A wire is bent in the form of an equilateral triangle PQR of side 10 cm and carries a current of 5 A. It is placed in a magnetic field B of magnitude 2 T directed perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle and also the net magnetic force on the loop.

Dear student, 


Magnetic force is given by, $F=IBl\sin \theta$
Magnetic field is perpendicular to the length of wire, so $\theta =90°$
Magnetic force on side QR,
$F_1=IBl\sin 90°=5\times 2\times 0.1=1 \mathrm{N}$ (direction shown in figure)
Magnetic force on side PR,
$F_2=IBl\sin 90°=5\times 2\times 0.1=1 \mathrm{N}$ (direction shown in figure)
Magnetic force on side PQ,
$F_3=IBl\sin 90°=5\times 2\times 0.1=1 \mathrm{N}$ (direction shown in figure)
Net magnetic force,
$$\begin{gathered} \mathrm{Resultant} \mathrm{of} {\overrightarrow{F}}_1 \mathrm{and} {\overrightarrow{F}}_2 \\ F\text{'}=\sqrt{{F_1}^2+{F_2}^2+2F_1{F}_2\cos 120°}=\sqrt{1+1-1}=1 \mathrm{N} \left(\mathrm{direction} \mathrm{shown} \mathrm{in} \mathrm{figure}\right) \\ \mathrm{Resultant} \mathrm{of} \overrightarrow{\mathrm{F}}\text{'} \mathrm{and} {\overrightarrow{\mathrm{F}}}_3 \\ F=1-1=0 \end{gathered}$$
So, net magnetic force the loop is zero.

Regards.