Class Wiki - User contributions [en]

Electronics Receiver

2010-04-05T15:26:20Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t)</math>
 

<math> = x(t)cos(w_c*t)-y(t)sin(w_c*t) </math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:25:52Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t)</math>
 

<math>= x(t)cos(w_c*t)-y(t)sin(w_c*t) </math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:25:06Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t)
 

= x(t)cos(w_c*t)-y(t)sin(w_c*t) </math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:24:13Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t) 

= x(t)cos(w_c*t)-y(t)sin(w_c*t) </math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:23:29Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t) 

= x(t)cos(w_c*t)-y(t)sin(w_c*t) </math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:22:48Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t) 

= x(t)cos(w_c*t)-y(t)sin(w_c*t) <\math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:22:05Z

Kevin.Starkey:

Electronics Receiver Explanation:

When a Signal is transmitted they take the original baseband m(t) signal and send it through a bandpass processor that splits the signal into a real (x(t)) and an imaginary (y(t)) parts and then shifts the signal to +/- a designated frequency by mixing the signals with cos(w_c*t) for x(t) and -sin(w_c*t) for y(t). This is shown math mathematically below where v(t) is the output signal and w_c is the amount shifted.

<math> v(t) = Re[(x(t) + jy(t))*(cos(w_c*t)+jsin(w_c*t)) = Re[x(t)cos(w_c*t) + x(t)jsin(wc_t) + jy(t)cos(w_c*t) -y(t)sin(w_c*t) 

= x(t)cos(w_c*t)-y(t)sin(w_c*t) <\math>

[[Image:Comunications2.png]]

The purpose of your receiver is to retrieve this signal. This is done by first taking the transmitted signal and mixing it with cos(w_c*t)(to get x(t) and -sin(w_c*t) (to get y(t) also referred to as the quadrature signal) and filtering those two signals using a low pass filter. After you have those signals you can send them through a baseband processor (in the computer in this case) and it will process the signal and send the results to your speakers.

[[Image:Comunications.png]]

Electronics Receiver

2010-04-05T15:09:03Z

Kevin.Starkey:

Electronics Receiver

2010-04-05T04:38:42Z

Kevin.Starkey:

Electronics Receiver

2010-04-05T04:35:20Z

Kevin.Starkey:

File:Comunications2.png

2010-04-05T04:34:42Z

Kevin.Starkey:

Electronics Receiver

2010-04-05T04:34:05Z

Kevin.Starkey:

Electronics Receiver

2010-04-05T03:47:06Z

Kevin.Starkey: New page: Image:Comunications.png

[[Image:Comunications.png]]

File:Comunications.png

2010-04-05T03:43:04Z

Kevin.Starkey:

Engineering Electronics

2010-04-05T03:40:35Z

Kevin.Starkey: /* Contributing Articles */

==Publish or Perish Game==
*[[Electronics Score Pages]]
*[[Rules]]
*[[Conference Deadlines]]

==Questions==

==Links==
*[http://www.dspguru.com/sites/dspguru//files/QuadSignals.pdf Quadrature Signals Explained]
* Software Defined Radio Links
**[http://people.wallawalla.edu/~Rob.Frohne/R2_DSP/9804x040.pdf R2 DSP (an early software defined radio using a dedicated DSP)]
**[http://www.nonstopsystems.com/radio/frank_radio_sdr.htm Softrock and Theory]
**[http://www.wb5rvz.com/sdr/ Softrock Build Instructions and Notes]
**[http://groups.yahoo.com/group/softrock40/ Softrock Yahoo Interest Group]
**[http://www.flex-radio.com/News.aspx?topic=publications This collection of Software Defined Radio publications is fantastic.]
**[http://www.sdradio.eu/sdradio/ SDRadio]
**[http://openhpsdr.org/ Open High Performance Software Defined Radio]

==2010 Contributors==

#[[Greg Fong|Fong, Greg]]
#[[Ben Henry|Henry, Ben]]
#[[Lau, Chris]]
#[[Shepherd,Victor]]
#[[Vier, Michael]]

==2010 Articles==
*[[Ideal vs. Nonideal Op Amps]]
*[[Chapter 1]]
*[[Chapter 2]]
*[[Basic Op Amp circuits]]
*[[Key Facts from Reading Chapter 1]]
*[[Golden Rules]]
*[[Integrator_Amplifier]] (by [[Ben Henry|Ben]])

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.

*[[Chapter 3 Problems]] by [[Ben Henry|Ben]]
* Disecting an Instrumentation Amplifier via [[Superposition]]
*[[Reading from Chapter 4]]

==Draft Articles awaiting review==
*[[Feedback in Amplifiers]]

==Contributing Articles==

*[[Generalized Transmitter]] (in progress, Luke)
*[[Generalized Receiver]] (in progress, Luke)
*[[Electronics Receiver]] (in progress, Kevin)

Engineering Electronics

2010-04-05T03:39:23Z

Kevin.Starkey: /* Contributing Articles */

Example Problems with Transformers

2010-01-18T04:08:22Z

Kevin.Starkey:

==Problems 1-3==

'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''

'''Problem 1.''' An ''ideal'' step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. If the input voltage is 1200V, what is the resulting output voltage?

'''Solution''' Using the equation <math> e_2 = \frac{N_2}{N_1}e_1 </math> we get 

<math> e_2 = \frac{2}{10}1200 = 240V </math>

Example Problems with Transformers

2010-01-18T04:07:23Z

Kevin.Starkey:

Example Problems with Transformers

2010-01-18T03:32:31Z

Kevin.Starkey:

==Problems 1-3==

'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''

Aric Vyhmeister

2010-01-18T03:30:30Z

Kevin.Starkey: New page: Article#1 Ohm's Law and Reluctance

Article#1 [[Ohm's Law and Reluctance]]

Ohm's Law and Reluctance

2010-01-18T03:29:17Z

Kevin.Starkey:

==Ohm's Law==

Electric circuits share many of the same characteristics as magnetic circuits. One of the most important and fundamental equations describing electric circuits is Ohm’s Law. Georg Ohm, a German physicist, published this famous equation in 1827, drawing significant influence from Fourier’s previous work in heat conduction.

Ohm’s Law states that the resistance of a conductor is constant and inversely proportional to the current running through the conductor and directly proportional to the voltage across it, or expressed mathematically as

<center><math>R=\frac{V}{I}</math></center>

The SI units for resistance are the Ohm (<math>\Omega</math>), voltage is expressed in volts (V) and current is measured in amperes (A).

Expressed differently, the current density J (A/<math>m^2</math>) is proportional to the product of conductivity sigma (V/m) and the electric field E (siemens/meter, s/m), or

<center><math>\displaystyle J=\sigma E</math></center>

==Reluctance==

Now, consider a material of very high permeability with flux <math>\Phi</math> running through it and separated by a very small gap of area A, where the flux flows through this gap as if it were free space. The flux density crossing this gap is B. Therefore

<center><math>\displaystyle \Phi=BA</math></center>

The magnetic field intensity can be approximated by <math>H=\frac{B}{\mu_0}</math>

The reluctance of this gap is analogous to the resistance described by Ohm’s Law, where it is the ratio of the magnetomotive force to the flux, or

<center><math>\R = \frac{F}{\phi}</math></center>

Example Problems with Transformers

2010-01-18T03:26:16Z

Kevin.Starkey: New page: ''Problems 1-3'' ==Kevin Starkey, Nick Christman, Aric Vyhmeister==

''Problems 1-3''

==Kevin Starkey, Nick Christman, Aric Vyhmeister==

Electromechanical Energy Conversion

2010-01-18T03:22:30Z

Kevin.Starkey: /* Draft Articles */

[[Rules]]

[[Class Roster]]

[[Points]]

==Questions==

What do we do when we are finished with the draft and ready to publish?
* If it's been approved by the reviewers, move it to the articles section

Does anyone know why my LaTEX stuff is changing sizes throughout my article? [[An Ideal Transformer Example]]

*(John Hawkins) As I understand it, the text is made full size (larger) if there is ever a function call, i.e. something starting with a backslash, excluding some things like greek letters. I have just put "\ " (the function call for a space) at the beginning of an equation and had it work. If you don't want to change anything about your equation but just want it displayed full size, type "\,\!" (small forward space and small backward space) somewhere in your equation.
*Thanks John!

==Announcements==

If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.
* [[Ferromagnetism]]
* [[Magnetic Circuits]]
* [[Gauss Meters]]
* [[Ampere's Law]]
* [[DC Motor]]
* [[AC vs. DC]]
* [[AC Motors]]
* [[Fringing]]
* [[Electrostatics]]
* [[Example Problems of Magnetic Circuits]]
* [[Magnetic Circuit]] (John Hawkins)
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]
* [[Example: Ideal Transformer Exercise]] (John Hawkins)

==Reviewed Articles==
These articles have been reviewed and submitted.
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)
* [[Magnetic Flux]] (Jason Osborne)
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)

Example Problems of Magnetic Circuits

2010-01-18T03:21:29Z

Kevin.Starkey: New page: =Problem 1 Kevin Starkey= Given a copper core with: Susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math> Length of core L = 1 m Gap length g = .01 m Cross sectional area A = ....

=Problem 1 Kevin Starkey=

Given a copper core with:

Susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math>

Length of core L = 1 m

Gap length g = .01 m

Cross sectional area A = .1 m

Current I = 10A

N = 5 turns

Find: <math> B_g </math> 
[[Image:EMEC1.png]]

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
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> 

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

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly, to get the reluctance of the gap 

<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> 

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> 
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> 
This is the magnetic field in the center of the gap due to the applied current

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

Kevin Starkey EMEC

2010-01-18T03:21:14Z

Kevin.Starkey:

=My Articles=

Article#1-[[Example Problems of Magnetic Circuits]]

Article#2-[[Example Problems with Transformers]]

Kevin Starkey EMEC

2010-01-18T03:20:05Z

Kevin.Starkey:

=My Articles=

Article#1-[[Example problems of magnetic circuits]]

Article#2-[[Example Problems with Transformers]]

Kevin Starkey EMEC

2010-01-18T03:12:57Z

Kevin.Starkey:

=My Articles=

Article#1-[[Example problems of magnetic circuits]]

Kevin Starkey EMEC

2010-01-18T03:12:32Z

Kevin.Starkey:

=My Articles=

Article#1-[[Example Problems of Magnetic Circuits]]

Example problems of magnetic circuits

2010-01-11T07:25:07Z

Kevin.Starkey:

Example problems of magnetic circuits

2010-01-11T07:24:20Z

Kevin.Starkey:

Given a copper core with:

Susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math>

Length of core L = 1 m

Gap length g = .01 m

Cross sectional area A = .1 m

Current I = 10A

N = 5 turns

Find: <math> B_g </math> 
[[Image:EMEC1.png]]

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
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> 

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

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly, to get the reluctance of the gap 

<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> 

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> 
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> 
This is the magnetic field in the center of the gap due to the applied current

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

File:EMEC1.png

2010-01-11T07:23:32Z

Kevin.Starkey: uploaded a new version of "Image:EMEC1.png"

File:EMEC1.png

2010-01-11T07:19:37Z

Kevin.Starkey:

Example problems of magnetic circuits

2010-01-11T03:16:59Z

Kevin.Starkey:

Given a copper core with:

Susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math>

Length of core L = 1 m

Gap length g = .01 m

Cross sectional area A = .1 m

Current I = 10A

N = 5 turns

Find: <math> B_g </math>

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
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> 

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

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly, to get the reluctance of the gap 

<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> 

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> 
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> 
This is the magnetic field in the center of the gap due to the applied current

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

Example problems of magnetic circuits

2010-01-11T02:11:45Z

Kevin.Starkey:

Given:

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

length of core L = 1 m

Gap length g = .01 m

cross sectional area A = .1 m

current I = 10A

N = 5 turns

Find: <math> B_g </math>

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
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> 

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

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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> 

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

Example problems of magnetic circuits

2010-01-11T02:09:57Z

Kevin.Starkey:

Given:

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

length of core L = 1 m

Gap length g = .01 m

cross sectional area A = .1 m

current I = 10A

N = 5 turns

Find: <math> B_g </math>

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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> 

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

Example problems of magnetic circuits

2010-01-11T02:09:30Z

Kevin.Starkey:

Given:

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

length of core L = 1 m

Gap length g = .01 m

cross sectional area A = .1 m

current I = 10A

N = 5 turns

Find: <math> B_g </math>

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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> 

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

Example problems of magnetic circuits

2010-01-11T02:09:03Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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> 

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

Example problems of magnetic circuits

2010-01-11T02:06:57Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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> 

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

Example problems of magnetic circuits

2010-01-11T02:04:28Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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> 

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

Example problems of magnetic circuits

2010-01-11T02:01:35Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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>

Example problems of magnetic circuits

2010-01-11T02:01:16Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<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 \time 10^{3} </math>

Example problems of magnetic circuits

2010-01-11T01:59:13Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<math> R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} = \frac {.01}{4 \times \pi \times 10^{-7} (\sqrt{.1} + .01)^2} = </math>

Example problems of magnetic circuits

2010-01-11T01:58:48Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<math> R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} = \frac {.01}{4 \times \pi (\sqrt{.1} + .01)^2} = </math>

Example problems of magnetic circuits

2010-01-11T01:56:00Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> 
Similarly to get the reluctance of the gap 

<math> R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} </math>

Example problems of magnetic circuits

2010-01-11T01:52:13Z

Kevin.Starkey:

Example problems of magnetic circuits

2010-01-11T01:51:58Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6}

Example problems of magnetic circuits

2010-01-11T01:51:21Z

Kevin.Starkey:

Example problems of magnetic circuits

2010-01-11T01:50:11Z

Kevin.Starkey:

Given:

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

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4 &times \pi &times 10^{-7}(1+-9.7 &times 10^{-6}) = 1.2566 &times 10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566x10^{-6}*.1} = 7.96x10^{6}

Example problems of magnetic circuits

2010-01-11T01:48:14Z

Kevin.Starkey:

Given:

A copper core with susceptibility <math> \chi_m = -9.7x10^{-6} </math>

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

Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <math> \mu = \mu_0 (1 + \chi_m)</math> 
Which yeilds <math> \mu = 4*\pi*10^{-7}(1+-9.7x10^{-6}) = 1.2566x10^{-6} </math> 
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: 

<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566x10^{-6}*.1} = 7.96x10^{6}

Example problems of magnetic circuits

2010-01-11T01:42:32Z

Kevin.Starkey: