Inductance of a Solenoid with a Half-Inserted Magnetic Core: 04-BS-9, December 2018, Q2

Question 2: Inductance of a Solenoid with a Half-Inserted Magnetic Core (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
Solenoid: turns $N$, length $L$, cross-section $A$ (long-solenoid limit, $A/L^2\ll1$)	base inductance $L_0=\mu_0N^2A/L$
Core: same cross-section $A$, same length $L$, relative permeability $\mu_r$	$\mu_r=20$
Core placement	length $L/2$ inside the solenoid, $L/2$ protruding outside

Find. The new inductance of the solenoid with the core partially inserted, in terms of the base formula $L_0$.

Side view: the solenoid winding spans length L (left rectangle); the magnetic core (red band, also length L) is inserted so only its right half (L/2) lies inside the winding, the other half protruding beyond the coil's right end.

Approach. Treat the winding's length as two magnetic reluctances in series along the axis — a core-filled half and an air-filled half — since the long-solenoid limit makes $B$ essentially uniform and axial, forcing $B$ (hence $\Phi/A$) to be continuous across the core/air interface while $H=B/\mu$ differs.

Two reluctances in series. Over the winding's length $L$, one half ($L/2$) is filled with the core and the other half ($L/2$) is air: $$R_{\text{core}}=\frac{L/2}{\mu_0\mu_rA},\qquad R_{\text{air}}=\frac{L/2}{\mu_0A}$$
Inductance from total reluctance. Ampere's law around the full winding gives $NI=\Phi(R_{\text{core}}+R_{\text{air}})$, so $$L_{\text{new}}=\frac{N\Phi}{I}=\frac{N^2}{R_{\text{core}}+R_{\text{air}}}=\frac{N^2\mu_0A}{(L/2)(1+1/\mu_r)}=\frac{2N^2\mu_0A\,\mu_r}{L(\mu_r+1)}$$
Express as a multiple of the base formula $L_0=\mu_0N^2A/L$. $$\frac{L_{\text{new}}}{L_0}=\frac{2\mu_r}{\mu_r+1}=\frac{2(20)}{21}=\frac{40}{21}$$ $$\boxed{L_{\text{new}}=\frac{40}{21}\,\frac{\mu_0N^2A}{L}\approx1.905\,L_0}$$

Quantity	Result
Reluctance ratio $R_{\text{core}}:R_{\text{air}}$	$1:20$ (core has $1/\mu_r$ the reluctance of air)
$L_{\text{new}}/L_0$	$40/21\approx1.905$
$L_{\text{new}}$	$\dfrac{40}{21}\,\mu_0N^2A/L$

Concept & Theory

Topic. Magnetic-circuit (reluctance) analysis of a partially-filled long solenoid.

Key relations. Reluctance $R=\text{length}/(\mu A)$; series reluctances add like series resistances; Ampère's law around the full winding gives $NI=\sum H\cdot(\text{length})$.

Why this works. In the long-solenoid limit the interior field is uniform and purely axial, and the core/air interface is perpendicular to that field, so $B$ (hence flux $\Phi=BA$) must be continuous across it while $H=B/\mu$ is not — exactly the condition a series-reluctance (magnetic-circuit) model captures, the direct magnetic analogue of resistors in series.

Common pitfall. Averaging the permeability ($\mu_{r,\text{avg}}=(1+20)/2=10.5$) and scaling $L_0$ by that — reluctances in series add like resistors, so the low-permeability (air) half dominates the total reluctance far more than a simple average would suggest.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (inductance, magnetic circuits); Hayt & Buck, Ch. 9.

Question 2 of 8: Inductance of a Solenoid with a Half-Inserted Magnetic Core

Question 2: Inductance of a Solenoid with a Half-Inserted Magnetic Core (20 marks)