A magnetic compass needle oscillates 30 times | vedclass

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Question

(D) The frequency of oscillation of a magnetic needle in a magnetic field $B$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B

Vedclass · Accepted Answer

$0.7$

Vedclass · Answer

(D) The frequency of oscillation of a magnetic needle in a magnetic field $B$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B \cos \delta$ is the horizontal component of the earth's magnetic field and $\delta$ is the angle of dip.Given $f_1 = 30 	ext{ oscillations/min}$ at $\delta_1 = 45^{\circ}$ and $f_2 = 40 	ext{ oscillations/min}$ at $\delta_2 = 30^{\circ}$.Thus,$f_1^2 \propto B_1 \cos 45^{\circ}$ and $f_2^2 \propto B_2 \cos 30^{\circ}$.Taking the ratio: $\frac{f_1^2}{f_2^2} = \frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}$.Substituting the values: $(\frac{30}{40})^2 = \frac{B_1 (1/\sqrt{2})}{B_2 (\sqrt{3}/2)}$.$\frac{9}{16} = \frac{B_1}{B_2} 	imes \frac{1}{\sqrt{2}} 	imes \frac{2}{\sqrt{3}} = \frac{B_1}{B_2} 	imes \sqrt{\frac{2}{3}}$.$\frac{B_1}{B_2} = \frac{9}{16} 	imes \sqrt{\frac{3}{2}} = 0.5625 	imes 1.2247 \approx 0.689 \approx 0.7$.

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