A current carrying closed loop in the form of a right angle isosceles triangle $A B C$ is placed in a…

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

1. $-\mathbf{F}$
2. $\mathbf{F}$
3. $\sqrt{2} \mathbf{F}$
4. $-\sqrt{2} \mathbf{F}$

Solution

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

Asked in: NEET 2011 (Screening)

Practice more Magnetic Fields due to Electric Current questions on Aicharya