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. $3\sqrt{{(F_3-F_1)}^2-F{{}_2}^2}$
• B. $4F_3-F_1-F_2$
• C. $5F_3-F_1-F_2$
• D. $\sqrt{{(F_3-F_1)}^2+F{{}_2}^2}$

Answer 1

The correct option is D $\sqrt{{(F_3-F_1)}^2+F{{}_2}^2}$

Total force on the current carrying closed loop

should be zero, if placed in uniform magnetic
field.
$F_{horizontal}=(F_3-F_1)$
$F_{vertical}=F_2$
Resultant of $\overrightarrow{F_1},\overrightarrow{F_2}and\overrightarrow{F_3}is\overrightarrow{F}$
where $F=\sqrt{(F_3-F_1{)}^2+F_2^2}$
Since total force = 0, hence force on QP is equal
to $\overrightarrow{F}$ in magnitude but opposite direction.
$F_{QP}=\sqrt{(F_3-F_1{)}^2+F_2^2}$