I have a confusion regarding MRI signal formation. So, as I understand it, we need to solve the Bloch equations for the excitation and the relaxation stages:
So, in the excitation stage, let us assume that we have an RF field excitation which is applied along with the slice selection gradient. The bloch equations are given by:
$$ \frac{dM}{dt} = \gamma M \times B - R $$
where $R$ is the vector accounting for the relaxation effects. Now, during excitation, the B1 field is on and as I understand it, the B
vector is given as:
\begin{align} B_x &= B_1 \cos(\omega t)\\ B_y &= B_1 \sin(\omega t)\\ B_z &= z G_z \end{align}
The $B_z$ field is basically the gradient times the position offset from the origin and the $B_x$ and $B_y$ fields are the real and imaginary parts of the B1 waveform.
Now, when the B1 field is turned off, I think $B_x$ and $B_y$ components should be 0. However, let us assume that we have a readout gradient on when we are sampling the MR signal.
What will contribute to the $B_z$ field now? Assuming that the RF excitation is with a 90 degree pulse and after the excitation the magnetization vector lies completely along the $Y$ axes.