A current carrying closed loop in the form of a right angle isosceles triangle $A B C$ is placed in a uniform magnetic field acting along $A B$. If the magnetic force on the $\operatorname{arm} B C$ is $\mathbf{F}$, the force on the arm $A C$ is

• Correct answer
  A

  $-\mathbf{F}$
• B

  $\mathbf{F}$
• C

  $\sqrt{2} \mathbf{F}$
• D

  $-\sqrt{2} \mathbf{F}$

Solution

$$ \begin{gathered} \overrightarrow{\mathbf{F}}_{A B}=0 \\ \overrightarrow{\mathbf{F}}_{A B}+\overrightarrow{\mathbf{F}}_{B C}+\overrightarrow{\mathbf{F}}_{C A}=0 \\ \overrightarrow{\mathbf{F}}_{B C}+\overrightarrow{\mathbf{F}}_{C A}=0 \\ \overrightarrow{\mathbf{F}}_{C A}=-\overrightarrow{\mathbf{F}}_{B C}=-\overrightarrow{\mathbf{F}} \end{gathered} $$

Hence, the correct answer is option: 'A'.