JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2022
- Hard
- MCQ
A compass needle of oscillation magnetometer oscillates \(20\) times per minute at a place \(P\) of dip \(30^{\circ}\). The number of oscillations per minute become \(10\) at another place \(Q\) of \(60^{\circ} dip\). The ratio of the total magnetic field at the two places \(\left(B_{Q}: B_{p}\right)\) is.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\sqrt{3}: 4\)
Step-by-step Solution
Detailed explanation
\(T =2 \pi \sqrt{\frac{ I }{ B _{ H } M }}\)
\(T _{1}=3\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ p } \cos 30^{\circ}\right) M }}\)
\(T _{2}=6\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ Q } \cos 60^{\circ}\right) M }}\)
\(\frac{3}{6}=\sqrt{\frac{ B _{ Q } / 2}{B _{ p } \frac{\sqrt{3}}{2}}}\)
\(\frac{3}{6}=\sqrt{\frac{ B _{ Q }}{\sqrt{3} B _{ p }}}\)
\(\frac{ B _{ Q }}{\frac{3}{4}}=\frac{ B _{ p }}{ B _{ Q }}=\sqrt{3}: 4\)
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