[NOTES/EM-07016] Force on a Line and Volume Element of a Wire

Force on a line element of a wire:

Suppose a wire of cross section area \(A\), and carries a steady current $I$. Let the density of charge carriers per unit length be \(\lambda\). Then we have

The total charge of charge carriers in element \(d\ell = \lambda d\ell\).

• Then total charge flowing across, by a point in time $\Delta t$ is $ \Delta q =\lambda (v\Delta t)$. This quantity, by definition of current is \(I\Delta t\). Therefore, the current \(I\) is related to the charge density \(\rho\) by \(I = \lambda v\)
• Also $\vec v (d\ell)= v \overrightarrow{dl}$ because both these quantities have the same magnitude and have the same direction at every point of a wire as that of the tangent to the wire. Therefore, \begin{equation} \lambda (v\Delta t) = I \Delta t \Rightarrow I=\lambda v \end{equation}

Using the above statements, we get the force, \(\vec F\), a on line element
\(\overrightarrow{d\ell}\) as
\begin{eqnarray}
\vec F &=& \Delta q\vec{v}\times\vec{B}\\
&=& (\lambda d\ell) (\vec v\times \vec B)\\
&=&(\lambda v\overrightarrow {d\ell})\times \vec B.
\end{eqnarray}
Therefore, we get the force as  \[\vec F = I\overrightarrow{dl} \times\vec{B}.\]

Force on a volume element:

\noindent For a volume distribution, we need on the force on a small volume $\Delta V$ is
\begin{equation}
d\vec F
= (\rho dV) \vec v \times \vec B\\
= \vec j \times \vec B \, (\Delta V)
\end{equation}
Thus the force per unit volume is \(\vec j \times \vec B\).