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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.

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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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