Torque on Bar Magnet in Magnetic Field:
Take a bar magnet NS of length 2l. Let ‘m’ be the strength of each pole.
These two forces are equal and opposite and they act at different points. Hence they form the couple which tends to rotate the magnet clockwise.
Draw NA perpendicular on SA produced.
|Torque acting on bar magnet (τ) will be given by|
τ = Force x Perpendicular Distance
τ = mB x NA ……….(i)
From △NSA, NA/NS = Sin θ
⇒ NA = NS Sin θ
⇒ NA = 2l Sin θ
Therefore, τ = mB x 2l Sin θ
⇒ τ = (m x 2l) B Sin θ
⇒ τ = MB Sin θ (where M = m x 2l = magnetic dipole moment)
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