Magnetic Effects & Magnetism

JEE Main level, single correct, medium

258 of 276

Question 258 of 276

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

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Answer: (C) T=2πmiBT=\sqrt{\dfrac{2\pi m}{iB}}

τ=MBsin⁡⁡θ=Iα\tau =MB\sin ⁡\theta =I\alpha
πR2iBθ=mR22α\pi {R}^{2}iB\theta =\dfrac{m{R}^{2}}{2}\alpha
ω=2πiBm=2πT;T=2πmiB\omega =\sqrt{\dfrac{2\pi iB}{m}}=\dfrac{2\pi }{T};T=\sqrt{\dfrac{2\pi m}{iB}}

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