energy radiated away. The bolometric bremsstrahlung luminosity of a CDAFs is written as: $$L_c=4\pi R^{3}_{s} \int_{1}^{R_{out}} F_{bol} R^{2} dR$$ by equating this relation to $L_c=\eta_c \epsilon_c \dot{M} c^{2}$ and after some calculations we obtain : $$\rho_0=0.37 (\frac{\eta_c \epsilon_c }{10^{-2}})\frac{10^{-3}}{m R^{2s-\frac{1}{2}}_{out}}\frac{(2s-\frac{1}{2})c_4}{c^{2}_0\sqrt{c_3}}$$

Therefor, by substituting the above expressions for the density $\rho_0$ and temperature of the gas in equation 56, we find the bremsstrahlung spectrum from a CDAFs as follows: $$\nu L_\nu=0.4 \times 10^{19}m  \frac{(\eta_c \epsilon_c)^{2}}{R^{4s-1}_{out}}\frac{c_4}{c^{2}_0 c^{\frac{3}{2}}_3}(2s-\frac{1}{2})^{2}$$ $$\int_{1}^{R_{out}} \nu R^{2s-\frac{1}{2}} \exp{(-\frac{1.7\times 10^{-23}\nu R}{c_3})} dR$$ We assume that the accretion flow extends from an outer radius $R_{out}$ down to an inner radius $R_{in}=1$. The bremsstrahlung emission can arises from all radii in the flow in contrast to synchrotron emission and modified Comptonization processes. As we have introduced above $\nu L_{\nu} \propto \nu \exp(\frac{-h\nu}{KT})$, by differentiating respect to $\nu$ from this expression, we can obtain the peak of the bremsstrahlung spectrum occurs at: $$h \nu_{peak}=K T=K \frac{T_0 c_3}{R}$$

As we can see in this equation, the peak of the spectrum is located in the X-ray frequency band.

=3.6in =2.3in

=3.6in =2.2in

=3.6in =2.2in

=3.6in =2.2in

=3.6in =2.2in

=3.6in =2.8in

=3.6in =2.8in

Results
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In this section, we will investigate numerically the role of magnetic 