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JEE Main 2017 · 2 Apr · Shift 1, Physics question 18

A magnetic needle of magnetic moment 6.7×10−26.7\times10^{-2} Am2^2 and moment of inertia 7.5×10−67.5\times10^{-6} kg m2^2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is :

Choose one option
Answer and worked solution

Official answer: A

T=2πImB=2π7.5×10−66.7×10−2×0.01=2π×0.1058=0.665 s.T=2\pi\sqrt{\frac I{mB}}=2\pi\sqrt{\frac{7.5\times10^{-6}}{6.7\times10^{-2}\times0.01}}=2\pi\times0.1058=0.665\text{ s}.
Ten oscillations take 6.65 s.

Topics:Magnetic dipole and bar magnetSimple harmonic motion
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