Question

A current carrying 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 $\overrightarrow{F}$, the force on the arm $AC$ is :

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

Answer 1

The correct option is B $-\overrightarrow{F}$

If the magnetic field along $AB$ is $B$, the force acting on $BC$,

$F=BIl$

Now the length of the side $AC$ is $\sqrt{2}l$.

Also the current reverses the direction in this section as compared to that in $BC$.

Hence, force on $AC=\sqrt{2}l(\overrightarrow{B}\times \overrightarrow{I})$

$=-BIl\sin {45}^{\circ }=-BIl=-\overrightarrow{F}$

Hence, $\left(B\right)$ is the correct answer.