Torque on Current carrying coil In magnatic field (pak physics notes)

Consider a rectangular coil of wire through which an electric current 'I' is being passed. The coil is placed in a uniform magnetic field of magnitude "B".

	The vertical side AB of the coil experiences a force F which is directed perpendicular into the paper. There is an equal and opposite force F on side CD. Since the plane of coil is parallel to the magnetic field therefore there is no force on either side AC and BD. Force on AB and CD will produce torque in these arms.
	In this case L perpendicular to B i.e. q = 90, hence
	F = BIL (1) F = BIL                     AB = b/2 (BIL)…………(1) Similarly             CD = b/2 (BIL)…………(2)
	These two torque result in a couple in the coil and the coil starts moving (rotating) in magnetic field. Since both the torque are trying to rotate the coil in the same direction, therefore the total torque produced in the coil will be the sum of all the individuals torque.
	=  AB +  CD                       = b/2 (BIL) + b/2 (BIL)     = 2/2 b (BIL)                = IB (Lb)
	but Lb = area of coil = A
	= IBA
	If there are "N" number of turns of wire in the coil then the torque will becomes N times
	= BINA
	When the coil makes an angle a with the direction of magnetic field, the torque will become
	= BINA Cosa

  

                                  (www.pak-physicsnotes.blogspot.com)

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