SKEDSOFT

Physics For Engineers - 2

Force on a Current Carrying Conductor: A conductor has free electrons which can move in the presence of a field. Since a magnetic field exerts a force $q\vec v\times\vec B$on a charge moving with a velocity $\vec v$, it also exerts a force on a conductor carrying a current.

Consider a conducting wire carrying a current $I$. The current density at any point in the wire is given by'

\begin{displaymath}\vec J = ne\vec v\end{displaymath}
where $n$is the number density of electrons having a charge $e$each and $\vec v$is the average drift velocity at that point.
Consider a section of length $dl$of the wire. If $A$is the cross sectional area of the section oriented perpendicular to the direction of $\vec J$, the force on the electrons in this section is
\begin{displaymath}d\vec F = dq\vec v\times\vec B\end{displaymath}
where $dq$is the amount of charge in the section
\begin{displaymath}dq = neAdl\end{displaymath}

Thus the force on the conductor in this section is  \begin{displaymath}d\vec F = neA dl \vec v\times\vec B\end{displaymath}

If $\vec{dl}$represents a vector whose magnitude is the length of the segment and whose direction is along the direction of $\vec v$, we may rewrite the above as

\begin{eqnarray*} d\vec F &=& neAv\vec{dl}\times\vec B\\ &=& JA \vec{dl}\times\vec B\\ &=& I \vec{dl}\times\vec B \end{eqnarray*}

The net force on the conductor is given by summing over all the length elements. If $\hat u_l$denotes a unit vector in the direction of the current, then  \begin{displaymath}\vec F = I \int \hat u_l\times\vec B dl\end{displaymath}