Example problems of magnetic circuits

Given:

A copper core with susceptibility $\chi _{m}=-9.7\times 10^{-6}$

length of core L = 1 m

Gap length g = .01 m

cross sectional area A = .1 m

current I = 10A

N = 5 turns

Find: $B_{g}$

Solution: First we need to find the permeability of copper $\mu$ given by the equation
$\mu =\mu _{0}(1+\chi _{m})$

Which yeilds $\mu =4\times \pi \times 10^{-7}(1+-9.7\times 10^{-6})=1.2566\times 10^{-6}{\frac {N}{A^{2}}}$

Now with this, the length and cross sectional area of the core we can solve for reluctance $R_{c}$ by:

$R_{c}={\frac {L}{\mu A}}={\frac {1}{1.2566\times 10^{-6}\times .1}}=7.96\times 10^{6}$

Similarly to get the reluctance of the gap

$R_{g}={\frac {g}{\mu _{0}({\sqrt {A}}+g)^{2}}}={\frac {.01}{4\times \pi \times 10^{-7}({\sqrt {.1}}+.01)^{2}}}=74.8\times 10^{3}$

Now Using $B_{g}={\frac {NI}{(R_{g}R_{c})(({\sqrt {A}}+g)^{2}}}$
Yields <math> B_g = \frac{5 \times 10}{74.8 \times 10^{3} \times 7.96 \times 10^{6} \times (\sqrt{.1} + .01)^2} = .789 \times 10^{-9}

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