Q.
A closed loop $PQRS$ carrying a current is placed in a uniform magnetic field. If the magnetic forces on segment PS, SR and RQ are $F_{1},F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is
A) $\sqrt{\left( F_{3} - F_{1} \right)^{2} - F_{2}^{2}}$
B) $F_{3} + F_{1} - F_{2}$
C) $F_{3} - F_{1} + F_{2}$
D) $\sqrt{\left( F_{3} - F_{1} \right)^{2} + F_{2}^{2}}$
Explanation
(d) Since all the given forces are lying in plane, so the given loop is in equilibrium
[image:4_image93.emf]
$F_{4}^{''} = F_{4}\cos\phi = F_{2}$
$F_{4}^{''''} = F_{4}\sin\phi = F_{3} - F_{1}$
$\Rightarrow F_{4}^{2} = F_{2}^{2} + \left( F_{3} - F_{1} \right)^{2}$
$\Rightarrow F_{4} = \sqrt{F_{2}^{2} + \left( F_{3} - F_{1} \right)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.