A current carrying closed loop in the form of a right angle 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
AIPMT 2011, Medium
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$\text { Here, } \vec{F}_{B C}=\vec{F}$
$\because \quad \vec{F}_{A B}=0$
The net magnetic force on a current carrying closed loop in a uniform magnetic field is zero.
$\therefore \vec{F}_{A B}+\vec{F}_{B C}+\vec{F}_{A C}=0$
$\Rightarrow \vec{F}_{A C}=-\vec{F}_{B C}$ $\left(\because \vec{F}_{A B}=0\right)$
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