Understanding Inductance: Formulas & Contradictions Explained

[yungman]

I am confused on the formulas of inductance.

In "Fields and Waves Electromagnetics" by David Cheng:

L = \frac{\Lambda}{I} \;\hbox{ where }\; \Lambda = N \Phi

N is the number of turns on the inductor, \Lambda is called flux linkage and

\Phi = \int_S \vec B \cdot d\vec l

\Rightarrow W = \frac 1 2 LI^2



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

L = \frac{\Phi}{ I} \;\hbox { instead of }\; \frac{\Lambda}{I}

During derivation of energy using magnetic field:

\frac {dW}{dt} = IV

-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

\Rightarrow\; L=\frac{\Phi}{I}

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 \Lambda in the former text and \Phi in the latter one are the same:
_ Engineers understand \Lambda=N\phi as flux linkage, where \phi is the flux through 1 round of the coil. So \Lambda is simply the total flux through the coil.
_ Scientists understand as \Phi as the TOTAL flux through the coil.
They are just different notations

 

[yungman]

Thanks for the reply. This make sense in:

L = \frac{\Lambda}{I} \;\hbox{ vs }\; L = \frac{\Phi}{I}

and in energy equation:

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

But then both books went on and defind:

\Phi = \int_S \vec B \cdot d\vec S = \int_C \vec A \cdot d\vec l =LI

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:

\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

But if you use:

\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)

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

 

[hikaru1221]

\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

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:
\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

\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)

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 \Phi_{engineer} then flux linkage \Lambda, scientists simply care about the net effective region which corresponds to the total flux \Phi_{scientist}, which turns out to be equal to \Lambda.

 

[yungman]

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

Have a nice day.

Alan