A compass needle of an oscillation magnetomet | vedclass

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(A) The time period of an oscillation magnetometer is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $B_H = B \cos \delta$ is the horizontal compone

Vedclass · Accepted Answer

$\sqrt{3}: 4$

Vedclass · Answer

(A) The time period of an oscillation magnetometer is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $B_H = B \cos \delta$ is the horizontal component of the Earth's magnetic field and $\delta$ is the angle of dip.Frequency $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{M B \cos \delta}{I}}$. Thus,$f \propto \sqrt{B \cos \delta}$.At place $P$: $f_P = 20 	ext{ oscillations/min}$,$\delta_P = 30^{\circ}$.At place $Q$: $f_Q = 10 	ext{ oscillations/min}$,$\delta_Q = 60^{\circ}$.Taking the ratio:$\frac{f_P}{f_Q} = \sqrt{\frac{B_P \cos 30^{\circ}}{B_Q \cos 60^{\circ}}}$$\frac{20}{10} = \sqrt{\frac{B_P (\sqrt{3}/2)}{B_Q (1/2)}}$$2 = \sqrt{\frac{B_P \sqrt{3}}{B_Q}}$Squaring both sides: $4 = \frac{B_P \sqrt{3}}{B_Q}$Therefore,$\frac{B_Q}{B_P} = \frac{\sqrt{3}}{4}$.

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