JEE Mains · Physics · STD 12 - 5. Magnetism and matter
A magnetic compass needle oscillates \(30\) times per minute at a place where the dip is \(45^o\), and \(40\) times per minute where the dip is \(30^o\). If \(B_1\) and \(B_2\) are respectively the total magnetic field due to the earth at the two places, then the ratio \(B_1/B_2\) is best given by
- A \(3.6\)
- B \(1.8\)
- C \(2.2\)
- D \(0.7\)
Answer & Solution
Correct Answer
(D) \(0.7\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}_{1}=\frac{1}{2 \pi} \sqrt{\frac{\mu \mathrm{B}_{1} \cos 45^{\circ}}{1}} \quad\) \(\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{\mathrm{B}_{1} \cos 45^{\circ}}{\mathrm{B}_{2} \cos 30^{\circ}}\)…
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