show that the coefficients $ a(\phi), b(\phi)$, corresponding to the electro-magnetic scatterers, in the conductivity formula (Eq.\[eq:sigma\]), are obtained as $$\begin{aligned}
\label{eq-13}
  a(\phi)&=& \frac{2}{3\pi K}\cos\phi, \nonumber\\
  b(\phi)&=& \frac{2}{\pi K}\sin\phi,\end{aligned}$$ where the prefactor $K$, can be obtained from the Fermi’s golden rule as

$$\begin{aligned}
\label{eq-38}
   K = w(\phi,\phi')|\langle \vec{k}'|\hat{V}/V_0| \vec{k}\rangle|^{-2}\,.\end{aligned}$$

in which $$\begin{aligned}
\label{eq-38}
  w(\phi,\phi')= \displaystyle
  \frac{1}{(2\pi)^2}\int_0^\infty k'\, dk' w(\vec{k},\vec{k}'),\end{aligned}$$

Finally from the calculated expressions of $ a(\phi)$ and $b(\phi)$ via Eq. \[eq-13\] and using the Boltzmann Eq. \[2dani\], we obtain the following form for the non-equilibrium distribution function;

$$\label{eq-14}
  f(\phi,\theta)\!=\!f_0 - ev{\cal E}(-\partial_\epsilon f_0) \frac{2}{\pi K}
  \bigg[\frac{1}{3}\cos\theta\cos\phi + \sin\theta\sin\phi\bigg]\,.$$

The Anisotropic Magnetoresistance (AMR) effect is the difference in electrical conductivities, depending on whether the scatterer’s magnetic moments are parallel or anti-parallel to the direction of the current, that is defined as follows

$$\label{eq-19}
  \mbox{AMR} = \frac{\sigma_{xx}-\sigma_{yy}}{\sigma_{xx}+\sigma_{yy}}$$

Interestingly, for each value of $\alpha$, the AMR is constant and is chemical potential independent for both $n$ and $p$ doped borophane. For example for $\alpha=0.5$, AMR is about $0.805$.

Numerical Results and Discussion {#sec:results}
================================

In this section, our numerical results for the thermoelectric transport in borophane are presented. It should be noted that, except for Fig. \[fig9\] (that is related to the 