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I know that the NMR signal depends on the static magnetic field strength $ B_0 $ (by Boltzmann equation) and its gyromagnetic factor. While NMR occurs when the RF pulse frequency matches the equation: $$ \nu = \frac{g}{h}\mu_{n}B_{0} $$ what about its amplitude? Does the amplitude of the RF pulse affect the NMR signal? For example, does it affect the signal strength?

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    $\begingroup$ Please be a little more clear. You write about "signal amplitude" and "pulse amplitude" in different sentences, but sometimes its not 100% clear which one you mean (such as immediately after the equation). $\endgroup$ – Buck Thorn May 6 at 11:20
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V. approx explanation. The rf pulse is at 90$^\text{o}$ to the static field $B_0$ and has the effect of moving the magnetisation (sum of spin vectors) away from alignment along $B_0$ (z-axis normally) into the xy plane where it rotates and cuts through detector coils and induces FID signal. The FID can only be detected with a magnetisation component in the xy plane. The size of the rf pulse's magnetic field $B_I$ is proprotional to angle $\theta$ moved away from z-axis as $B_I\sim \tan(\theta) $ and so via magnetisation's component in xy plane (arrow on axis in sketch) to the size of the FID.

nmr-fid

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