Physics - Geometrical Optics Question with Solution | TestHub

Force on a conductor of length $\ell$ due to a magnetic field of strength $\overrightarrow{\mathrm{B}}$ is given by $\begin{array}{l}\overrightarrow{\mathrm{F}}=\mathrm{I}(\vec{\ell }\times \overrightarrow{\mathrm{B}})\\ \therefore |\overrightarrow{\mathrm{F}}|=\mathrm{I}\ell \mathrm{B}\sin \theta \end{array}$ Here, $\theta =90^{\circ }$ $\therefore \mathrm{F}=\mathrm{I}\ell \mathrm{B}=\mathrm{I}\times \pi \mathrm{R}\times \mathrm{B}=\pi \mathrm{RIB}$ Therefore, a force of magnitude $(\pi RIB)$ will act on each wire. The direction of the forces on each wire will be same. Thus, total force on the wire $\mathrm{AB}=\pi \mathrm{RIB}+\pi \mathrm{RIB}$ $=2\pi \mathrm{RIB}$