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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 :
($F_1$ acts towards the left on PS, $F_3$ acts towards the right on RQ and $F_2$ acts downward on SR.)
($F_1$ acts towards the left on PS, $F_3$ acts towards the right on RQ and $F_2$ acts downward on SR.)
A
$\sqrt{(F_3 - F_1)^2 - F_2^2}$
B
$F_3 - F_1 - F_2$
C
$F_3 - F_1 + F_2$
D
$\sqrt{(F_3 - F_1)^2 + F_2^2}$
Detailed Solution
The total force on a current carrying closed loop placed in a uniform magnetic field is zero.
$\vec{F}_{PS} + \vec{F}_{SR} + \vec{F}_{RQ} + \vec{F}_{QP} = 0$
$F_1$ and $F_3$ are along the horizontal in opposite directions, so the horizontal resultant is $F_{horizontal} = (F_3 - F_1)$.
$F_2$ is along the vertical, so $F_{vertical} = F_2$.
The resultant of $\vec{F}_1$, $\vec{F}_2$ and $\vec{F}_3$ is $\vec{F}$, where $F = \sqrt{(F_3 - F_1)^2 + F_2^2}$ because the two components are perpendicular.
Since the total force is zero, the force on QP is equal to $\vec{F}$ in magnitude but opposite in direction.
$F_{QP} = \sqrt{(F_3 - F_1)^2 + F_2^2}$
$\vec{F}_{PS} + \vec{F}_{SR} + \vec{F}_{RQ} + \vec{F}_{QP} = 0$
$F_1$ and $F_3$ are along the horizontal in opposite directions, so the horizontal resultant is $F_{horizontal} = (F_3 - F_1)$.
$F_2$ is along the vertical, so $F_{vertical} = F_2$.
The resultant of $\vec{F}_1$, $\vec{F}_2$ and $\vec{F}_3$ is $\vec{F}$, where $F = \sqrt{(F_3 - F_1)^2 + F_2^2}$ because the two components are perpendicular.
Since the total force is zero, the force on QP is equal to $\vec{F}$ in magnitude but opposite in direction.
$F_{QP} = \sqrt{(F_3 - F_1)^2 + F_2^2}$