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A rectangular coil pqrs of N turns length L and breadth B and carrying a current i is placed in a uniform magnetic field B with its longest side perpendicular and shorter sides parallel to the field as shown in the figure .compute the torque on the coil .the normal to its plane makes an angle Alpha with the field

A rectangular coil pqrs of N turns length L and breadth B and carrying a current i is placed in a uniform magnetic field B with its longest side perpendicular and shorter sides parallel to the field as shown in the figure .compute the torque on the coil .the normal to its plane makes an angle Alpha with the field

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Grade:12

1 Answers

Arun
25763 Points
2 years ago
The area of cross-section of the coil is A = lb
 
Now, the magnetic moment of the coil is
 
m = IA = Ilb
 
Now, we know that the torque acting on the coil is given as
 
T = m x B = mb sin theta
 
Here, θ is the angle between the magnetic field and the normal to the plane of the coil.
 
Now, the magnetic field is directed perpendicular to the coil. Also, the normal to the coil will be perpendicular to the coil.
 
So, the angle between the normal and the field is 0.
 
Hence, the torque acting on the coil is zero.

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