Question

A current carrying closed loop, in the form of a right-angled isosceles triangle $\text{ABC}$, is placed in a uniform magnetic field directed alongside $\text{AB}$. If the magnetic force on the arm $\text{BC}$ is $F$, the force on the arm $\text{AC}$ is

• A. $-F$
• B. $F$
• C. $F\sqrt{2}$
• D. $-F\sqrt{2}$

Answer 1

The correct option is A $-F$

Given:
${\overrightarrow{F}}_{CB}=F$

The force on $\text{AB}$ will be zero as field and current are in the same direction.

$\therefore {\overrightarrow{F}}_{AB}=0$

As we know that, net force on a closed loop (of any shape) due to external magnetic field is zero. So,

${\overrightarrow{F}}_{BA}+{\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{CB}=0$

${\overrightarrow{F}}_{AC}+{\overrightarrow{F}}_{CB}=0(\because {\overrightarrow{F}}_{AB}=0)$

$\therefore {\overrightarrow{F}}_{AC}=-{\overrightarrow{F}}_{CB}=-F$

Therefore, option $\left(A\right)$ is correct.