Inductance of a Solenoid on a Cored Winding: 04-BS-9, December 2013, Q6

Question 6: Inductance of a Solenoid on a Cored Winding (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: a "relative permittivity" has no role in an inductance calculation — only the core's magnetic property (relative permeability $\mu_r$) sets $L$. The value 100 is read here as $\mu_r=100$. The core's overall length (10 cm) exceeds the winding's own length (3 cm); only the wound length enters the standard solenoid formula, since the field is generated by, and confined to, the turns themselves.

Given.

Quantity	Value
Wound (coil) length $l$	$3$ cm $=0.03$ m
Number of turns $N$	$50$
Core diameter	$5$ mm $\Rightarrow$ radius $r=2.5\times10^{-3}$ m
Core relative permeability $\mu_r$	$100$ (see check note)

Find. The self-inductance $L$ of the wound solenoid.

Tightly-wound solenoid on a circular core of relative permeability 100 (permittivity value in the source read as permeability — see check note).

Approach. Use the standard long-solenoid inductance formula $L=\mu_0\mu_r N^2A/l$, with $A$ the core's cross-sectional area and $l$ the coil's own wound length.

Cross-sectional area. $$A=\pi r^2=\pi(2.5\times10^{-3})^2=1.963\times10^{-5}\ \text{m}^2$$
Inductance. $$L=\frac{\mu_0\mu_r N^2 A}{l}=\frac{(4\pi\times10^{-7})(100)(50)^2(1.963\times10^{-5})}{0.03}$$ $$L=\boxed{2.056\times10^{-4}\ \text{H}=0.2056\ \text{mH}}$$

Quantity	Result
Core area $A$	$1.963\times10^{-5}$ m²
Inductance $L$	$0.2056$ mH

Concept & Theory

Topic. Self-inductance of a cored solenoid.

Key relations. $B=\mu_0\mu_r nI$ inside a long solenoid ($n=N/l$); flux linkage $N\Phi=N B A$; inductance $L=N\Phi/I=\mu_0\mu_r N^2A/l$.

Why this works. A magnetic core raises the flux for the same current by the factor $\mu_r$, since $B=\mu_0\mu_r H$ inside the material; because $L$ is proportional to the flux linked per unit current, the core multiplies the air-core inductance by exactly $\mu_r$.

Common pitfall. Using the core's full physical length (10 cm) instead of the coil's wound length (3 cm) in the formula — only the length actually carrying turns generates the confined field the formula assumes.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (Magnetic Forces, Materials and Devices — inductance and inductors).

Question 6 of 8: Inductance of a Solenoid on a Cored Winding

Question 6: Inductance of a Solenoid on a Cored Winding (20 marks)