A small circular loop of conducting wire has radius a and carries current $I.$ It is placed in a uniform magnetic field $B$ perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period $T.$ The mass of the loop is $m$ then:

• A

  $T=\sqrt{\frac{2m}{iB}}$
• B

  $T=\sqrt{\frac{\pi m}{2iB}}$
• Correct answer
  C

  $T=\sqrt{\frac{2\pi m}{iB}}$
• D

  $T=\sqrt{\frac{\pi m}{iB}}$

Hence, the correct answer is option: 'C'.