question_answer

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 F, the force on the arm AC is              [AIPMT (S) 2011]

A)  \[-\mathbf{F}\]             

B)       \[\mathbf{F}\] C)  \[\sqrt{2}\mathbf{F}\]    D)       \[-\sqrt{2}\mathbf{F}\] Correct Answer: A Solution : [a]

\[{{\mathbf{F}}_{AB}}=0\]

\[{{\mathbf{F}}_{AB}}+{{\mathbf{F}}_{BC}}+{{\mathbf{F}}_{CA}}=0\]

\[{{F}_{BC}}+{{F}_{CA}}=0\]

\[{{F}_{CA}}=-{{F}_{BC}}=-F\]