Example problems of magnetic circuits - Revision history

Kevin.Starkey at 07:25, 11 January 2010

2010-01-11T07:25:07Z

← Older revision		Revision as of 00:25, 11 January 2010
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			=Problem 1 Kevin Starkey=

	Given a copper core with:		Given a copper core with:

Kevin.Starkey at 07:24, 11 January 2010

2010-01-11T07:24:20Z

← Older revision		Revision as of 00:24, 11 January 2010
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	Find: <math> B_g </math>		Find: <math> B_g </math> <br>
			[[Image:EMEC1.png]]

	Solution:		Solution:

Kevin.Starkey at 03:16, 11 January 2010

2010-01-11T03:16:59Z

← Older revision		Revision as of 20:16, 10 January 2010
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	Given:		Given a copper core with:

	~~A copper core with susceptibility~~ <math> \chi_m = -9.7 \times 10^{-6} </math>		Susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math>

	~~length~~ of core L = 1 m		Length of core L = 1 m

	Gap length g = .01 m		Gap length g = .01 m

	~~cross~~ sectional area A = .1 m		Cross sectional area A = .1 m

	~~current~~ I = 10A		Current I = 10A

	N = 5 turns		N = 5 turns
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	Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br>		Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br>

	Now ~~with this,~~ the length ~~and~~ cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br>		Now using the length, cross sectional area, and permeability of the core we can solve for reluctance <math> R_c </math> by: <br>

	<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br>		<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br>
	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 recall the equation for the magnetic field of a gap as seen in class <math> B_g = \frac{NI}{(R_g R_c)((\sqrt{A} + g)^2} </math> <br>
	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>		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> <br>
			This is the magnetic field in the center of the gap due to the applied current

Nicholas.Christman at 02:56, 11 January 2010

2010-01-11T02:56:10Z

← Older revision		Revision as of 19:56, 10 January 2010
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	----		----
	[[Nick Christman]]:		[[Nick Christman]]:
	* 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.		* I would change "Given: (...)[list] A copper core with..." to "Given a copper core with: [list]" 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, <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>		* 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>
			* This sentence is kind of strange, "Now with this, the length and cross sectional area of the core we can solve for reluctance..." Maybe make it, "With the permeability, length, and cross sectional area of the copper core we can now solve for the reluctance..." Something like that might flow a little better."
			* Below that, I think you need a comma after "Similarly."
			* You might want to add some more words to the last two lines... Instead of saying "Now using..." say something like, "Recall that the equation for the magnetic field of the gap is..." or something to that effect.
			* Lastly, you should think of some sort of conclusion... what exactly does this mean?

Nicholas.Christman at 02:47, 11 January 2010

2010-01-11T02:47:56Z

← Older revision		Revision as of 19:47, 10 January 2010
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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>
	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>

	==Reviewers==
			===Reviewers:===
	----		----
	[[Nick Christman]]:		[[Nick Christman]]:
			* 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, <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>

Nicholas.Christman at 02:42, 11 January 2010

2010-01-11T02:42:23Z

← Older revision		Revision as of 19:42, 10 January 2010
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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>
	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}

			==Reviewers==
			----
			[[Nick Christman]]:

Kevin.Starkey at 02:11, 11 January 2010

2010-01-11T02:11:45Z

← Older revision		Revision as of 19:11, 10 January 2010
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	Solution:		Solution:
	First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br>		First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br>
	Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> <br><br>		Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br>

	Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br>		Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br>

Kevin.Starkey at 02:09, 11 January 2010

2010-01-11T02:09:57Z

← Older revision		Revision as of 19:09, 10 January 2010
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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>
	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}

Kevin.Starkey at 02:09, 11 January 2010

2010-01-11T02:09:30Z

← Older revision		Revision as of 19:09, 10 January 2010
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	Find: B		Find: <math> B_g </math>

	Solution:		Solution:

Kevin.Starkey at 02:09, 11 January 2010

2010-01-11T02:09:03Z

← Older revision		Revision as of 19:09, 10 January 2010
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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>
	Yields <math> B_g = \frac{5 \times 10}{74.8 \times 10^{3} \times 7.96 \times 10^{6} \times (\sqrt{.1} + .01)^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}