Tip angle in NMR

Question

Tipping the magnetisation $\vec{M}$ with a $\vec{B}_1$ field for the time $\tau$, results in a tip angle of $\alpha = \omega \tau$ where $\omega$ is the frequency of the $\vec{B}_1$ field. I think I am mixing up, something but I am not sure what. Where is the condition, that $\omega = \omega_0$ for tipping $\vec{M}$ in the xy-plane? This would be reached by a $\frac{\pi}{2}$ pulse.($\omega_0$ is the frequency of the static $\vec{B}_0$ field).

And what would be the tip angle in CW (continuous wave NMR)? Is there a tip angle or does it always tip in the xy plane?

Answer 1

I'm not sure I fully understand your question, but the following may be useful.

In NMR experiments, the $B_{1}$ field needs to actually be at the radio frequency of $\omega_{0}$. This frequency $\omega_{0}$ is determined by the static $B_{0}$ field, in particular $\omega_{0}=\gamma B_{0}$. Once you have the $B_{1}$ field operating at the frequency $\omega_{0}$, you are able to "tip" the magnetization since you are in the resonance condition. The tipping rate will be proportional to the magnitude of $B_{1}$ but not the frequency of $B_{1}$. The tip angle itself is then proportional to the product of the magnitude of $B_{1}$ and the time $\tau$.

I hope this helps.