Mean Magnetic Path Length of the Magnetic core of a transformer

[janu203]

For the Image given, i have to calculate reluctances of air gap and core which depends on respective mean path lengths.
For the left hand core, the mean path length calculated in the solution manual is 1.11 meter . However what I don't understand is that why is the length of the air gap included in calculating mean path length of the core?

Don't we have to subtract the air gap length from the core so that true mean path length of the left core can be calculated?

 

[Charles Link]

A previous homework problem that was posted on Physics Forums along with the discussion might be helpful in answering any questions you might have with the above: https://www.physicsforums.com/threads/absolute-value-of-magnetization.915111/ $\\$ Also, in the future please use the homework template for any homework questions. The staff has the option of deleting the thread for not using the homework template.

 

[janu203]

A previous homework problem that was posted on Physics Forums along with the discussion might be helpful in answering any questions you might have with the above: https://www.physicsforums.com/threads/absolute-value-of-magnetization.915111/ $\\$ Also, in the future please use the homework template for any homework questions. The staff has the option of deleting the thread for not using the homework template.

I will be mindful in future but that thread doesn't answer my question either.

 

[Charles Link]

And an additional comment: From what I can tell, this method of solution, (see post 2 including the "link" to the Feynman lectures solution), needs an addition which isn't presented in the Feynman lectures, but is necessary when the problem contains a branch like the above. In that case, you need a 3rd region, in the middle section, where $H_{middle}=H_{left}+H_{right}$. I think with that addition, the problem should be entirely workable. Also in each of the two gaps, you need $H_{gap \, left}$, and $H_{gap \, right}$.

 

[Charles Link]

And to answer your question, in $\oint H \cdot dl$, for each section of $H$, the integral simply involves the approximate path through the material. It's not high precision. In this case, I believe you actually have two separate sections of the path in the material: $\int H_{middle} \cdot dl +\int H_{right} \cdot dl +\int H_{right \, gap} \cdot dl=NI$ is what the equation reads for the right loop. Similarly for the left loop. That along with $H_{middle}=H_{right}+H_{left}$.(The reason for this is that $B_{middle}=B_{left}+B_{right}$ because the lines of flux of $B$ are conserved and continuous). $\\$ Also: $\int H_{middle} \cdot dl=H_{middle} L_{middle}$, etc. $\\$ The Feynman type solution also assumes a continuous $B$ across the gap so that $H_{right}=B_{right}/\mu$ and $H_{right \, gap}=B_{right}/\mu_o$, etc. with $B_{right}$ being the same in both expressions.

 

[janu203]

And to answer your question, in $\oint H \cdot dl$, for each section of $H$, the integral simply involves the approximate path through the material. It's not high precision. In this case, I believe you actually have two separate sections of the path in the material: $\int H_{middle} \cdot dl +\int H_{right} \cdot dl +\int H_{right \, gap} \cdot dl=NI$ is what the equation reads for the right loop. Similarly for the left loop. That along with $H_{middle}=H_{right}+H_{left}$.(The reason for this is that $B_{middle}=B_{left}+B_{right}$ because the lines of flux of $B$ are conserved and continuous). $\\$ Also: $\int H_{middle} \cdot dl=H_{middle} L_{middle}$, etc. $\\$ The Feynman type solution also assumes a continuous $B$ across the gap so that $H_{right}=B_{right}/\mu$ and $H_{right \, gap}=B_{right}/\mu_o$, etc. with $B_{right}$ being the same in both expressions.

yeah now i can get a feel for it
tx a bundle