Example problems of magnetic circuits: Difference between revisions

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Similarly to get the reluctance of the gap <br>
Similarly to get the reluctance of the gap <br>


<math> 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} </math> <br> <br>
<math>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} </math> <br> <br>


Now Using <math> B_g = \frac{NI}{(R_g R_c)((\sqrt{A} + g)^2} </math> <br>
Now Using <math> B_g = \frac{NI}{(R_g R_c)((\sqrt{A} + g)^2} </math> <br>

Revision as of 18:09, 10 January 2010

Given:

A copper core with susceptibility

length of core L = 1 m

Gap length g = .01 m

cross sectional area A = .1 m

current I = 10A

N = 5 turns


Find:

Solution: First we need to find the permeability of copper given by the equation


Which yeilds

Now with this, the length and cross sectional area of the core we can solve for reluctance by:



Similarly to get the reluctance of the gap



Now Using
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}