Register to reply 
Question on energy in magnetic field. 
Share this thread: 
#1
Dec1410, 10:19 AM

P: 3,883

I am refering to the book "Introduction to Electrodynamics" by Griffiths p317, Energy in magnetic field.
[tex]\Phi = \int_S \vec B \cdot d \vec a = \int_C \vec A \cdot d \vec l [/tex] [tex] LI = \int_C \vec A \cdot d \vec l [/tex] But the inductance in defined as the flux linkage divid by the current that create the flux or: [tex] L = \frac {\Lambda}{I} [/tex] If the inductor is N turn, [itex] \Lambda = N\Phi[/itex] I want to verify according to the book that: [tex] LI = \int_C \vec A \cdot d \vec l \Rightarrow W = \frac {1}{2} I \int_C \vec A \cdot d \vec l \Rightarrow W = \frac {1}{\mu_0}[\int_v B^2 d \tau  \int_S (\vec A X \vec B) \cdot d\vec a][/tex] Only apply to a single loop inductor or a straight wire type that N=1. Because for multi turn N inductor, [itex] \Lambda = N\Phi[/itex] 


#2
Dec1510, 01:42 AM

P: 799

The notation S under the integral sign denotes the TOTAL effective surface. If there are N turns of area A, then S corresponds to NA, not A alone.



#3
Dec1510, 10:22 PM

P: 3,883

[tex]W_m = \frac 1 2 \sum_{k=1}^M \Delta I_k \Lambda_k = \frac 1 2 \sum_{k=1}^M N_k \int_{C_k} \vec A \cdot \vec J dv_k[/tex] The book went on to combine all the M parts together give [tex]W_m = \int_v B^2 dv [/tex] As you can see this cannot be done if the inductors have different turns ie [itex] N_1 [/itex] not equal to [itex] N_2 [/itex] not equal to [itex] N_3 [/itex] ......etc. You can integrate the loop, but not if they have different turns. So they equation must stop at: [tex]W_m = \frac 1 2 \sum_{k=1}^M\; I_k \; \Lambda_k \;=\; \frac 1 2 \sum_{k=1}^M \sum_{j=1}^M [L_{jk} \; I_j\; I_k] \;\;\hbox { where }\;\; \Lambda_k = \sum_{j=1}^M N_j \; I_j \; L_{jk}\; [/tex] Let me know I am right or not. Thanks Alan 


#4
Dec1610, 12:24 AM

P: 799

Question on energy in magnetic field.
By the way, which book were you referring to in post #3? 


#5
Dec1610, 01:29 AM

P: 3,883

I use what I got from you this morning and modify the equations in the book to get this. Lets derive the formula: [tex] W_m\;=\; \frac 1 2 \sum_{k=1}^M I_k \; \Lambda_k \;=\; \frac 1 2 \sum_{k=1}^M I_k \; N_k \;\Phi_k \;=\; \frac 1 2 \sum_{k=1}^M I_k \; N_k \; \sum_{j=1}^M \int_{S_k} \Vec B_j \cdot d\vec {S_k} \;=\; \frac 1 2 \sum_{j=1}^M \frac 1 2 \sum_{k=1}^M \mu I_j I_k N_j N_k S_k [/tex] This forumla is consistance with what the book gave in terms of self inductance and mutual inductance. Below is the formula derive from inductance, if you expand it out, it will become my formula above: [tex] W_m\;=\; \frac 1 2 \sum_{j=1}^M \sum_{k=1}^M L_{jk}\;I_j \; I_k [/tex] For example for two inductor of [itex]N_1 \;\hbox { and }\; N_2 [/itex] turns coupling together, M=2: [tex] W_m\;=\; \frac 1 2 \sum_{j=1}^2 \frac 1 2 \sum_{k=1}^2 \mu I_j I_k N_j N_k S_k \;=\; \frac 1 2 \mu I_1^2 N_1^2 S_1 \;+\; \frac 1 2 \mu I_1 \;I_2 N_1 \; N_2 S_2 \;+\;\frac 1 2 \mu I_1 \;I_2 N_1 \; N_2 S_1 \;+\; \frac 1 2 \mu I_2^2 N_2^2 S_2[/tex] If [itex] S_1=S_2[/itex] then: [tex] W_m\;=\; \frac 1 2 \mu I_1^2 N_1^2 S_1 \;+\; \mu I_1 \;I_2 N_1 \; N_2 S_2 \;+\; \frac 1 2 \mu I_2^2 N_2^2 S_2[/tex] The book I am using is "Fields and Waves Electromagnetics" By David K Cheng. I have 8 EM books including Griffiths, Ulaby, Hyat & Buck, Popovic, Schwarz, Kraus and JD Jackson. This and Griffiths are the best. Cheng cover a lot of engineering EM, phasor, smith chart, transmission line, antenna, cavity. Griffiths is more on materals and retarded potential. 


#6
Dec1610, 10:07 AM

P: 3,883

Anyway, I just don't see
[tex] Wm = \int_v B^2 dv [/tex] contain the informations of number of turns of each individual inductors, the surface of each inductors etc. Unless the formulas assumes all surface are same, all turns are the same. That is a big assumption. This is the reason I still resort back to the very primitive formula with the summation of individual inductor as shown in post #5. 


#7
Dec1710, 12:28 AM

P: 799

For post #5: Is there any inconsistency?
For post #6: I think you should check the formula; I remember there is some constant. Anyway, no big deal. The formula is true for magnetic field in general. I don't have the book by David Cheng, but there is another free book provided by MIT OCW namely Electromagnetic Fields and Energy, where you may find the relevant information. Check out page 12, chapter 11, section 11.3 of the book: http://ocw.mit.edu/courses/electrica...emonstrations/ The (3b) equation in the page shows the partial energy density contributed by magnetic component of the EM field in linear medium: [tex]w_m=\frac{1}{2}\mu H^2 = \frac{1}{2\mu}B^2[/tex] . The derivation of this equation, as you can see, is not deduced from the consideration of any setup or inductors, but any general EM field (so EM waves are included). This is also how scientists work. I believe some engineers also work in this way; but sometimes they choose [tex]W_m=LI^2/2[/tex] for example for the ease of analyzing their application. That means, this formula doesn't care about the setup, e.g. how many inductors are there or whether they are coupling. This formula applies whenever there is magnetic field, either when the field is stored inside some setup or when the field propagates as waves. The number of turns, the dimensions (area & length), etc, of inductors in particular and storage devices in general are included in the volume in which you calculate the integral [tex]\int_V B^2dV[/tex]. Those parameters define the volume (and also the magnitude of B) in which the magnetic field is stored. 


#8
Dec1710, 01:08 PM

P: 3,883

Hey Hikaru1221
Thanks for all your help. I have to spend some time read this material first. What is the inconsistency in #5, that is very important for me.....as an engineer?!!! 


#9
Dec1710, 02:26 PM

P: 799

Important to all, I think.



#10
Dec1710, 03:08 PM

P: 3,883

BTW, what is your background? Obviously my background is engineer. I just want to cross into Physics in my old age!!! 


#11
Dec1710, 03:21 PM

P: 799

I guess I'm lost in some sort of confusion. I thought post #6 of yours backed up some point in post #5, so there should be some inconsistency, but I didn't find it. About my background, I would say I'm neither engineer nor scientist I am a bit of scientist and a bit of engineer. Anyway I find it hard to define "I" 


#12
Dec1710, 04:03 PM

P: 3,883

What do you do in your career then? Or what are you planning to do if you are a student. 


#13
Dec1710, 04:28 PM

P: 799

It seems ok to me (Sorry, I haven't checked the math; anyway, the idea behind the equations seems ok )
I'm college student, so I'm currently unemployed I'm studying Electrical Engineering. 


#14
Dec1710, 06:27 PM

P: 3,883

Thanks for you help. What EM book are you using? Take a look at Cheng's book, I have 8 and this one clearly the best for engineer. Go very deep into phasor, transmission lines, smith chart. From that you can get into designing filters, matching networks with distributed elements. I spent a lot of time doing that kind of design and simulations. Cheng's book is as good as Griffiths for scientist, these are the two must have for EM. They really compliment each other. I would say they are about 50% cross over. 


#15
Dec1810, 12:24 AM

P: 799

I'm still in my first year in college, so I haven't taken EM course yet. Anyway, Cheng's book is the compulsory one for engineering EM course in my university, so I'll definitely go through it, hopefully next semester. Thank you for the advice and warning



#16
Dec1810, 02:08 AM

P: 3,883

What university you going? Cheng's book is on the difficult side for undergrad. I have seen easier books. The one used by San Jose State is a much easier version of this. I actually looked into the advanced electromagnetic course offer in Santa Clara university ( which is consider a good school) , They are using Hyat and Buck which is no more difficulty than Cheng's book, and that one is lousy. I think you would enjoy Cheng's book. Get the Griffiths book also, it will answer a lot of questions you have on Cheng's book. This is because Cheng tried not to go into the physics side and move a little fast on the begining part and go much more detail in field and wave. The two really compliment each other. You seems to be very interested in EM and you sure know a lot, you should keep at it. I heard so many people afraid of EM and they just craw through it!!! Where did you learn at the EM stuff, can't be from the freshman physics!!! You need to know at very minimum vector calculus and differential equation to really understand it. I actually studied partial differential equation and now restudy the whole thing again plus the Griffiths book. I found design distributed element transmission line circuit very interesting and that is the part of EM I am strong at. I design multipole filters, matching networks using just traces on pc board. If you look at the circuit board, it just has a pattern of traces looking like a maze and very few capacitors and resistors. This is all done after you study Cheng's book on transmission lines and smith chart, then you get into microwave design and get deeper into this kind. I use to have a lot of fun designing these stuffs. Good luck on your study. Alan 


#17
Jan911, 05:31 PM

P: 1,394

I find the connection between #8 and #9 hilarious



#18
Jan1011, 11:15 AM

P: 188

The derivation by Griffiths is a bit handwavy as I recall. The final formula is, as pointed out earlier in the thread, a general statement about the energy contained in a magnetic field. The derivation is restricted to the oneloop case though. I haven't seen it myself, but I think Wangsness derives the formula in a more general way in "Electromagnetic Fields". The result is the same obviously, but I think the reasoning is more general.
I don't really know what's true for a loop with N turns, but isn't it true then that [tex] N\Phi = LI[/tex]? Therefore, shouldn't one simly replace [tex] \int_C \mathbf{A} \cdot d \mathbf{l} [/tex] with [tex] LI/N[/tex], and not [tex]LI[/tex], as is done in Griffiths derivation? 


Register to reply 
Related Discussions  
Magnetic Field , Selfinductance & energy Question  Introductory Physics Homework  6  
Electric field from magnetic field and kinetic energy  Introductory Physics Homework  0  
Energy of magnetic field created by magnetic dipoles in a shphere.  Classical Physics  1  
Deriving the energy of a magnetic dipole in a magnetic field  Introductory Physics Homework  8  
Deriving the energy of a magnetic dipole in a magnetic field  Classical Physics  1 