Energy Stored in a Magnetic Field — Question 6

Question 6

Increasing inductance has different energy consequences at fixed current and fixed linkage

A lossless linear inductor is changed slowly from L0 to 2L0 by moving part of its geometry. Initially its current is I0 and energy U0. Compare a current-controlled case with an isolated zero-resistance case having conserved linkage.

1. Find final current and energy in each case.

2. For the fixed-current case, find electrical work and mechanical work delivered by the magnetic system.

Question 6 - Solution

Increasing inductance has different energy consequences at fixed current and fixed linkage

(a) Method and calculation

$$I=I_0:\quad U_f=\tfrac12(2L_0)I_0^2=\boxed{2U_0}.$$$$\Lambda=L_0I_0:\quad I_f=I_0/2,\quad U_f=\Lambda^2/(4L_0)=\boxed{U_0/2}.$$

(b) Interpretation and further calculation

$$W_{\mathrm{elec}}=\int I_0^2dL=L_0I_0^2=2U_0.$$$$\boxed{W_{\mathrm{mech,out}}=W_{\mathrm{elec}}-\Delta U=U_0}.$$

Physical check

A moving geometry allows mechanical energy exchange. Using only the change in stored energy misses the source contribution at fixed current.