Question

A closed-loop $PQRS$ carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments $PS$ , $SR$ 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. $\sqrt{\left(F_{3} - F_{1}\right)^{2} + F_{2}^{2}}$

Answer 1

Answer: (D)

$\because $ Net force on current carrying closed loop is zero.
$\overset{ \rightarrow }{F}_{1}+\overset{ \rightarrow }{F}_{2}+\overset{ \rightarrow }{F}_{3}+\overset{ \rightarrow }{F}_{P Q}=0$
$\Rightarrow \left(\overset{ \rightarrow }{F}\right)_{P Q}=-\left(\left(\overset{ \rightarrow }{F}\right)_{1} + \left(\overset{ \rightarrow }{F}\right)_{2} + \left(\overset{ \rightarrow }{F}\right)_{3}\right)$
$\overset{ \rightarrow }{F}_{1}+\overset{ \rightarrow }{F}_{3}$ is perpendicular to $\overset{ \rightarrow }{F}_{2}$ .
$\therefore F_{P Q}=\sqrt{\left(F_{3} - F_{1}\right)^{2} + F_{2}^{2}}$