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(B) The net magnetic force on a current-carrying closed loop in a uniform magnetic field is always zero.Let the forces on the arms $AB,$ $BC,$ and $AC

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$-\vec F$

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(B) The net magnetic force on a current-carrying closed loop in a uniform magnetic field is always zero.Let the forces on the arms $AB,$ $BC,$ and $AC$ be $\vec{F}_{AB},$ $\vec{F}_{BC},$ and $\vec{F}_{AC}$ respectively.According to the principle of superposition,$\vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{AC} = 0.$Since the magnetic field is acting along the arm $AB,$ the angle between the current element $I\vec{dl}$ of arm $AB$ and the magnetic field $\vec{B}$ is $0^\circ$ or $180^\circ.$Therefore,the magnetic force on arm $AB$ is $\vec{F}_{AB} = I(\vec{L}_{AB} 	imes \vec{B}) = 0.$Substituting this into the net force equation: $0 + \vec{F}_{BC} + \vec{F}_{AC} = 0.$Given that $\vec{F}_{BC} = \vec{F},$ we get $\vec{F} + \vec{F}_{AC} = 0.$Thus,$\vec{F}_{AC} = -\vec{F}.$

List-$I$	List-$II$
$(A)$ Permeability of free space	$(I) \ [M L^2 T^{-2}]$
$(B)$ Magnetic field	$(II) \ [M T^{-2} A^{-1}]$
$(C)$ Magnetic moment	$(III) \ [M L T^{-2} A^{-2}]$
$(D)$ Torsional constant	$(IV) \ [L^2 A]$

$A$ current-carrying closed loop in the form of a right-angled isosceles triangle $ABC$ is placed in a uniform magnetic field acting along $AB.$ If the magnetic force on the arm $BC$ is $\vec F,$ the force on the arm $AC$ is

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