Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Self-Inductance of a Long Solenoid

4.3.2

Self-Inductance of a Long Solenoid

Applying the general definition of self-inductance to the concrete geometry of a long solenoid of length l, cross-sectional area A, and turn density n (turns per unit length) gives an explicit formula for L. Passing current i through the solenoid produces a nearly uniform field inside it, $B=\mu_0 n i$ , directed along the axis; the flux through each single turn is $\Phi_B=BA=\mu_0 n i A$ (since the field is along the normal, $\theta=0$ ), and the total flux linkage of all $N=nl$ turns is $N\Phi_B = (nl)(\mu_0 n i A) = \mu_0 n^2 A l\, i$ . Comparing with $N\Phi_B=Li$ from equation (4.8) gives the self-inductance of a long solenoid,

L = \mu_0 n^2 A l \qquad (4.10)

showing that L depends purely on the solenoid's own GEOMETRY (turn density n, cross-sectional area A, length l) and on the medium filling it -- if the core is a dielectric or magnetic material of relative permeability $\mu_r$ , this becomes $L=\mu_0\mu_r n^2 A l$ , with a high-permeability iron core boosting the inductance far above the air-core value.

Energy stored in an inductor. Building up a current i in an inductor (with negligible resistance) requires external work to be done against the self-induced back-emf $\varepsilon=-L\,di/dt$ at every instant. Moving a small charge $dq=i\,dt$ against this opposition requires work $dW=-\varepsilon\,dq=-\varepsilon\,i\,dt$ ; substituting $\varepsilon=-L\,di/dt$ gives $dW=Li\,di$ . Integrating this from zero current up to a final current i gives the total work done in establishing that current,

W = \int_0^i Li\,di = \dfrac{1}{2}Li^2

and since this work is stored entirely as magnetic potential energy (with no resistance to dissipate it), the energy stored in an inductor is

$U_B = \dfrac{1}{2}Li^2 \qquad (4.11)$ …

Figure 4.20Self-inductance of a long solenoid

What this figure shows. A long solenoid of length l, cross-sectional area A, and N turns (turn density $n=N/l$ ) carries a current i, and is drawn with its magnetic field $\vec B$ shown as a set of parallel arrows running along the solenoid's own axis, uniform everywhere well inside the winding. The figure anchors the derivation of $L=\mu_0 n^2 A l$ : because the field inside a long solenoid is essentially uniform and given by $B=\mu_0 n i$ , the flux through each single turn is $BA$ , and multiplying by the total number of turns N (and using $N=nl$ ) gives the total flux linkage $N\Phi_B = \mu_0 n^2 A l\, i$ , from which …

Misc Example 4.10Average emf in an iron-core solenoid as current ramps up

Worked out. A solenoid of 500 turns, wound on an iron core of relative permeability 800, has length 40 cm and radius 3 cm, and the average emf induced is required as the current changes from 0 to 3 A in 0.4 s. First the self-inductance is found using $L=\mu_0\mu_r n^2 A l = \mu_0\mu_r(N/l)^2(\pi r^2)l = \mu_0\mu_r\pi r^2 N^2/l$ ; substituting the given numbers yields $L\approx1.77$ H. The average induced emf is then $\varepsilon = L\,\dfrac{di}{dt} = 1.77\times\dfrac{3}{0.4} \approx 13.275$ V. The example shows how dramatically an iron core (through its large relative permeability $\mu_r=800$ ) boosts a solenoid's inductance compared with an identical air-core coil, and hence boosts the …

Misc Example 4.11Relative permeability from the ratio of iron-core to air-core inductance

Worked out. An air-core solenoid has self-inductance 4.8 mH, and replacing its core with an iron core raises the self-inductance to 1.8 H; the relative permeability of the iron is required. Because $L_{air}=\mu_0 n^2 A l$ and $L_{iron}=\mu_0\mu_r n^2 A l$ share every geometric factor ( $n$ , A, l) in common, their ratio isolates $\mu_r$ directly: $\mu_r = L_{iron}/L_{air} = 1.8/(4.8\times10^{-3}) = 375$ . This is a clean illustration of how comparing two inductance measurements -- with and without a magnetic core, but with everything else about the coil unchanged -- is a simple and practical way to measure a material's relative permeability experimentally, without needing to know the coil's exact t …

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