Question 6 of 8: Self-inductance of a solenoid with a partial magnetic core

Question 6: Self-inductance of a solenoid with a partial 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. A long, thin, tightly wound solenoid whose bore is only partly filled by a magnetic rod of the same cross-section; the rest of the bore is air.

Find. A closed-form expression for the self-inductance $L$ in terms of $N,A,l,d$ and $\mu_0$.

Approach. Use Ampère's law on an axial path to fix $H$, form $B=\mu H$ separately in the cored and air sections, and sum the flux linked by the turns lying over each section; $L=\Lambda/I$ follows.

1. Fix the magnetising field from Ampère's law. For a long thin solenoid the return path outside contributes negligibly, so an axial path of length $l$ closing far outside gives $$\oint\mathbf{H}\cdot d\mathbf{l}=H\,l=NI\;\Rightarrow\;H=\frac{NI}{l},$$ the same value in both sections. This is precisely what the premise $l^{2}\gg A$ licenses: end effects and radial spreading are negligible.
2. Flux density in each section. The constitutive relation differs where the rod sits: $$B_{core}=\mu_0\mu_r H=\mu_0\mu_r\frac{NI}{l},\qquad B_{air}=\mu_0\frac{NI}{l}.$$ The flux through a single turn is $\Phi=BA$ with the same $A$ in both sections, since the rod fills the bore.
3. Count the turns over each section. The winding is uniform with $n=N/l$ turns per unit length, so $Nd/l$ turns lie over the core and $Ns/l$ over the air gap.
4. Sum the flux linkage. $$\Lambda=\frac{Nd}{l}\left(\mu_0\mu_r\frac{NI}{l}A\right)+\frac{Ns}{l}\left(\mu_0\frac{NI}{l}A\right)=\frac{\mu_0AN^{2}I}{l^{2}}\left(\mu_r d+s\right).$$
5. Divide by the current. $$L=\frac{\Lambda}{I}\;\Rightarrow\;\boxed{L=\frac{\mu_0AN^{2}}{l^{2}}\left(s+\mu_r d\right)=\frac{\mu_0AN^{2}}{l^{2}}\left(l+9d\right)}$$ on substituting $\mu_r=10$ and $s=l-d$.
6. Check the limits. With $d\to 0$ this reduces to the textbook air-cored solenoid $L=\mu_0AN^{2}/l$, and with $d\to l$ to the fully filled value $10\,\mu_0AN^{2}/l$. The result can be written as an enhancement factor on the empty solenoid, $L=L_{air}\left(1+9d/l\right)$, which makes the linear dependence on the fill fraction obvious.
7. Numerical illustration. Taking $N=500$, $A=4\ \text{cm}^{2}$, $l=20\ \text{cm}$ (so $l^{2}/A=100$, comfortably satisfying the premise) and $d=8\ \text{cm}$: $L_{air}=0.628\ \text{mH}$ and the enhancement factor is $1+9(0.08)/0.20=4.6$, giving $L=2.89\ \text{mH}$.

The linear dependence on $d$ is characteristic of the open magnetic circuit. Because the solenoid's flux returns through free space rather than through iron, the rod does not have to carry the whole flux, and a long thin rod of modest permeability magnetises essentially to $\mu_0\mu_rH$ with a negligible demagnetising correction. Each turn then links whatever flux crosses it locally, and the total inductance is simply the length-weighted average of the cored and empty values. This is the model that describes real ferrite-rod inductors, and it is the reason a partially inserted slug is used as a practical tuning element: the inductance tracks the insertion depth almost linearly.