The MRI Signal and Reciprocity

Need to understand the MRI signal well enough to understand how we get an image, what it relates to, what properties will the signal have. Faraday's Law tells us that if you know that something is producing a magnetic field you can calculate how much flux it will generate through a coil, and measure a voltage. In MRI we measure the signal from the coil, and need to determine the magnetic fields (in our case spin distribution) that produces such a signal. How do we do that ???

• Consider two identical loops of wire ...

  
  
  
  
  
  
  
  
  
  
  
  
  
  

  
  

  $$\phi_{ji} I_{j} = \phi_{ij}I_{i}$$ (7.1)

  
  
  see reciprocity. If coil A can produce flux through coil B, then the reciprocal is also true that coil B can produce a flux through coil A.

• start from Faraday's Law, writing down what magnetic field the spins would produce, if we knew their distribution

• start with flux only, worry about time derivative later
  $$\begin{eqnarray*} \Phi_{M} &=& \int_{area\;of\;coil} \vec{B}_{M} \cdot d\vec{S}... ...M})}{\vec{x}_{coil}-\vec{x}_{M}} \cdot d\vec{S}_{coil} \right) \end{eqnarray*}$$
  
  
  
  
  

• this is still not so exciting, since we knew this, but through the miracle of mathematics (provided in the text) , we can turn this problem around to give the following
  $$\begin{eqnarray*} \Phi_{M}(t) = \int_{sample} d^{3}r \vec{{\cal B}}^{receive}(\vec{r}) \cdot \vec{M}(\vec{r}, t) \end{eqnarray*}$$
  
  
  
  
  

  
  
  
  
  
  
  
  

  often refer to the ability to invert such electromagnetic problems as ``principle of reciprocity''

• now have the signal as if the field produced by our detection coils is passing through the magnetization, giving rise to the flux. imagine, for a moment that field of the detection coil is uniform, and we can ignore the integral, ..., flux now looks like the magnetization.

• finally, to find the MR signal, need to find the emf, by taking the time derivative
  $$\begin{eqnarray*} emf &=& - \frac{d}{dt} \Phi_{M}(t) \nonumber\\ &=& - \frac {d... ...^{3} r \vec{M}(\vec{r},t) \cdot \vec{\cal B}^{receive}(\vec{r}) \end{eqnarray*}$$
  
  
  
  
  
  Note: $\cal B$ implies magnetic field/current, or magnetic field per unit current (IN-CLASS QUESTION)