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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
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The correct answer is:
$\sqrt{\left(F_3-F_1\right)^2+F_2^2}$
Since net force on current carrying loop in uniform magnetic field is zero therefore force on remaining segment will be equal and oppsoite to $\mathrm{F}$.
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