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

The length element $dl = Rd\theta$. Since the current direction opposite to the direction of the tangential unit vector $\hat u_\theta$, the vector $\vec{dl}$is

\begin{displaymath}\vec{dl} = -Rd\theta\hat u_\theta\end{displaymath}

Thus the force is

\begin{eqnarray*} d\vec F &=& I\vec{dl}\times\vec B\\ &=& -IRd\theta \hat u_\theta\times (-B\hat k)\\ &=& IRB\hat u_r d\theta \end{eqnarray*}

The force on the semicircular arc is obtained by integrating the above. By symmetry, the force on a current element situated at a point Q on the second quadrant, situated symmetrically with P, has the same magnitude and is directed along the radial direction at Q. On resolving the forces along the x and y directions, the components parallel to the x-axis cancel for such symmetric pairs while the component parallel to the y- axis add up. Thus

\begin{displaymath}\vec F = 2IRB\int_0^{\pi/2}\hat\jmath \sin\theta = 2IRB\hat\jmath\end{displaymath}

The magnitude of the force is seen to be equal to the product of $IB$with the diameter of the semi-circle.

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