Energy Stored in a Magnetic Field — Question 7

Question 7

Nonlinear magnetic energy is the integral of current against linkage change

A lossless, nonhysteretic coil has linkage $\Lambda(i)=ai+bi^3$, where a and b are positive constants. The current starts at zero and the geometry is fixed.

1. Derive the stored energy U(I).

2. Compare it with one-half Lambda I and with the complementary integral $W_c=\int_0^I\Lambda(i)di$.

Question 7 - Solution

Nonlinear magnetic energy is the integral of current against linkage change

(a) Method and calculation

$$dU=i\,d\Lambda=i(a+3bi^2)di.$$$$\boxed{U(I)=aI^2/2+3bI^4/4}.$$

(b) Interpretation and further calculation

$$\tfrac12\Lambda I=aI^2/2+bI^4/2\Rightarrow\boxed{U-\tfrac12\Lambda I=bI^4/4}.$$$$W_c=aI^2/2+bI^4/4\Rightarrow\boxed{U+W_c=I\Lambda}.$$

Physical check

The familiar one-half L I squared formula assumes constant linear inductance; the supplied nonlinear model requires the work integral.