MyOpenMath/Solutions/Toroid inductance

https://openstax.org/books/university-physics-volume-2/pages/14-2-self-inductance-and-inductors#CNX_UPhysics_31_02_Toroid

is a link to a section OpenStax University Physics that discusses the rectangular toroid. This discussion reinforces a number of concepts that are important for scientists and engineers to grasp. One concept is the construction of an integral over a mathematically defined volume. In the textbook the magnetic field was integrated. Here we reinforce that skill by asking students to calculate the volume of a toroid. The serious student should strive to understand every step in this derivation, because methods like this might be required in more advanced courses.



The toroidal inductor gives us the opportunity to:
 * 1) Learn how calculus can be used to find the volume and area of objects.
 * 2) Use Amphere's law to find the magnetic field for certain geometries with high symmetry.
 * 3) Calculate the self inductance of certain coils.

Volume
This is a good example of how a double integral can be used to find the volume of shapes with a high degree of symmetry. Let $$V$$ denote a volume and $$A$$ denote an area, with the understanding that $$dV$$ denote a volume and $$dA$$ are differentials that can be integrated as the limit of Reimmann sums. The figure to the right resembles that of a rectangular washer that had its top surface shaved down. The rectangular area shaded in yellow is an area differential:


 * $$dA=h\,dr$$

where $$h=h(r)$$ is a function of $$r$$. For example, the function might be $$h(r)=Cr^{-\alpha},$$ where $$C$$ and $$\alpha$$ are constants. This function is defined with a range defined by two other constants $$(r_1,r_2)$$:


 * $$r_1<r<r_2$$

The figure shows two identical cross-sectional areas that can be integrated as, $$\int_{r_1}^{r_2}Cr^{-\alpha}\,dr$$, but this step would be premature. Instead we sweep the rectangle of differential (small) area $$h\,dr$$ around the circumference $$2\pi$$ to obtain the differential volume:


 * $$dV=2\pi r\,dA = 2\pi r\,h(r)dr$$


 * $$V=2\pi C\int_{r_1}^{r_2}r^{-\alpha+1}dr$$

Note how this technique offers a simple proof for the area of a circle, assuming that the circumference is known to be 2&pi;r.