Question

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 $ \mathop F\limits^ \to $ , the net force on the remaining three arms of the loop is
(A) 3 $ \mathop F\limits^ \to $
(B) - $ \mathop F\limits^ \to $
(C) -3 $ \mathop F\limits^ \to $
(D) $ \mathop F\limits^ \to $

Answer 1

Hint : A current carrying square loop is placed in a uniform, magnetic field, experiences a torque but no net force. The force on the two arms parallel to the field is zero. So we need to calculate the force on the arms perpendicular to the field.

Formula Used: In this solution we will be using the following formula,
$\Rightarrow \mathop F\limits^ \to = I\left( {\mathop {dl}\limits^ \to \times \mathop B\limits^ \to } \right) $ where $ \mathop F\limits^ \to $ is the force acting on a current carrying conductor, $ I $ is the current, $ \mathop {dl}\limits^ \to $ is the displacement and $ \mathop B\limits^ \to $ is the magnetic field.

Complete step by step answer
According to right hand rule, the direction of force exerted on a current carrying wire placed in a magnetic field is perpendicular to the magnetic field and current direction.
If the index finger is oriented along the magnetic field and middle finger is oriented along the current direction, and both of them are held mutually, then the thumb gives the direction of force on the wire, which is perpendicular to the two.
 $\Rightarrow \mathop F\limits^ \to = I\left( {\mathop {dl}\limits^ \to \times \mathop B\limits^ \to } \right) $ where $ \mathop F\limits^ \to $ is the force acting on a current carrying conductor, $ I $ is the current, $ \mathop {dl}\limits^ \to $ is the displacement and $ \mathop B\limits^ \to $ is the magnetic field.
From here, we know that $ I $ and $ \mathop {dl}\limits^ \to $ are in the same direction.
For a square loop carrying current, suspended in a uniform magnetic field, acting in the plane of the loop, we know that the force acting on one arm of the loop is $ \mathop F\limits^ \to $ .
On the upper and lower arms of the square, force acts upwards and downwards, respectively. Since the same amount of force acts on the opposite arms in opposite directions, thus the net effect is zero.
For the right arm, directly opposite to the arm on which the force is $ \mathop F\limits^ \to $ , the force acts in the opposite direction and is given by $ - \mathop F\limits^ \to $ .
 $ \therefore $ The net force acting on the three remaining arms is $ - \mathop F\limits^ \to $ .
The correct option is Option B.

Note
Alternatively, if we keep in mind that the net force acting on the loop should be zero, the solution shall be simpler.
If the force on one arm of the loop is $ \mathop F\limits^ \to $ the net force on the remaining three arms of the loop should be $ - \mathop F\limits^ \to $ .