A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow F$ the net force on the remaining three arms of the loop is
AIPMT 2010, Medium
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Let $ABCD$ is a square loop carrying current placed in a uniform magnetic field,
Due to this field the conductors of the loop will experience force.
The direction and magnitude of the force is directly proportional to the current carried in the conductor.
Given,
Force in one arm $AB = F$
since opposite branches of this loop carry current in opposite direction
$==>$ force in the opposite arm $CD = -F$
Similarly The force in in the branches $AD$ and $BC$ will be opposite in direction and cancel out when added.
Hence net force on the remaining three arms of the loop will be $(b)$ $-F$
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