The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:
  • A$16 \times 10^{-2} \mathrm{~J}$
  • B $8 \times 10^{-2} \mathrm{~J}$
  • C$4 \times 10^{-2} \mathrm{~J}$
  • D
    Zero
JEE MAIN 2024, Diffcult
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