Answers and Discussions for

## Problem 1, part a

A developed diagram approach is used, "unrolling" the machine such that increasing $\phi$ (counterclockwise as drawn) is to the left.

Magnetomotive force (MMF) for a machine with concentrated windings may be found as $\mathcal{F} = \int_{\phi = 0}^{360^\circ} N(\phi) i \, d\phi$. The current is $i_{fd} = 1 \, \textrm{A}$.

It is given that $N_{fd1} = \mp 5$ at an inferred position of $\phi_r = 90^\circ$ and $\phi_r = 270^\circ$, $N_{fd2} = \mp 3$ at an inferred position of $\phi_r = 60^\circ$ and $\phi_r = 240^\circ$, and $N_{fd3} = \mp 3$ at an inferred position of $\phi_r = 120^\circ$ and $\phi_r = 300^\circ$.

Because $\oiint_S \vec{B} \cdot d\vec{S} = 0$, including on the interior surface of the stator and ends of the machine, the average value in the developed diagram of $B(\phi_r)$ must be 0. The average value of MMF must also be 0 because the airgap is constant, meaning that $\mathcal{F}(\phi_r)$ and $B(\phi_r)$ are proportional to each other in the airgap from $B(\phi_r) = \frac{\mu_0 \mathcal{F}(\phi_r)}{g}$. Thus, the offset of -5.5 A is removed.

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