JEE Main 2022 · 27 Jul · Shift 2 · Physics Q42

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PhysicsSingle correct+4 / -1Question 42 of 90

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places $(B_Q:B_P)$ is :

\sqrt3:4

4:\sqrt3

\sqrt3:2

2:\sqrt3

Answer and worked solution

Official answer: A

Oscillation magnetometer. Frequency squared is proportional to the horizontal field: $\frac{B_P\cos30^\circ}{B_Q\cos60^\circ}=\left(\frac{20}{10}\right)^2=4$ .

$\frac{B_Q}{B_P}=\frac{\sqrt3}4$ . (A)

Topics:Earth's magnetism Magnetic dipole and bar magnet

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JEE Main 2022 · 27 Jul · Shift 2, Physics question 42