Solenoid Inductance With a Partial Magnetic Slug: 04-BS-9, May 2014, Q6

Question 6: Solenoid Inductance With a Partial Magnetic Slug (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.

Check: the paper prints the air-core inductance as "$6.28\times10^6$ H," an implausible exponent for a small air-core coil. Recomputing from the given $N$, length and area with the standard formula gives $L=6.283\times10^{-6}$ H — matching the value to 4 significant figures with a flipped sign on the exponent, confirming $10^{-6}$ H is the intended reading, used throughout below.

Given.

Quantity	Value
Turns $N$	$100$
Solenoid length $l$	$5$ cm $=0.05$ m
Cross-section $A$	$0.25\ \text{cm}^2=2.5\times10^{-5}\ \text{m}^2$
Air-core inductance $L_{\text{air}}$ (given/verified)	$6.283\times10^{-6}$ H
Slug: $\mu_r=10$, length $l_s=2$ cm, same cross-section	inserted into the core

Find. The new inductance $L$ with the slug in place.

100-turn air-core solenoid, 5 cm long, with a 2 cm magnetic slug (μr=10) inserted over part of its length.

Approach. The flux density $B$ (not $H$) must be uniform along the axis (same cross-section, same flux path), so apply Ampere's law with two series reluctance segments — one air, one slug — sharing the same winding and current.

Ampere's law with two segments in series. With $H_{\text{air}}=B/\mu_0$ over the remaining air length $l-l_s=0.03$ m and $H_{\text{slug}}=B/(\mu_0\mu_r)$ over $l_s=0.02$ m: $$NI=H_{\text{air}}(l-l_s)+H_{\text{slug}}\,l_s=\frac{B}{\mu_0}\left[(l-l_s)+\frac{l_s}{\mu_r}\right]$$
Effective magnetic length and new inductance. This is exactly the air-core formula with an effective length $$l_{\text{eff}}=(l-l_s)+\frac{l_s}{\mu_r}=0.03+\frac{0.02}{10}=0.032\ \text{m}$$ so, since $L\propto 1/l$ for fixed $N,A,\mu_0$: $$L_{\text{new}}=L_{\text{air}}\cdot\frac{l}{l_{\text{eff}}}=(6.283\times10^{-6})\left(\frac{0.05}{0.032}\right)$$ $$L_{\text{new}}=\boxed{9.82\times10^{-6}\ \text{H}}$$

Quantity	Result
Effective length $l_{\text{eff}}$	$0.032$ m
$L_{\text{new}}$	$9.82\times10^{-6}$ H

Concept & Theory

Topic. Magnetic circuits with series reluctances (partial-core solenoid).

Key relations. $L=\mu_0N^2A/l$ (air core); for a composite core, reluctances in series add: $\mathcal{R}=\sum l_i/\mu_i A$, giving an effective length $l_{\text{eff}}=\sum l_i/\mu_{r,i}$.

Why this works. The flux density $B$ must be the same everywhere along the axis (one continuous flux tube, same cross-section, no leakage), so $H$ differs between the air and slug regions in inverse proportion to their permeabilities; Ampere's law over the whole winding then behaves exactly like an equivalent single-material solenoid whose length is shortened by the slug's $1/\mu_r$ factor.

Common pitfall. Scaling the whole inductance by $\mu_r$ (as if the entire core were filled with the magnetic material) — here the slug fills only $2$ of the $5$ cm, so the enhancement is much more modest.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (Magnetic circuits, reluctance, inductance).

Question 6 of 8: Solenoid Inductance With a Partial Magnetic Slug

Question 6: Solenoid Inductance With a Partial Magnetic Slug (20 marks)