Force On A Current-carrying Conductor

From NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), a straight conductor of length $L$ carrying current $I$ in a magnetic field $B$ experiences a force $F=BIL\sin \theta$, where $\theta$ is the angle between the current direction and the field. This macroscopic force is the collective effect of the Lorentz force on all the moving charges inside the wire. Since the wire is perpendicular to the field, $\theta =90^{\circ }$ and $\sin \theta =1$. Substituting $B=0.6$ T, $I=5$ A, $L=0.4$ m: $F=(0.6)(5)(0.4)=1.2$ N. The value $0.6$ N halves the answer, as if $L=0.2$ m. The value $12$ N is ten times too large from a decimal-point error. The value $0.12$ N is ten times too small for the same reason. Plausibility check: the units combine as $\text{T}\cdot \text{A}\cdot \text{m}=\text{N}$, and a force of about a newton on a short current-carrying wire in a strong field is entirely reasonable.