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Question
Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.
Solution
Consider a rectangular loop ABCD carrying current I.
Case I - The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop.
No force is exerted by the magnetic field on the arms AD and BC.
Magnetic field exerts a force F1 on arm AB.
∴F1 = IbB
Magnetic field exerts a force F2 on arm CD.
∴F2 = IbB = F1
Net force on the loop is zero.
The torque on the loop rotates the loop in anti-clockwise direction.
Torque, τ = `f_1 a/2 + f_2 a/2`
`= IbBa/2 +IbBa/2`
= I(ab)B
τ = BIA
If there are ‘n’ such turns the torque will be nIAB
where, b → Breadth of the rectangular coil
a → Length of the rectangular coil
A = ab → Area of the coil
Case II - Plane of the loop is not along the magnetic field, but makes angle with it.
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
Magnitude of torque on the loop as in the figure:
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