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let's consider this picture in which a cylindrical resonant cavity is excited through a waveguide.

enter image description here

I have some questions:

1) Why is it called "Magnetic Coupling"? I thought this excitation simply was the propagation (through the slot) of the wave already present in the waveguide towards the resonant cavity

2) Why is there a magnetic loop in the last two pictures? what are its terminals connected to? Should be present also in the first two pictures?

Reference: here.

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  1. It is called magnetic coupling because the field that is present at the slot into the waveguide is primarily via the magnetic field. There may be some energy coupled into the resonator by the E-field, but it will be significantly less than energy coupled by the magnetic field.
  2. Look at the heading for the bottom two pictures. There are loops in the bottom two pictures because that's about how you magnetically couple a coaxial cable to a resonator, it's no longer about coupling a resonator to a waveguide.
  3. Look at the two pictures on the bottom, that show that the loop is connected to the center conductor and shield of the coax.
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  • \$\begingroup\$ Regarding 2, so does it mean that the loop is connected to the inner conductor and the shield of the coaxial cable? \$\endgroup\$
    – Kinka-Byo
    Sep 30, 2019 at 16:38
  • \$\begingroup\$ just to mention, the slot is very small compared to the wavelength and the dimensions of either the guide or resonator, just as the loop is very small. This is why it isn't simply propagation of the wave. The loop is connected as you say. I guess it's so obvious from the way it's drawn, the author doesn't bother to mention it. \$\endgroup\$
    – Neil_UK
    Sep 30, 2019 at 16:39
  • \$\begingroup\$ Do you mean that there are some phenomena like diffraction? \$\endgroup\$
    – Kinka-Byo
    Sep 30, 2019 at 16:48
  • \$\begingroup\$ I won't say that diffraction cannot be a useful concept here, but it would be difficult to make it meaningful. Basically, in a typical microwvave waveguide (and especially in coax) the dimensions are all fractions of a wavelength. There is no free-space transmission at all, so wedging what's actually happening into an analysis that appeals to diffraction would be exceedingly difficult. \$\endgroup\$
    – TimWescott
    Sep 30, 2019 at 18:48

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