Inductance and stored energy of a partly cored solenoid: 22-Elec-A7 Electromagnetics, May 2016, Q5

Question 5: Inductance and stored energy of a partly cored solenoid (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. A long, thin, uniformly wound solenoid is half filled by a ferromagnetic rod that exactly fills the bore over its own length and leaves the remaining half empty. The stated “relative permittivity 20” is the paper’s slip for relative permeability: permittivity has no bearing on the inductance of a magnetic circuit, and a ferromagnetic insert is specified by its $\mu_r$.

Given data
Quantity	Symbol	Value
Solenoid length	$l$	$10\ \text{cm}$
Number of turns	$N$	$500$
Bore diameter	$D$	$5\ \text{mm}$
Rod length	$d$	$5\ \text{cm}$
Empty length	$s = l - d$	$5\ \text{cm}$
Relative permeability of the rod	$\mu_r$	$20$
Winding current	$I$	$10\ \text{mA}$

Find. The inductance of the part-cored solenoid and the magnetic energy it stores at the stated current.

Figure 5.1 — the solenoid with the rod inserted over half its length. The winding is uniform, so half the turns link the cored section and half the empty one.

Approach. Use Ampère’s law to fix $H$ along the bore, compute the different flux densities in the cored and empty sections, add the flux linked by the turns over each section, and divide by the current.

Find the bore area. $$A = \frac{\pi D^{2}}{4} = \frac{\pi(5\times10^{-3})^{2}}{4} = 1.9635\times10^{-5}\ \text{m}^{2}$$
Fix the magnetic field intensity. For a long solenoid Ampère’s law gives a field that depends only on the current sheet, not on what fills the bore: $$H = \frac{NI}{l} = \frac{500 \times 0.010}{0.10} = 50.0\ \text{A/m}$$
Compute the two flux densities. The same $H$ produces very different $B$ in the two sections: $$B_{air} = \mu_0 H = 62.83\ \mu\text{T}, \qquad B_{core} = \mu_r\mu_0 H = 1.2566\ \text{mT}$$
Add the flux linkages. The winding is uniform, so the cored length carries $Nd/l$ turns and the empty length $Ns/l$. Summing $\Lambda = \sum N_i \Phi_i$ and dividing by $I$: $$L = \frac{\mu_0 A N^{2}\,(s + \mu_r d)}{l^{2}}$$
Substitute. With $s + \mu_r d = 0.05 + 20(0.05) = 1.05\ \text{m}$, $$L = \frac{(4\pi\times10^{-7})(1.9635\times10^{-5})(500)^{2}(1.05)}{(0.10)^{2}} = \boxed{0.6477\ \text{mH}}$$
Store the energy. $$W = \tfrac{1}{2}LI^{2} = \tfrac{1}{2}(6.477\times10^{-4})(0.010)^{2} = \boxed{32.38\ \text{nJ}}$$

A compact way to see the same result is to write it against the empty solenoid. The bare coil would have $L_{air} = \mu_0 A N^{2}/l = 61.69\ \mu\text{H}$, and the rod multiplies this by $1 + (\mu_r - 1)d/l = 1 + 19(0.5) = 10.5$, giving the same $0.6477\ \text{mH}$. Written this way the model has an immediate physical reading: inductance grows linearly with how far the rod is pushed in, which is exactly the behaviour of a ferrite slug tuner. The two limits are also correct by inspection — at $d = 0$ it returns $L_{air}$ and at $d = l$ it returns $\mu_r L_{air}$.

The energy can be confirmed independently by integrating the field energy density $\tfrac{1}{2}BH$ over the two sections: $\tfrac{1}{2}A(s B_{air} + d B_{core})H = 32.38\ \text{nJ}$, identical to $\tfrac{1}{2}LI^{2}$. Roughly 95 % of that energy sits inside the rod, which is the point of fitting one.

Check: a partly filled magnetic circuit admits two defensible models and they disagree substantially. The uniform-$H$ model used above treats the solenoid as an open, leaky structure whose return path is through air, so Ampère’s law holds $H = NI/l$ everywhere and the sections add in parallel as flux contributors. The alternative series-reluctance model treats the bore as a closed circuit with no radial leakage, giving $L = \mu_0AN^{2}/(s + d/\mu_r) = 0.1175\ \text{mH}$ — a factor of 5.5 smaller. Both reduce correctly at $d = 0$ and $d = l$, so the limits do not discriminate. The uniform-$H$ model is chosen here because the solenoid is long and thin with a modest $\mu_r$, the demagnetising factor of such a rod is small, and the measured behaviour of slug-tuned coils is the linear one. The paper’s note 1 invites exactly this kind of stated assumption.

Final results
Quantity	Value
Bore area	$A = 19.635\ \text{mm}^{2}$
Field intensity	$H = 50.0\ \text{A/m}$
Flux density, empty section	$B_{air} = 62.83\ \mu\text{T}$
Flux density, cored section	$B_{core} = 1.2566\ \text{mT}$
Inductance without the rod	$L_{air} = 61.69\ \mu\text{H}$
Inductance with the rod	$\mathbf{L = 0.6477\ \text{mH}}$
Stored magnetic energy at 10 mA	$\mathbf{W = 32.38\ \text{nJ}}$
Alternative series-reluctance value (see callout)	$0.1175\ \text{mH}$

Topic. Inductance of a solenoid with a partial magnetic core, and the distinction between the field intensity $H$, which the current alone fixes, and the flux density $B$, which the material sets.

Key relations. $H = NI/l$ for a long solenoid; $B = \mu_r\mu_0 H$; $L = \Lambda/I$ with $\Lambda = \sum N_i\Phi_i$; $L = \mu_0AN^{2}(s + \mu_r d)/l^{2} = L_{air}[1 + (\mu_r-1)d/l]$; $W = \tfrac{1}{2}LI^{2} = \tfrac{1}{2}\int BH\,dV$.

Why this works. Ampère’s law relates $H$ to the enclosed current and to nothing else, so inserting a rod does not change $H$ in an open solenoid — it changes $B$, and therefore the flux, and therefore the linkage. Because the winding is uniform, each section's contribution is weighted by the number of turns lying over it, which is what produces the $1/l^{2}$ signature: one factor of $l$ comes from $H = NI/l$ and the second from the turn density $N/l$.

Common pitfall. Reading “relative permittivity” literally and hunting for a capacitance. The other trap is applying the series-reluctance formula by reflex: it is the right model for a closed toroidal or gapped core, but for an open rod in air it badly understates the inductance because the flux is not forced to follow the bore.

Source. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed., ch. 11 (transmission lines) and ch. 12 (waveguides). W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed., ch. 10–11 (transmission lines, uniform plane waves).

Question 5 of 8: Inductance and stored energy of a partly cored solenoid

Question 5: Inductance and stored energy of a partly cored solenoid (20 marks)