Thus, the magnetic force on that section of wire will be N times the force on a single electron: \[\begin{aligned} \vec F = N\vec F_e = nAl (-e \vec v_d \times \vec B)=-nAle \vec v_d \times \vec B...Thus, the magnetic force on that section of wire will be N times the force on a single electron: →F=N→Fe=nAl(−e→vd×→B)=−nAle→vd×→B Recall the microscopic model of current to relate the drift velocity to the conventional current in the wire: I=−nAevd where the minus sign indicates that negative electrons flow in the opposite direction from the conventional current.
