Understanding Inductance: Formulas & Contradictions Explained

[yungman]

I am confused on the formulas of inductance.

In "Fields and Waves Electromagnetics" by David Cheng:

[tex]L = \frac{\Lambda}{I} \;\hbox{ where }\; \Lambda = N \Phi [/tex]

N is the number of turns on the inductor, [itex]\Lambda [/itex] is called flux linkage and

[tex]\Phi = \int_S \vec B \cdot d\vec l [/tex]

[tex]\Rightarrow W = \frac 1 2 LI^2 [/tex]



But when derive energy of inductor in "Introduction to Electrodynamics" by Griffiths. p317 and also later part of Cheng's book gave.

[tex] L = \frac{\Phi}{ I} \;\hbox { instead of }\; \frac{\Lambda}{I} [/tex]

During derivation of energy using magnetic field:

[tex]\frac {dW}{dt} = IV[/tex]

[tex] -V=\int_C \vec E \cdot d\vec l =\int_S \vec B \cdot d\vec S = -\frac {\partial \Phi}{\partial t} \;\Rightarrow\; W=\frac 1 2 I\phi [/tex]

[tex]\Rightarrow\; L=\frac{\Phi}{I}[/tex]

So the two are contradicting and I don't know how to make of it. Can anyone help explain this?

Thanks

Alan

 

[hikaru1221]

Hello yungman,
I believe engineers and scientists sometimes *like* talking at different frequencies The [tex]\Lambda[/tex] in the former text and [tex]\Phi[/tex] in the latter one are the same:
_ Engineers understand [tex]\Lambda=N\phi[/tex] as flux linkage, where [tex]\phi[/tex] is the flux through 1 round of the coil. So [tex]\Lambda[/tex] is simply the total flux through the coil.
_ Scientists understand as [tex]\Phi[/tex] as the TOTAL flux through the coil.
They are just different notations

 

[yungman]

Thanks for the reply. This make sense in:

[tex] L = \frac{\Lambda}{I} \;\hbox{ vs }\; L = \frac{\Phi}{I} [/tex]

and in energy equation:

[tex] W=\frac 1 2 \sum_{k=1}^ N LI^2 \;\hbox { vs }\; W=\frac 1 2 \sum_{k=1}^ N I\Phi [/tex]

But then both books went on and defind:

[tex]\Phi = \int_S \vec B \cdot d\vec S = \int_C \vec A \cdot d\vec l =LI[/tex]

Lets look at the calculation of self inductance of a long coil that has radius = a and N turn per unit length. To calculate inductance per unit length using this formula:

[tex] \vec B = \mu NI \;\Rightarrow\; \Phi = \int _S \vec B \cdot d \vec S = \mu N I \pi a^2 \;\Rightarrow\; L =\mu N \pi a^2 [/tex]

But if you use:

[tex]\Phi = \int_S \vec B \cdot d\vec S \;\hbox{ and }\; \Lambda = N\Phi = \mu N^2I(\pi a^2) \;\Rightarrow L=\mu N^2 (\pi a^2) [/tex]

So you see you cannot assume the physics book treat [itex] \Phi [/itex] as [itex] \Lambda[/itex] in engineering book.

 

[hikaru1221]

[tex] \vec B = \mu NI \;\Rightarrow\; \Phi = \int _S \vec B \cdot d \vec S = \mu N I \pi a^2 \;\Rightarrow\; L =\mu N \pi a^2 [/tex]

The region S (under the integral notation) in this case is actually the total region formed by N turns. This is how scientists work. So the correct calculation for this is:
[tex] \Phi = \int _S \vec B \cdot d \vec S = \mu N I \times N \pi a^2 \;\Rightarrow\; L =\mu N^2 \pi a^2 [/tex]

[tex]\Phi = \int_S \vec B \cdot d\vec S \;\hbox{ and }\; \Lambda = N\Phi = \mu N^2I(\pi a^2) \;\Rightarrow L=\mu N^2 (\pi a^2) [/tex]

And in this case, S is the region of just 1 turn. This is how engineers work.

So while engineers go from magnetic flux of 1 turn [tex]\Phi_{engineer}[/tex] then flux linkage [tex]\Lambda[/tex], scientists simply care about the net effective region which corresponds to the total flux [tex]\Phi_{scientist}[/tex], which turns out to be equal to [tex]\Lambda[/tex].

 

[yungman]

I see, thanks for your help. I guess I am the only odd ball engineer here also!

Have a nice day.

Alan