# Tip angle in NMR

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?

## 1 Answer

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.