# Does the amplitude of the RF pulse affect the NMR signal?

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?

• 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). May 6 at 11:20

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.