Question 8.15: In the magnetic circuit of Figure 8.27, calculate the curren...

The magnetic circuit of Figure 8.27 is analogous to the electric circuit of Figure 8.28. In Figure 8.27,\Re_{1},  \Re_{2},  \Re_{3}, and \Re_{a} are the reluctances in paths 143, 123, 35 and 16, and 56 (air gap), respectively. Thus

\Re_{1}=\Re_{2}=\frac{\ell}{\mu_{o}\mu_{r}S}=\frac{30\times 10^{-2}}{(4\pi\times10^{-7})(50)(10\times 10^{-4})}=\frac{3\times 10^{8}}{20\pi}

\Re_{3}=\frac{9\times 10^{-2}}{(4\pi\times10^{-7})(50)(10\times 10^{-4})}=\frac{0.9\times 10^{8}}{20\pi}

\Re_{a}=\frac{1\times 10^{-2}}{(4\pi\times10^{-7})(1)(10\times 10^{-4})}=\frac{5\times 10^{8}}{20\pi}

We combine \Re_{1} and \Re_{2} as resistors in parallel. Hence

\Re_{1}\parallel\Re_{2}=\frac{\Re_{1}\Re_{2}}{\Re_{1}+\Re_{2}}=\frac{\Re_{1}}{2}=\frac{1.5\times10^{8}}{20\pi}

The total reluctance is

\Re_{T}=\Re_{a}+\Re_{3}+\Re_{1}\parallel\Re_{2}=\frac{7.4\times10^{8}}{20\pi}

The mmf is

?=NI=\Psi_{a}\Re_{T}

But \Psi_{a}=\Psi=B_{a}S. Hence

I=\frac{B_{a}S\Re_{T}}{N}=\frac{1.5\times 10\times 10^{-4}\times 7.4\times 10^{8}}{400\times 20\pi}=44.16A