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A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle 180°, then the work done is ______. (M= magnetic dipole moment of the bar magnet)

Asked in GUJCET 2011 · Dipole in a uniform field

Answer: (2) 2MB

Step-by-step solution

Given: the magnet starts in stable equilibrium, so ⃗M is along ⃗B and θ₁=0°; it is turned to θ₂=180°.

Idea: the potential energy of a dipole in a uniform field is U=-MB cos θ, and the work done in turning it is the rise in this energy, W=U₂-U₁=MB(cos θ₁-cos θ₂).

W=MB(cos 0°-cos 180°)=MB(1-(-1)).

W=2MB.

The magnet is left antiparallel to ⃗B, in unstable equilibrium - the highest-energy orientation, U=+MB, reached from the lowest, U=-MB.

Why the other options are wrong

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