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NEET PHYSICSMedium

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 F1F_1, F2F_2 and F3F_3 respectively and are in the plane of the paper and along with the directions shown, the force on the segment QP is

A

F3+F1−F2F_3 + F_1 - F_2

B

(F3−F1)2+F22\sqrt{(F_3 - F_1)^2 + F_2^2}

C

(F3−F1)2−F22\sqrt{(F_3 - F_1)^2 - F_2^2}

D

F3−F1+F2F_3 - F_1 + F_2

Step-by-Step Solution

  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 dF⃗=I(dl⃗×B⃗)d\vec{F} = I (d\vec{l} \times \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: F⃗QP+F⃗PS+F⃗SR+F⃗RQ=0\vec{F}_{QP} + \vec{F}_{PS} + \vec{F}_{SR} + \vec{F}_{RQ} = 0
  3. Calculation: We need to find the force on segment QP (F⃗QP\vec{F}_{QP}). Rearranging the equilibrium equation: F⃗QP=−(F⃗PS+F⃗SR+F⃗RQ)\vec{F}_{QP} = - (\vec{F}_{PS} + \vec{F}_{SR} + \vec{F}_{RQ}) F⃗QP=−(F⃗1+F⃗2+F⃗3)\vec{F}_{QP} = - (\vec{F}_1 + \vec{F}_2 + \vec{F}_3) The magnitude of F⃗QP\vec{F}_{QP} is equal to the magnitude of the resultant of the other three forces. Based on the options provided, the expression (F3−F1)2+F22\sqrt{(F_3 - F_1)^2 + F_2^2} implies that the forces F1F_1 and F3F_3 act along the same line but in opposite directions (likely vertical components), and F2F_2 acts perpendicular to them (likely horizontal). Thus, the magnitude is the vector sum of these components: ∣F⃗QP∣=(Fvertical)2+(Fhorizontal)2=(F3−F1)2+F22|\vec{F}_{QP}| = \sqrt{(F_{vertical})^2 + (F_{horizontal})^2} = \sqrt{(F_3 - F_1)^2 + F_2^2}
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