A magnetic compass needle oscillates 30 times per minute at a place… | JEE Main 2019 (12 Apr, Shift 1)

Q13JEE Main 2019 · 12 APR · SHIFT 1Difficult

Vibration Magnetometer

A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45°, and 40 times per minute where the dip is 30°. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio $\frac{B_1}{B_2}$ is best given by:

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Answer: (c)

Frequency $n \propto \sqrt{B_H}$ with $B_H = B\cos\delta$ , so $n^2 \propto B\cos\delta$ .
$\dfrac{30^2}{40^2} = \dfrac{B_1\cos 45^\circ}{B_2\cos 30^\circ} \Rightarrow \dfrac{9}{16} = \dfrac{B_1}{B_2}\cdot\dfrac{0.707}{0.866}$
$\dfrac{B_1}{B_2} = 0.5625 \times 1.225 \approx 0.69$
So the answer is (c) 0.7.

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Vibration Magnetometer question from JEE Main 2019 (12 Apr, Shift 1)

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