A current in the OUTER solenoid produces a uniform field $B=\mu_0(N_{\text{out}}/L)I$ everywhere inside its bore, including inside the inner solenoid; the flux linking the inner solenoid's $N_{\text{in}}$ turns uses the SMALLER (inner) cross-sectional area, since that is the area actually enclosed by each inner turn.

Field produced by the outer solenoid (per unit current). $$B=\mu_0\frac{N_{\text{out}}}{L}I$$
Flux linkage with the inner solenoid. Each inner turn encloses area $\pi r_{\text{in}}^2$ (the smaller radius, since the field is uniform out to $r_{\text{out}}$ but only $\pi r_{\text{in}}^2$ of that area is actually inside the inner coil): $$M=\frac{N_{\text{in}}\Phi_{\text{in}}}{I}=\frac{\mu_0N_{\text{out}}N_{\text{in}}\pi r_{\text{in}}^2}{L}$$ $$M=\frac{(4\pi\times10^{-7})(100)(50)\pi(10^{-3})^2}{0.05}$$ $$M=\boxed{3.948\times10^{-7}\ \text{H}=394.8\ \text{nH}}$$

Quantity	Result
Mutual inductance $M$	$3.948\times10^{-7}$ H ($394.8$ nH)

Concept & Theory

Topic. Mutual inductance of two coaxial long solenoids of unequal radius (a Helmholtz-style flux-linkage problem, computed via the ideal-solenoid uniform-field approximation).

Key relations. $B_{\text{solenoid}}=\mu_0(N/L)I$ (uniform, ideal/long solenoid); $M=N_2\Phi_2/I_1$; here $M=\mu_0N_{\text{out}}N_{\text{in}}\pi r_{\text{in}}^2/L$ — the SMALLER radius squared appears regardless of which coil carries the current, which is exactly what reciprocity ($M_{12}=M_{21}$) requires.

Why this works. An ideal solenoid's field is uniform inside its own bore and (to the ideal approximation) zero outside it; whichever solenoid carries the current, the flux that actually links the OTHER coil's turns is limited by whichever cross-section is smaller — the field only exists (or only matters) over the smaller of the two areas either way, so both directions of calculation give the identical $r_{\text{in}}^2$ result.

Common pitfall. Using the outer radius $r_{\text{out}}$ (or an average) instead of the inner (smaller) radius in the flux-linkage area, which overstates $M$ by a factor of $(r_{\text{out}}/r_{\text{in}})^2=4$ here.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (mutual inductance, coupled coils).

Question 6 of 8: Mutual Inductance of Two Coaxial Solenoids

Question 6: Mutual Inductance of Two Coaxial Solenoids (20 marks)