A closed-loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segment

Question

A closed-loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RO (RQ) are $F_1$, $F_2$ and $F_3$ respectively and are in the plane of the paper and along with the directions shown, the force on the segment QP is

Accepted Answer

1. **Net Force Principle:** The net magnetic force on any closed current-carrying loop placed in a uniform magnetic field is zero. This is a fundamental result derived from the integration of the force element $d\vec{F} = I (d\vec{l} 	imes \vec{B})$ over a closed loop .
2. **Vector Sum:** For the loop PQRS, the vector sum of the forces on all segments must be zero:
   $$\vec{F}_{QP} + \vec{F}_{PS} + \vec{F}_{SR} + \vec{F}_{RQ} = 0$$
3. **Calculation:** We need to find the force on segment QP ($\vec{F}_{QP}$). Rearranging the equilibrium equation:
   $$\vec{F}_{QP} = - (\vec{F}_{PS} + \vec{F}_{SR} + \vec{F}_{RQ})$$
   $$\vec{F}_{QP} = - (\vec{F}_1 + \vec{F}_2 + \vec{F}_3)$$
   The magnitude of $\vec{F}_{QP}$ is equal to the magnitude of the resultant of the other three forces. Based on the options provided, the expression $\sqrt{(F_3 - F_1)^2 + F_2^2}$ implies that the forces $F_1$ and $F_3$ act along the same line but in opposite directions (likely vertical components), and $F_2$ acts perpendicular to them (likely horizontal). Thus, the magnitude is the vector sum of these components:
   $$|\vec{F}_{QP}| = \sqrt{(F_{vertical})^2 + (F_{horizontal})^2} = \sqrt{(F_3 - F_1)^2 + F_2^2}$$

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