A small circular loop of conducting wire has radius $a$ and carries current $I$. It is placed in a uniform magnetic field $\mathrm{B}$ perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period $T$. If the mass of the loop is $m$ then 
  • A$\mathrm{T}=\sqrt{\frac{\pi \mathrm{m}}{2 \mathrm{IB}}}$
  • B$\mathrm{T}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}$
  • C$\mathrm{T}=\sqrt{\frac{\pi \mathrm{m}}{\mathrm{IB}}}$
  • D$\mathrm{T}=\sqrt{\frac{2 \mathrm{m}}{\mathrm{IB}}}$
JEE MAIN 2020, Medium
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