A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in

   a uniform magnetic field action along AB. If the magnetic force on the arm BC is F, the force

     on the arm AC is

When a current carrying closed loop is placed in a uniform magnetic field, it

            experiences no net force${\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{CA}=0$.But${\overrightarrow{F}}_{AB}=0$,as the side AB is parallel to magnetic  field hence it will experience no force due to magnetic field.

Hence ${\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{CA}=0\Rightarrow {\overrightarrow{F}}_{BC}=-{\overrightarrow{F}}_{CA}=\text{F},$

$\therefore {\overrightarrow{F}}_{\text{CA}}=-\text{F}$