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