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JEE Mains · Physics · STD 12 - Magnetism and matter
  • JEE Mains 2017
  • Medium
  • MCQ

A magnetic needle of magnetic moment \(6.7 \times 10^{-2} \) \(Am^2\) and moment of inertia \(7.5 \times  10^{-6}\) \( kgm^2\) is performing simple harmonic oscillations in a magnetic field of \(0.01\ T\). Time taken for \(10\) complete oscillations is.....\(s\)

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Answer & Solution

Correct Answer

(A) \(6.65\) 

Step-by-step Solution

Detailed explanation

Given : Magnetic moment, \(M=6.7 \times 10^{-2}\, \mathrm{Am}^{2}\)

Magnetic field, \(\mathrm{B}=0.01\, \mathrm{T}\)

Moment of inertia, \(I=7.5 \times 10^{-6}\, \mathrm{Kgm}^{2}\)

\(\text { Using, } T =2 \pi \sqrt{\frac{I}{M B}} \)

\(=2 \pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}=\frac{2 \pi}{10} \times 1.06\, s\)

Time taken for \(10\) complete oscillations

\(t=10\, T=2 \pi \times 1.06 \)

\(=6.6568 \approx 6.65\, \mathrm{s}\)

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