Inductance and Stored Energy of a Partially Ferromagnetic-Cored Solenoid: 17-Phys-A3 Electromagnetics, May 2016, Q5

Question 5: Inductance and Stored Energy of a Partially Ferromagnetic-Cored Solenoid (equal value)

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 "ferromagnetic material" is characterized by its relative permeability, not permittivity — permittivity has no bearing on magnetic inductance. This is read as a wording slip in the source and the value 20 is used as the core's relative permeability, $\mu_r=20$, which is the only interpretation that makes the stated calculation possible.

Given. Solenoid length $l=10$ cm, $N=500$ turns, coil diameter 5 mm; a 5 cm long, 5 mm diameter ferromagnetic slab ($\mu_r=20$) fills the coil's cross-section over half its length; drive current $I=10$ mA.

Given data
Quantity	Symbol	Value
Solenoid length	$l$	10 cm
Turns	$N$	500
Coil diameter	$d$	5 mm
Core length	$l_{core}$	5 cm
Core relative permeability	$\mu_r$	20
Drive current	$I$	10 mA

Find. Total solenoid inductance $L$ and stored magnetic energy $W_m$ at $I=10$ mA.

Approach. Treat the solenoid as two series sections of equal turn density $n=N/l$, one air-cored (5 cm) and one core-filled (5 cm), each with $L_i=\mu_0\mu_{r,i}n^2 A\,l_i$; add them, then apply $W_m=\tfrac12 LI^2$.

Turn density and cross-sectional area. $$n=\frac{N}{l}=\frac{500}{0.10}=5000\ \text{turns/m},\qquad A=\pi\left(\frac{d}{2}\right)^2=\pi(0.0025)^2=1.9635\times10^{-5}\ \text{m}^2.$$
Inductance of each 5 cm section. Since both sections share the same turn density and area, $L_i=\mu_0\mu_{r,i}n^2A\,l_i$ (using $N_i=n\,l_i$ so $N_i^2/l_i=n^2l_i$): $$L_{air}=\mu_0(1)n^2A(0.05)=4\pi\times10^{-7}(5000)^2(1.9635\times10^{-5})(0.05)=30.84\ \mu\text{H},$$ $$L_{core}=\mu_0(20)n^2A(0.05)=20\times L_{air}\big|_{\mu_r=1\text{ per length}}=616.85\ \mu\text{H}.$$
Total inductance. $$L=L_{air}+L_{core}=30.84+616.85=\boxed{647.7\ \mu\text{H}}.$$
Stored magnetic energy at $I=10$ mA. $$W_m=\tfrac12 LI^2=\tfrac12(647.7\times10^{-6})(0.01)^2=\boxed{3.24\times10^{-8}\ \text{J}}=32.4\ \text{nJ}.$$

Final results
Quantity	Value
Turn density $n$	5000 turns/m
Air-section inductance	30.84 μH
Core-section inductance	616.85 μH
Total inductance $L$	647.7 μH
Stored magnetic energy $W_m$ (at 10 mA)	32.4 nJ

Topic: Solenoid inductance with a partial-length magnetic core.

Key relations: $L=\mu_0\mu_r n^2 A l$ for a uniformly-cored solenoid; a partially cored solenoid is two series inductors, one per material region, each keeping the same turn density and area; $W_m=\tfrac12 LI^2$.

Why this works: With the core diameter matching the coil's, the field inside each 5 cm region is uniform and set only by that region's $\mu_r$, so the two halves act as independent solenoids in series — their inductances simply add, exactly as two inductors in series would.

Common pitfall: Applying $\mu_r=20$ to the whole 10 cm length instead of only the cored half (over-predicting $L$ by roughly 60%), or using $N=500$ directly in each half-length term instead of the correctly-scaled $N_i=n\,l_i=250$.

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

Question 5 of 8: Inductance and Stored Energy of a Partially Ferromagnetic-Cored Solenoid

Question 5: Inductance and Stored Energy of a Partially Ferromagnetic-Cored Solenoid (equal value)