JEE Mains · Physics · STD 12 - 5. Magnetism and matter
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.
- A \(\sqrt{3}: 4\)
- B \(4: \sqrt{3}\)
- C \(\sqrt{3}: 2\)
- D \(2: \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3}: 4\)
Step-by-step Solution
Detailed explanation
\(T =2 \pi \sqrt{\frac{ I }{ B _{ H } M }}\) \(T _{1}=3\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ p } \cos 30^{\circ}\right) M }}\) \(T _{2}=6\,sec =2 \pi \sqrt{\frac{ I }{\left( B _{ Q } \cos 60^{\circ}\right) M }}\)…
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