QJEEAdv15Ph7f41: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph7f41 Solution: A conductor (shown in the figure) carrying constant current I…

The force on a conducting element of length $d\overrightarrow l$ , carrying a current I in a magnetic field $\overrightarrow B$ , is given by d $\overrightarrow F$ = I d $\overrightarrow l$ $\times$ $\overrightarrow B$ . If the field is uniform, the total force on the conductor is given by

$\overrightarrow F = \int_a^g {Id\overrightarrow l \times \overrightarrow B = I\left( {\int_a^g {d\overrightarrow l } } \right) \times \overrightarrow B = I\,\overrightarrow {ag} \times \overrightarrow B }$

Note that $\overrightarrow B$ is taken out of the integral sign because it is constant. The vector $\overrightarrow {ag} = 2(L + R)\widehat x$ . The magnetic forces on the conductor in the given cases are

Case (A) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat z)$

$= - 2IB(L + R)\widehat y$

Case (B) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat x) = \overrightarrow 0$

Case (C) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat y)$

$= 2IB(L + R)\widehat z$