Question 8.16: A U-shaped electromagnet shown in Figure 8.29 is designed to...

The tractive force across the two air gaps must balance the weight. Hence

F=2\frac{(B_{a}^{2}S)}{2\mu_{o}}=mg

or

B_{a}^{2}=\frac{mg\mu_{o}}{S}=\frac{400\times 9.8\times 4\pi\times 10^{-7}}{40\times 10^{-4}}

B_{a}=1.11Wb/m^{2}

But

?=NI=\Psi(\Re_{a}+\Re_{i})

\Re_{a}=\frac{\ell_{a}}{\mu S}=\frac{2\times 0.1\times10^{-3}}{4\pi\times 10^{-7}\times 40\times 10^{-4}}=\frac{6\times 10^{6}}{48\pi}

\Re_{i}=\frac{\ell_{i}}{\mu_{o}\mu_{r}S}=\frac{50\times 10^{-2}}{4\pi\times 10^{-7}\times 3000\times 40\times 10^{-4}}=\frac{5\times 10^{6}}{48\pi}

?_{a}=\frac{\Re_{a}}{\Re_{a}+\Re_{i}}?=\frac{6}{6+5}NI=\frac{6}{11}NI

Since

?_{a}=H_{a}\ell_{a}=\frac{B_{a}\ell_{a}}{\mu_{o}}

N=\frac{11}{6}\frac{B_{a}\ell_{a}}{\mu_{o}I}=\frac{11\times 1.11\times 0.1\times 10^{-3}}{6\times 4\pi\times 10^{-7}\times 1}

N=162