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 \(9.85\) ]
Pick an option to check your answer.
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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