Spin Polarization of ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$ in the Heavy-Ion Reaction ${}_{}{}^{100}{}_{}{}^{}\mathrm{Mo}$(${}_{}{}^{14}{}_{}{}^{}\mathrm{N}$, ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$)${}_{}{}^{102}{}_{}{}^{}\mathrm{Ru}$

Polarization of product ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$ in the reaction ${}_{}{}^{100}{}_{}{}^{}\mathrm{Mo}$(${}_{}{}^{14}{}_{}{}^{}\mathrm{N}$, ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$)${}_{}{}^{102}{}_{}{}^{}\mathrm{Ru}$ at 90 MeV was measured as a function of product kinetic energy at the reaction angle of 20°. The polarization was found antiparallel to the vector ${\overrightarrow{\mathrm{k}}}_f\times {\overrightarrow{\mathrm{k}}}_i$ for higher-energy ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$, while parallel for those with lower energy. The zero crossing occurred at around maximum in the energy spectrum.