Coulomb scattering probability for ions is much larger than fusion.

Deflection due to scattering can lead to bremsstrahlung radiation power losses, lowering plasma temperature.

Differential Scattering Cross-section

Force equation: $F_c = \frac{1}{4\pi\varepsilon_0} \frac{q_a q_b}{r^2}$
To specify cross-section, use impact parameter $r_0$ and scattering angle $\theta_s$.
In center of mass system, scattering is simple antiparallel motion.
In laboratory system, $\theta_s = \theta_c$.
Area of axisymmetric ring: $dA = 2\pi r_0 dr_0$
Solid angle of scattering: $d\Omega^* = 2\pi \sin(\theta_c) d\theta_c$
Differential scattering cross-section: $\sigma_s(\theta_c) = \frac{d\sigma}{d\Omega^*}$
Total scattering cross-section: $\sigma_s = \int_0^{4\pi} \sigma_s(\theta_c) d^2\Omega = \int_0^\pi \sigma_s(\theta_c) d\Omega^*$
Relationship between $r_0$ and $\theta_c$: $\tan\left(\frac{\theta_c}{2}\right) = \frac{q_a q_b}{4\pi\varepsilon_0 m_r v_r^2 r_0}$
Differential cross-section for Coulomb scattering:
$\sigma_s(\theta_c) = \frac{K^2}{4\sin^4(\theta_c/2)} = \left(\frac{q_a q_b}{4\pi\varepsilon_0 m_r v_r^2}\right)^2 \frac{1}{\sin^4(\theta_c/2)}$
As $r_0 \rightarrow 0$ (head-on collision), $\sigma_s(\theta_c) \rightarrow K^2/4$.

Debye Length

Plasma is globally neutral but has local charge variations.
These variations establish electric potential and field.
Thermal motion of ions and electrons influences these effects.
The spatial extent of this effect is represented by Debye length.
Plasma self-shields from applied electric fields; effects are confined to immediate surroundings.
In hot plasma, ions/electrons have kinetic energy, allowing some particles to escape.
Local electric field due to charge density: $\nabla \cdot \mathbf{E} = \frac{\rho^c}{\varepsilon_0}$
Scalar potential function: $\mathbf{E} = -\nabla\Phi$
Poisson's equation: $\nabla \cdot (-\nabla\Phi) = \frac{\rho^c}{\varepsilon_0}$
Introducing definition: $\rho^c = \sum_{j=e,i} q_j N_j$
For electrons, $N_e = C \exp\left(-\frac{q_e\Phi}{kT_e}\right)$ (Boltzmann equation)
Simplified differential equation: $\nabla^2\Phi - \frac{\Phi}{\lambda_D^2} = 0$
Debye length ($\lambda_D$): $\lambda_D = \sqrt{\frac{\varepsilon_0 k T_e}{q_e^2 N}}$
For shielding ions by electrons: $\lambda_{De} = \sqrt{\frac{\varepsilon_0 k T_e}{q_e^2 N_e(r\rightarrow\infty)}}$
For screening electrons by ions: $\lambda_{Di} = \sqrt{\frac{\varepsilon_0 k T_i}{q_i^2 N_i(r\rightarrow\infty)}}$
In equilibrium, for p, d, or t: $T_e = T_i = T$ and $N_i(r\rightarrow\infty) = N_e(r\rightarrow\infty) = N$.
Simplified Debye length: $\lambda_D = \sqrt{\frac{\varepsilon_0 k T}{q_e^2 N}}$
Potential in plasma: $\Phi_{plasma} \propto \frac{\exp(-r/\lambda_D)}{r}$
Free space potential: $\Phi_{free\ space} \propto \frac{1}{r}$
$\Phi$ is attenuated in plasma depending on $\lambda_D$.

Scattering Limit

To avoid singularity at $\theta_c \rightarrow 0$, consider a finite non-zero bound $\theta_{min}$.
The value of $\theta_{min}$ must be chosen such that Coulomb scattering is relatively small.
For plasma, $r_{0,max} = \lambda_D$.
Minimum angle: $\theta_{min} = 2 \tan^{-1}\left(\frac{K}{\lambda_D}\right)$
The scattering cross-section is always higher than fusion cross-section for d-t for any given particle energy.
Thus, the more likely reaction between d and t is scattering, not fusion.

Bremsstrahlung Radiatio