- Magnetic Susceptibilities

Magnetic Susceptibilities

We have seen that an assembly of nuclear magnets in a steady magnetic field $H_{_{0}}$ absorbs power from a suitably applied RF field.

We know from basic E&M theory (macroscopic formulation of Maxwell's equations) that absorption is associated with the imaginary part of the susceptibility $\chi$. In our case, $\chi =\chi ^{'}+i\chi ^{''}$ is the complex nuclear magnetic susceptibility, where $\chi ^{'}$ is the real part and is associated with dispersion while $\chi ^{''}$ is the imaginary part and is associated with absorption.

---

We will first derive the static (no transverse RF field) magnetic susceptibility $\chi _{_{0}}$.

Consider an assembly of identical weakly interacting nuclei of spin number $I$, in thermal equilibrium at a spin temperature $T_{_{S}}$ in a steady magnetic field $H_{_{0}}$. The nuclei having quantum number $m$ are found to be in the energy level $-m\mu H_{_{0}}/I$. The population of this level is then weighted by the Boltzmann factor $e^{\tfrac {m\mu H_{_{0}}}{IkT_{_{S}}}}\approx 1+{\tfrac {m\mu H_{_{0}}}{IkT_{_{S}}}}$. This approximation is very good for most practical conditions. Hence the population of each level is given by

$N(m)={\tfrac {N}{2I+1}}(1+{\tfrac {m\mu H_{_{0}}}{IkT_{_{S}}}})$.

The total magnetic moment, the magnetization ${\mathcal {M}}$, is given by

${\mathcal {M}}=\sum _{-I}^{I}N(m)m\mu /I={\tfrac {N\mu ^{2}H_{_{0}}}{I^{2}(2I+1)kT_{_{S}}}}\sum _{-I}^{I}m^{2}={\tfrac {N\mu ^{2}H_{_{0}}(I+1)}{3kT_{_{S}}I}}$,

since $\sum _{-I}^{I}m^{2}={\tfrac {1}{3}}I(I+1)(2I+1)$.

The static magnetic susceptibility is then given by

$\chi _{_{0}}={\tfrac {\mathcal {M}}{H_{_{0}}}}={\tfrac {N\mu ^{2}(I+1)}{3kT_{_{S}}I}}$.

---

Now consider the RF magnetic susceptibility (the susceptibility with the transverse field applied). Let's find the power absorbed by the system in terms of $\chi ^{''}$.

Consider a transverse magnetic field $H=2H_{_{1}}cos(2\pi \nu t)$.