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