question_answer

Consider a triangular loop ABC carrying current\[I\]. The triangle is equilateral with edge length a. A uniform magnetic field B exits in a direction parallel to AB. The forces on the arms AB, BC and CA will be respectively

A)  \[0,\,\sqrt{3}aBI,\sqrt{3}aBI\]

B)  \[0,\,\frac{\sqrt{3}aBI}{2},\frac{\sqrt{3}aBI}{2}\]

C)  \[0,\,0,\,\sqrt{3}aBI\]

D)  \[0,\,0,\,0\]

E)  \[0,\,0,\,0,\sqrt{2}aBI\]

Correct Answer: B

Solution :

                Situation can be diagrammed as shown below The force on wire AB is \[{{F}_{1}}=I\,AB\times B=0\]    \[(\because \,AB||B)\] The force on wire BC \[{{F}_{2}}=I\,BC\times B\] \[|{{F}_{2}}|=I\,aB\sin 120{}^\circ \]        \[=\frac{\sqrt{3}}{2}IaB\] Also force on wire AC is \[{{F}_{3}}=\frac{\sqrt{3}}{2}IaB\]