JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2019
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(0.7\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}_{1}=\frac{1}{2 \pi} \sqrt{\frac{\mu \mathrm{B}_{1} \cos 45^{\circ}}{1}} \quad\)
\(\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{\mathrm{B}_{1} \cos 45^{\circ}}{\mathrm{B}_{2} \cos 30^{\circ}}\)
\({{\text{f}}_2} = \frac{1}{{2\pi }}\sqrt {\frac{{\mu {{\text{B}}_2}\cos {{30}^\circ }}}{1}} \)
\(\therefore \frac{\mathrm{B}_{1}}{\mathrm{B}_{2}}= 0.7\)
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