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. $-\sqrt2\, \vec{F}$
• B. $-\vec{F}$
• C. $\vec{F}$
• D. $\sqrt2 \,\vec{F}$

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.
$\vec{ F _{\text {net }}}=\vec{ F }$ 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 $-\vec{ F }$