A closed-loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS , SR and RQ are F1 , F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is

Question

A closed-loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS , SR and RQ are F1 , F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xvdfqrjdlphpskdp.jpg" /> (A) √(F3 - F1)2 - F22 (B) F3-F1+F2 (C) F3-F1-F2 (D) √(F3 - F1)2 + F22

Tardigrade Admin · Accepted Answer

∵ Net force on current carrying closed loop is zero. overset arrow F1+ overset arrow F2+ overset arrow F3+ overset arrow FP Q=0 ⇒ ( overset arrow F)P Q=-(( overset arrow F)1 + ( overset arrow F)2 + ( overset arrow F)3) overset arrow F1+ overset arrow F3 is perpendicular to overset arrow F2 . ∴ FP Q=√(F3 - F1)2 + F22

Q. A closed-loop $PQRS$ carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments $PS$ , $SR$ and $RQ$ are $F_{1}$ , $F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown, the force on the segment $QP$ is

$(F_{3} - F_{1})^{2} - F_{2}^{2}$

$F_{3} - F_{1} + F_{2}$

$F_{3} - F_{1} - F_{2}$

$(F_{3} - F_{1})^{2} + F_{2}^{2}$

$∵$ Net force on current carrying closed loop is zero.
$F \to_{1} + F \to_{2} + F \to_{3} + F \to_{PQ} = 0$
$\Rightarrow (F \to)_{PQ} = - ((F \to)_{1} + (F \to)_{2} + (F \to)_{3})$
$F \to_{1} + F \to_{3}$ is perpendicular to $F \to_{2}$ .
$∴ F_{PQ} = (F_{3} - F_{1})^{2} + F_{2}^{2}$

Q. A closed-loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS , SR and RQ are F1​ , F2​ and F3​ respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is

(F3​−F1​)2−F22​​

F3​−F1​+F2​

F3​−F1​−F2​

(F3​−F1​)2+F22​​