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Step-by-Step Explanation
Step 1: Understanding the Physical Situation
A square loop carrying a steady current is placed in a uniform magnetic field such that the field lies in the plane of the loop. Each side of the loop then experiences a magnetic force that depends on the current in that side, the length of the side, and the direction of the magnetic field.
Step 2: Forces on the Opposite Arms of the Loop
For a current-carrying conductor in a uniform magnetic field, the magnetic force is given by
$$\overrightarrow{F} = I \, \overrightarrow{L} \times \overrightarrow{B},$$
where
$I$ is the current,
$\overrightarrow{L}$ is the vector representing the length and direction of the current in the conductor,
$\overrightarrow{B}$ is the magnetic field.
Opposite sides of the square loop carry currents in opposite directions. Because the field is in the plane of the loop, the forces on two opposite arms are equal in magnitude but opposite in direction.
Step 3: Net Force on the Loop
When you consider all four sides together, the sum of their forces adds up to zero, implying the loop as a whole experiences no net linear force. However, the loop can experience a torque because the lines of action of certain pairs of forces do not coincide.
Step 4: Relation of Forces on One Arm Versus the Remaining Three Arms
The problem states that the magnetic force on one specific arm of the loop is
$$\overrightarrow{F}.$$
To keep the net force on the entire loop at zero, the total force on the other three arms must be
$$-\overrightarrow{F},$$
thereby canceling the effect of $\overrightarrow{F}$ on that single arm. This ensures that the loop does not accelerate translationally, even though it may experience a net torque.
Conclusion
Hence, if the force on one arm of the square loop is $\overrightarrow{F}$, the net force on the remaining three arms is
$$ -\overrightarrow{F}.$$
This balances the force on that one arm, resulting in an overall net force of zero on the loop.
