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JEE Mains · Physics · STD 12 - Magnetism and matter
  • JEE Mains 2021
  • Medium
  • Numerical

In a uniform magnetic field, the magnetic needle has a magnetic moment \(9.85 \times 10^{-2} \,{A} / {m}^{2}\) and moment of inertia \(5 \times 10^{-6} \,{kgm}^{2}\). If it performs \(10\) complete oscillations in \(5\, seconds\) then the magnitude of the magnetic field is \(....\,mT.\) [ Take \(\pi^{2}\) as \(9.85\) ]

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Answer & Solution

Correct Answer

(A) \(8\)

Step-by-step Solution

Detailed explanation

\(T =2 \pi \sqrt{\frac{ I }{ MB }}\)

\(\frac{5}{10} =2 \pi \sqrt{\frac{5 \times 10^{-6}}{9.85 \times 10^{-2} \times B}}\)

\(B =8 \times 10^{-3} T\)

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