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Question
A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field `vecB`. Prove that the torque `vectau`acting on the loop is give by `vectau =vecm xx vecB,`where `vecm` is the magnetic moment of the loop.
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Solution
Plane of the loop is at an angle with the direction of the magnetic field,
Let the angle between the field and the normal is θ. The 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, thus
F1 = F2 = IbB
Magnitude of torque on the loop as in the figure:
`because tau =F_1 1/2 sintheta +F_2 1/2 sintheta =IlBsintheta`
or, τ = IAB sin θ (Where area,A = lb)
If there are ‘n’ such turns the torque will be nIAB sinθ
Magnetic moment of the current, m = IA
`vectau=vecm xx vecB`
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- Assertion (A): The deflecting torque acting on a current-carrying loop is zero when its plane is perpendicular to the direction of the magnetic field.
- Reason (R): The deflecting torque acting on a loop of the magnetic moment `vecm` in a magnetic field `vecB` is given by the dot product of `vecm` and `vecB`.
