An equilateral triangular loop ADC of side l carries a current i in the direction shown in figure. The loop is kept in a uniform horizontal magnetic field $\overrightarrow{\mathrm{B}}$as shown in figure. Net force on the loop is

The magnetic forces acing on the three sides of the triangle are as follows:
${\mathrm{F}}_{\mathrm{CD}}=0 \mathrm{as} \mathrm{CD} \mathrm{is} \mathrm{parallel} \mathrm{to} \overrightarrow{\mathrm{B}}$
${\mathrm{F}}_{\mathrm{AC}}=\mathrm{ilBsin} {60}^{\circ }=\frac{\sqrt{3}}{2}\mathrm{ilB}$ (Perpendicular to the plane of diagram and outwards) as the side AC makes an angle 60° with $\overrightarrow{\mathrm{B}}$.
${\mathrm{F}}_{\mathrm{AD}}=\mathrm{ilBsin} {120}^{\circ }=\frac{\sqrt{3}}{2}\mathrm{ilB}$ (Perpendicular to the plane of diagram and outwards) as the side AD makes an angle 120° with $\overrightarrow{\mathrm{B}}$.
Therefore, net force on the loop is
$\mathrm{F}=0+\frac{\sqrt{3}}{2}\mathrm{ilB}+\frac{\sqrt{3}}{2}\mathrm{ilB}=\sqrt{3}\mathrm{ilB}$ perpendicular to the plane of the loop, outwards.