Question

In a vibration magnetometer, the time period of a bar magnet oscillating in the horizontal component of earth’s magnetic field is 2 sec. When another magnet is brought near and parallel to it, then the time period reduces to 1 sec. The ratio of the horizontal component of the earth’s field and the field due to magnet will be
$
  (a)\dfrac{1}{3} \\
  (b){\text{ 3}} \\
  {\text{(c) }}\sqrt 3 \\
  (d){\text{ }}\dfrac{1}{{\sqrt 3 }} \\
$

Answer 1

Hint: In this question use the direct formula for the time period of a bar magnet in terms of the horizontal component of earth’s magnetic field that is $T = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} $. When another magnet is brought near now the horizontal component will be the resultant horizontal magnetic field and it will get added up with the horizontal component of another magnet that is being brought near.

Formula used –As we know that the relation between time period of a bar magnet in terms of horizontal component of earth magnetic field is given as,
$T = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} $, where T = time period of bar magnet.
                                              I = moment of inertia of bar magnet.
                                             M = Magnetic moment of bar magnet.
                                            ${B_H}$ = Horizontal component of earth magnetic field.

Complete Step-by-Step solution:
Now it is given that the time period of this is 2 sec.
 $ \Rightarrow T = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} = 2$.................... (1)
Now when another magnet is brought near and parallel to it the resultant horizontal component of the magnetic field got added.
Let the horizontal component of another magnet = B
So the resultant horizontal magnetic field = ${B_H} + B$
And the time period $\left( {{T_1}} \right)$ of this combination is given which is 1 sec.
$ \Rightarrow {T_1} = 2\pi \sqrt {\dfrac{I}{{M\left( {{B_H} + B} \right)}}} = 1$........................ (2)
Now divide equation (1) and (2) we have,
$ \Rightarrow \dfrac{{2\pi \sqrt {\dfrac{I}{{M{B_H}}}} }}{{2\pi \sqrt {\dfrac{I}{{M\left( {{B_H} + B} \right)}}} }} = \dfrac{2}{1}$
Now simplify this equation we have,
$ \Rightarrow \dfrac{{\sqrt {{B_H} + B} }}{{\sqrt {{B_H}} }} = \dfrac{2}{1}$
Now take square on both sides we have,
$ \Rightarrow \dfrac{{{B_H} + B}}{{{B_H}}} = {\left( {\dfrac{2}{1}} \right)^2} = 4$
$ \Rightarrow {B_H} + B = 4{B_H}$
$ \Rightarrow 3{B_H} = B$
$ \Rightarrow \dfrac{{{B_H}}}{B} = \dfrac{1}{3}$
So this is the required ratio of horizontal component of earth’s field and the field due to magnet.
Hence option (A) is the correct answer.

Note – There are three components that are responsible for the magnitude as well as the direction of the earth’s magnetic field. Magnetic declination, magnetic inclination or the angle of dip and the horizontal component of the earth’s magnetic field. The intensity of magnetic field is greater near the magnetic poles and the intensity is weakest near the equator.