Example problems of magnetic circuits: Difference between revisions
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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> |
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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} |
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}</math> |
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==Reviewers== |
===Reviewers:=== |
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[[Nick Christman]]: |
[[Nick Christman]]: |
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* I would change "Given: ... A copper core with" to "Given a copper core with:" to make it a little more consistent or even take all the information you have and make it into a complete sentence/paragraph. |
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* This looks strange to me, <math>\mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6})</math> maybe make it <math>\mu = 4 \pi \times 10^{-7}(1-9.7 \times 10^{-6})</math> or <math>\mu = 4 \pi \times 10^{-7}(1+(-9.7 \times 10^{-6}))</math> |
Revision as of 18:47, 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
Reviewers:
- I would change "Given: ... A copper core with" to "Given a copper core with:" to make it a little more consistent or even take all the information you have and make it into a complete sentence/paragraph.
- This looks strange to me, maybe make it or