Mutual Induction — Question 5

Question 5

Positive magnetic energy imposes a bound on mutual inductance

For two fixed linear coils, $U=\tfrac12L_1i_1^2+\tfrac12L_2i_2^2+Mi_1i_2$, with positive self-inductances. A proposed model uses L1=2.00 H, L2=8.00 H, and M=5.00 H.

1. Complete the square and derive the physical bound on M.

2. Test the proposed model and exhibit currents giving negative energy if the model is invalid.

Question 5 - Solution

Positive magnetic energy imposes a bound on mutual inductance

(a) Method and calculation

$$U=\frac{L_2}{2}\left(i_2+\frac{M}{L_2}i_1\right)^2+\frac12\left(L_1-\frac{M^2}{L_2}\right)i_1^2.$$$$\boxed{M^2\le L_1L_2,\quad |k|\le1}.$$

(b) Interpretation and further calculation

$$M_{\max}=\sqrt{(2)(8)}=4.00\,H<5.00\,H;\quad i_1=1\,A,\ i_2=-5/8\,A.$$$$\boxed{U=\tfrac12(2-25/8)=-.5625\,J}.$$

Physical check

A passive lossless magnetic model cannot store negative energy relative to its zero-current state.