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 respectively is \(\frac{4}{√x}\) The value of x is
Correct Answer: 243
Solution and Explanation
Solution:
The frequency of oscillation of a compass needle in a magnetic field is given by: $$ f = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}} $$ where:
$f$ is the frequency of oscillation
$M$ is the magnetic moment of the needle
$B_H$ is the horizontal component of the Earth's magnetic field
$I$ is the moment of inertia of the needle
Since $M$ and $I$ are constant, we have:
$$ f \propto \sqrt{B_H} $$ $$ f^2 \propto B_H $$ Also, $B_H = B \cos \delta$, where $B$ is the total magnetic field and $\delta$ is the dip angle. So, $f^2 \propto B \cos \delta$. Let $f_1 = 20$ oscillations/minute and $\delta_1 = 30^\circ$.
Let $f_2 = 30$ oscillations/minute and $\delta_2 = 60^\circ$. Then, $$ f_1^2 \propto B_1 \cos \delta_1 $$ $$ f_2^2 \propto B_2 \cos \delta_2 $$ Therefore, $$ \frac{f_1^2}{f_2^2} = \frac{B_1 \cos \delta_1}{B_2 \cos \delta_2} $$ Substituting the given values: $$ \frac{20^2}{30^2} = \frac{B_1 \cos 30^\circ}{B_2 \cos 60^\circ} $$ $$ \frac{400}{900} = \frac{B_1 (\sqrt{3}/2)}{B_2 (1/2)} $$ $$ \frac{4}{9} = \frac{B_1}{B_2} \sqrt{3} $$ $$ \frac{B_1}{B_2} = \frac{4}{9\sqrt{3}} $$ We are given that $\frac{B_1}{B_2} = \frac{4}{\sqrt{x}}$. Therefore, $$ \frac{4}{\sqrt{x}} = \frac{4}{9\sqrt{3}} $$ $$ \sqrt{x} = 9\sqrt{3} $$ $$ x = (9\sqrt{3})^2 = 81 \cdot 3 = 243 $$ Thus, $x = 243$.
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