The magnetic force on a current
Now we are ready to find the force on a current-carrying wire in a magnetic field. The current consists of charged particles moving with the velocity $\FLPv$ along the wire. Each charge feels a transverse force \begin{equation*} \FLPF=q\FLPv\times\FLPB \end{equation*}  (Fig. 13–5a). If there are $N$ such charges per unit volume, the number in a small volume $\Delta V$ of the wire is $N\,\Delta V$. The total magnetic force $\Delta\FLPF$ on the volume $\Delta V$ is the sum of the forces on the individual charges, that is, \begin{equation*} \Delta\FLPF=(N\,\Delta V)(q\FLPv\times\FLPB). \end{equation*} But $Nq\FLPv$ is just $\FLPj$, so \begin{equation} \label{Eq:II:13:9} \Delta\FLPF=\FLPj\times\FLPB\,\Delta V \end{equation}  (Fig. 13–5b). The force per unit volume is $\FLPj\times\FLPB$. If the current is uniform across a wire whose cross-sectional area is $A$, we may take as the volume element a cylinder with the base area $A$ and the length $\Delta L$. Then \begin{equation} \label{Eq:II:13:10} \Delta\FLPF=\FLPj\times\FLPB A\,\Delta L. \end{equation}  Now we can call $\FLPj A$ the vector current $\FLPI$ in the wire. (Its magnitude is the electric current in the wire, and its direction is along the wire.) Then \begin{equation} \label{Eq:II:13:11} \Delta\FLPF=\FLPI\times\FLPB\,\Delta L. \end{equation}  The force per unit length on a wire is $\FLPI\times\FLPB$. This equation gives the important result that the magnetic force on a wire, due to the movement of charges in it, depends only on the total current, and not on the amount of charge carried by each particle—or even its sign! The magnetic force on a wire near a magnet is easily shown by observing its deflection when a current is turned on, as was described in Chapter 1 (see Fig. 1–6).