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in JEE by (450 points)

Same current I = 2A is flowing in a wire frame as shown in figure. The frame is a combination of two equilateral triangles ACD and CDE of side 1m. It is placed in uniform magnetic field B = 4T acting perpendicular to the plane of frame. The magnitude of magnetic force acting on the frame is :

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As the applied magnetic field (B=4T) is directed perpendicular to the plane of the frame, it will make angle

(α = 90o ) with each edge of length, L=1m . Here each edge carries current, I=2A.Hence the force on each edge will be

F =BILsinα  =4×2×1×sin90o N=8N

The forces on AC and AD will be inclined at an angle 120o . So their resultant will be 

=  Sqrt (F2+F2+2.F..F.cos120o )=F =8N

Similarly the resultant of two forces on EC and ED will also be F =8N.

These two resultants will act in the same direction EA as the direction of force F on CD.

So the net force on the frame will be =3F=3×8=24N.

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