Module 3 : MAGNETIC FIELD
Lecture 14 : Force on a Current Carrying Conductor

\includegraphics{fig3.5.eps}

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}

Example 1         Example 2         Example 3

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