Question

A current carrying closed 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 $\vec{F}$, the force on the arm $AC$ is

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

Answer 1

Answer: (B)

Force along $AB$ is zero as magnetic field is along $AN$.
The net force on a current carrying loop is always zero.
$\overrightarrow{F_{\text{net }}}=\vec{F}$ on $AB+\vec{F}$ on $BC+\vec{F}$ on $AC$
$0=0+\vec{F}+(\vec{F}\text{ on }AB)$
we get, $\vec{F}$ on $AC$ is $-\vec{F}$