JEE Mains · Physics · STD 12 - 5. Magnetism and matter
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 \(9.85\) ]
- A \(8\)
- B \(10\)
- C \(12\)
- D \(14\)
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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