Question

A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45$^\circ $, and 40 times per minute where the dip is 30$^\circ $. If ${B_1}$ and ${B_2}$are respectively the total magnetic field due to the earth at the two places, then the ratio ${B_1}/{B_2}$ is best given by:
A. 2.2
B. 1.8
C. 0.7
D. 3.6

Answer 1

Hint: The frequency of oscillation of a magnetic needle at a place is directly related to the square root of the horizontal component of the earth’s magnetic field at that place. By comparing the magnetic fields obtained at two given places, we can get the required ratio of magnetic fields.

Formula used:
The relation between the frequency of oscillation of a magnetic needle and the magnetic field at a place is given by the following expression:
$\nu = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{\mu B\cos \theta }}{I}} $

Complete step by step answer:
The frequency of oscillations of a magnetic needle in earth’s magnetic field is given as
$\nu = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{\mu {B_H}}}{I}} $
Here $\mu $ is the magnetic moment of the magnetic needle, ${B_H}$ represents the horizontal component of earth’s magnetic field while I represents the moment of inertia of the magnetic needle about the axis about which it oscillates. The horizontal component of earth’s magnetic field is given as follows:
${B_H} = B\cos \theta $
Here $\theta $ is known as the angle of dip which the magnetic field makes with the horizontal axis. Therefore, we have
$\nu = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{\mu B\cos \theta }}{I}} $
We are using the same magnetic needle so the value of magnetic moment and the moment of inertia will remain the same at the two places. So, we can write that
$\nu \propto \sqrt {B\cos \theta } $
Now for the magnetic needle at first place, we are given that
$
  {\nu _1} = 30{\min ^{ - 1}} \\
  {\theta _1} = 45^\circ \\
 $
For this place, we have
${\nu _1} \propto \sqrt {{B_1}\cos {\theta _1}} $ …(i)
For the magnetic needle at second place, we are given that
$
  {\nu _2} = 40{\min ^{ - 1}} \\
  {\theta _2} = 30^\circ \\
 $
For this place, we have
${\nu _2} \propto \sqrt {{B_2}\cos {\theta _2}} $ …(ii)
Dividing equation (i) by equation (ii), we get
$\dfrac{{{\nu _1}}}{{{\nu _2}}} = \sqrt {\dfrac{{{B_1}\cos {\theta _1}}}{{{B_2}\cos {\theta _2}}}} $
Now on inserting the known values, we get
$\dfrac{{30}}{{40}} = \sqrt {\dfrac{{{B_1}\cos 45^\circ }}{{{B_2}\cos 30^\circ }}} $
Squaring both sides, we get
$
  \dfrac{{900}}{{1600}} = \dfrac{{{B_1} \times \dfrac{1}{{\sqrt 2 }}}}{{{B_2} \times \dfrac{{\sqrt 3 }}{2}}} \\
   \Rightarrow \dfrac{9}{{16}} = \sqrt {\dfrac{2}{3}} \times \dfrac{{{B_1}}}{{{B_2}}} \\
   \Rightarrow \dfrac{{{B_1}}}{{{B_2}}} = \dfrac{9}{{16}} \times \sqrt {\dfrac{3}{2}} \\
   \therefore \dfrac{{{B_1}}}{{{B_2}}} = 0.68 \simeq 0.7 \\
 $
Hence, the correct answer is option C.

Note:
The angle of dip of earth’s magnetic field represents the angle that the earth’s magnetic field vector makes with the horizon at a particular place on earth. The horizontal component of the magnetic field is the effective magnetic field at the given place on earth.