Lenz's Law — Question 6

Question 6

Opposite lobes of a figure-eight circuit can add rather than cancel induction

A figure-eight circuit has equal lobe areas A, wound so positive traversal is counterclockwise in the left lobe and clockwise in the right. The left field is $+B(t)\hat z$ and the right field is $-B(t)\hat z$, with B increasing. Total resistance is R and self-inductance is negligible.

1. Find the signed total flux and induced current.

2. Find the induced field direction in each lobe and verify opposition to each applied change.

Question 6 - Solution

Opposite lobes of a figure-eight circuit can add rather than cancel induction

(a) Method and calculation

$$\Phi=A(+B)+(-A)(-B)=2AB.$$$$\boxed{i=-2A\dot B/R<0}.$$

(b) Interpretation and further calculation

$$\text{left: negative traversal is clockwise}\Rightarrow\vec B_{\mathrm{ind,left}}\parallel-\hat z.$$$$\text{right: negative traversal is counterclockwise}\Rightarrow\boxed{\vec B_{\mathrm{ind,right}}\parallel+\hat z}.$$

Physical check

Reversing both the field and the winding sense makes the two flux contributions add.