\rightarrow \infty \right) $, because ${\it \Pi }_{{\it 0}}=0$, it is straightforwardly obtained from (3.24) that $C_{5}=0$ and no long-range mode propagates.

At $T\neq 0$, taking into account that $C_{5}=0$ along with eq. (3.25) in the magnetic field solution (3.21), we can write the magnetic field penetration as

$$B\left( x\right) =\overline{B}_{1}\left( T\right) \left( e^{-(x+L)\xi
_{1}}+e^{\left( x-L\right) \xi _{1}}\right) +\overline{B}_{2}\left( T\right)
\left( e^{-(x+L)\xi _{2}}+e^{\left( x-L\right) \xi _{2}}\right)  \eqnum{4.4}$$

where,

$$\overline{B}_{1}\left( T\right) =\frac{\gamma _{1}}{\gamma _{1}-\gamma _{2}}%
\overline{B},\text{ \qquad \quad }\overline{B}_{2}\left( T\right) =\frac{%
\gamma _{2}}{\gamma _{2}-\gamma _{1}}\overline{B}  \eqnum{4.5}$$

For densities $n_{e}\ll m^{2}$, the coefficients $\overline{B}_{1}$and $%
\overline{B}_{2}$ can be expressed, in the low temperature approximation $%
\left( T\ll \omega _{c}\right) $, as

$$B_{1}\left( T\right) \simeq -\frac{\left( \pi n_{e}\right) ^{3/2}}{m^{2}}%
\left[ 1/m+\frac{5}{2}\beta \exp -\left( \frac{\pi n_{e}\beta }{2m}\right)
\right] \overline{B},\qquad  \eqnum{4.6}$$

$$B_{2}\left( T\right) \simeq \left[ 1+\frac{5\pi n_{e}}{2m^{2}}\sqrt{\pi n_{e}%
}\beta \exp -\left( \frac{\pi n_{e}\beta }{2m}\right) \right] \overline{B} 
\eqnum{4.7}$$

Hence, in the infinite-strip sample the applied magnetic field is totally screened within the anyon fluid on two different scales, $\lambda _{1}=1/\xi
_{1}$ and $\lambda _{2}=1/\xi _{2}$. At $T=200K$, for the density value considered above, the penetration lengths are given by $\lambda _{1}\simeq
0.6\times 10^{-8}cm$ and $\lambda _{2}\simeq 0.1\times 10^{-4}cm$ . Moreover, taking into account that $\xi _{1}$ 