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=IBlsinθ
Magnetic field is perpendicular to the length of wire, so θ=90°
Magnetic force on side QR,
F1=IBlsin90°=5×2×0.1=1 N (direction shown in figure)
Magnetic force on side PR,
F2=IBlsin90°=5×2×0.1=1 N (direction shown in figure)
Magnetic force on side PQ,
F3=IBlsin90°=5×2×0.1=1 N (direction shown in figure)
Net magnetic force,
Resultant of F1 and F2F'=F12+F22+2F1F2cos120°=1+1-1=1 N direction shown in figureResultant of F' and F3F=1-1=0
So, net magnetic force the loop is zero.


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