PhysicsEasy49×since 2003Q7617The magnetic moment of a bar magnet is 0.5 Am20.5 \mathrm{~Am}^20.5 Am2. It is suspended in a uniform magnetic field of 8×10−2 T8 \times 10^{-2} \mathrm{~T}8×10−2 T. The work done in rotating it from its most stable to most unstable position is:A4×10−2 J4 \times 10^{-2} \mathrm{~J}4×10−2 JB16×10−2 J16 \times 10^{-2} \mathrm{~J}16×10−2 JC8×10−2 J8 \times 10^{-2} \mathrm{~J}8×10−2 JDZeroCheck answerSkip