Query about a protons magnetic moment precession during excitation in MRI I am trying to learn more about the physics behind an MRI and am getting stuck trying to understand the process behind the excitation of the transverse component of the magnetisation.
From what I understand so far, you have a very strong magnetic field $B_0$ on the order of teslas which aligns the spins in a patient. Due to QM spins can never completely align with the field and so are stuck at an angle, therefore they precess at the larmor frequency.
You can apply an external field $B_1$ which is alternating at the larmor frequency and allow the protons to transition to the next highest energy level (the spin will point in the opposite orientation).
In the article I was reading online it said that the spin precesses about this new field $B_1$. Why is this? $B_1$ is much weaker than $B_0$? And I would've thought that the spin would only precess about the net external field, not just a particular component?
I am confused about why the transition to the next energy state looks like a downward spiral ( the first gif in this article ):
https://www.imaios.com/en/e-Courses/e-MRI/NMR/Excitation
 A: If you consider B0 acting along the z axis then we know the there will be precession about this axis due to the component of the magnetic moment in the x y plane. This precession occurs at the larmor frequency.
If you apply a B1 field that acts orthogonal to B0 and rotates at the same rate w about the z axis, then the component of the magnetic moment along the z axis will give rise to a torque that causes precession about B1. This is why there is a downwards spiral motion.
There is a precession about X-Y due to B0 and there is precession about B0 due to the fact the torque is coming from the longitudinal component of the magnetic moment.
A: We're are dealing with magnetic fields which are vector quantities so you have to consider also the direction and not only the module. The excitation field, $B_1$, must be orthogonal to the direction of the $\mathbf{B_1} × \mathbf{μ} \ne 0$, to generate transitions of the nuclear spin state.
To convince you that a small magnetic field is able to excite the spins consider the system in a rotating frame.
In this frame the axes rotate with an angular velocity $\omega_{rot}$. The apparent frequency of the Larmor precession in such a frame will be
$\omega_0 - ω_{rot} = \Omega$.
Seen by this rotating frame the original external field $B_0$ is reduced to $B_{red}$
$B_{red} = -\frac{\Omega}{\gamma}$
If we take an $\omega_{rot} = \omega_{tx}$, (tx= transmitter) $B_1$ appears to be stationary.
We've to consider both fields. The vector sum of $B_{red}$ and $B_1$ is called effective field $B_{eff}$
$B_{eff} = \sqrt{B_1 + B_{red}}$
The key fact is that, although $B_0$ is very much larger than $B_1$, we can eliminate the effect of the $B_0$ field by
setting the transmitter frequency close to the Larmor frequency i.e. by making the offset small. With this condition the reduced field $B_{red}$ is small and it is then possible for the small $B_1$ field to begin to exert an influence. In the limit that the offset is zero ($\Omega = 0$), $B_{red}$ disappears and the $B_1$ field is the only one left in the rotating frame.


*What is depicted in the animation seems to be the decay of the magnetization vector after a 180° pulse with $B_0$ oriented along the negative $z$. The spiral is caused by the decay of the $M_{xy}$ component of the magnetization with a time constant $T_2$ and the $M_z$ component with a time constant $T_1$
