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प्रश्न
Derive the expression for the torque acting on a current-carrying loop placed in a magnetic field.
उत्तर
The plane of the loop is not along the magnetic field but makes an angle with it.
Let the dimension of the rectangular coil ABCD, be AB × BC = a × b
The angle between the field and the normal is θ.
Forces on BC and DA are equal and opposite and they cancel each other as they are collinear.
Force on AB is F1 and force on CD is F2.
F1 = F2 = IbB
The magnitude of the torque on the loop as in the figure:
`tau = F_1 a/2 sinθ + F_2 a/2 sinθ`
= IabBsinθ
If there are ‘n’ such turns the torque will be sin nIab sinθ
The magnetic moment of the current m = IA
`vectau = vecm xx vecB`
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