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Question

(A) For a closed loop carrying a current in a uniform magnetic field,the net magnetic force on the loop is zero.Let the force on segment $QP$ be $F_4$

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$\sqrt {{{\left( {F_3 - F_1} \right)}^2} + F_2^2} $

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(A) For a closed loop carrying a current in a uniform magnetic field,the net magnetic force on the loop is zero.Let the force on segment $QP$ be $F_4$.Since the loop is in equilibrium under the influence of these forces,the vector sum of all forces must be zero:$\vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} = 0$$\vec{F_4} = -(\vec{F_1} + \vec{F_2} + \vec{F_3})$Resolving the forces into horizontal and vertical components:Let the horizontal direction be along the $x$-axis (positive towards the right) and the vertical direction be along the $y$-axis (positive upwards).$F_{4x} = -(F_{3x} + F_{2x} + F_{1x}) = -(F_3 + 0 - F_1) = F_1 - F_3$$F_{4y} = -(F_{3y} + F_{2y} + F_{1y}) = -(0 - F_2 + 0) = F_2$The magnitude of the force $F_4$ is given by:$F_4 = \sqrt{F_{4x}^2 + F_{4y}^2} = \sqrt{(F_1 - F_3)^2 + F_2^2} = \sqrt{(F_3 - F_1)^2 + F_2^2}$

$A$ 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

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