Verified Solution • 2025–26 Session moderate

QJEEM20Ph63b9: JEE Main 2020 Physics - Class 12 JEEM20Ph63b9 Solution: A small circular loop of conducting wire has radius a and…

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. If the mass of the loop is m then :

Multiple Choice Options

  • A T=2mIBT = \sqrt {{{2m} \over {IB}}}
  • B T=πmIBT = \sqrt {{{\pi m} \over {IB}}}
  • C T=πm2IBT = \sqrt {{{\pi m} \over {2IB}}}
  • D T=2πmIBT = \sqrt {{{2\pi m} \over {IB}}}

Step-by-Step Verified Solution

τ\tau = - MBsin θ\theta

I α\alpha = - MBsin θ\theta

for small θ\theta ,

α\alpha = −MBIθ- {{MB} \over I}\theta

∴\therefore ω2{\omega ^2} = MBI{{MB} \over I}

⇒\Rightarrow ω\omega = I(πR2)BmR22\sqrt {{{I\left( {\pi {R^2}} \right)B} \over {{{m{R^2}} \over 2}}}} = 2IπBm\sqrt {{{2I\pi B} \over m}}

∴\therefore T = 2πω{{2\pi } \over \omega } = 2πmIB\sqrt {{{2\pi m} \over {IB}}}

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