JEE MainPhysicsModerateMoving charges 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 $\\mathrm{m}\$ then

Show answer and solution

\tau = MBsin\theta = I\alpha

\pi R^{2}IB\theta = \frac{{mR}^{2}}{2}\alpha

\omega = \sqrt{\frac{2\pi IB}{m}} = \frac{2\pi}{T};T = \sqrt{\frac{2\pi m}{IB}}

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx

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