shows a triangular loop PQR carrying a current i. The triangle is equilateral with edge-length i. A uniform magnetic field B exists in a direction parallel to PQ. Find the forces acting on the three wirs PQ, QR and RP separately.

Explanation

The force on the wire PQ is
`vec F_1 = i vec PQ xx vec B= 0`.
as the field `vecB` is parallel to `vecPQ`.
the force on QR is
`vecF _2 = I vec QR X vec B`
or `F_2 = ilb sin 120@`
`= (sqrt3 )/(2) ilB`
. From the rule of vector product, this force
is perpendicular to the plane of the diagram and is coming out of it.