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Question: A dip needle initially in the magnetic meridian when it shows an angle of dip \(\theta \) at a place...

A dip needle initially in the magnetic meridian when it shows an angle of dip θ\theta at a place. The dip circle is rotated through an angle xx in the horizontal plane and then it shows an angle of dip θ\theta '. Then, tanθtanθ\dfrac{{\tan \theta }}{{\tan \theta '}} will be
A. cosx\cos x
B. 1cosx\dfrac{1}{{\cos x}}
C. 1sinx\dfrac{1}{{\sin x}}
D. 1tanx\dfrac{1}{{\tan x}}

Explanation

Solution

The angle of dip or simply dip is an angle made by the total magnetic field of the Earth, with its surface (or horizontal). The dip needle is the device used for measuring dip at a certain place. The change in the dip needle alters the horizontal component, as the dip circle is placed on a horizontal surface.

Formula used:
tanθ=BVBH\tan \theta = \dfrac{{{B_V}}}{{{B_H}}}

Complete answer:
The dip needle or dip circle is a device used to measure the dip angle at a given place. It is perfectly balanced about the horizontal axis. So that the needle is free to swing along the magnetic meridian.

The dip angle is said to be the angle between the earth’s magnetic field and the horizontal. It varies from place to place. Mathematically it is given by,
tanθ=BVBH\tan \theta = \dfrac{{{B_V}}}{{{B_H}}}
Where,
θ\theta is the dip angle
BV{B_V} it the vertical component of Earth’s magnetic field
BH{B_H} is the horizontal component of Earth’s magnetic field
According to the condition given in the question, let us say that initially the dip angle is θi=θ{\theta _i} = \theta . Then,
\eqalign{ & \tan {\theta _i} = \tan \theta = \dfrac{{{B_V}}}{{{B_H}}} \cr & \cr} .
Once the dip circle is rotated through an angle xx is the vertical component of the earth’s magnetic field remains the same. But, the horizontal component changes to BHcosx{B_H}\cos x. If we assume that the angle now is θf=θ{\theta _f} = \theta ', then
\eqalign{ & \tan {\theta _f} = \tan \theta ' = \dfrac{{{B_V}}}{{{B_H}\cos x}} \cr & \cr}
From the above equations, we can write
\eqalign{ & \dfrac{{\tan {\theta _i}}}{{\tan {\theta _f}}} = \dfrac{{\tan \theta }}{{\tan \theta '}} = \dfrac{{\dfrac{{{B_V}}}{{{B_H}}}}}{{\dfrac{{{B_V}}}{{{B_H}\cos x}}}} = \dfrac{{{B_V}}}{{{B_H}}} \times \dfrac{{{B_H}\cos x}}{{{B_V}}} = \cos x \cr & \therefore \dfrac{{\tan \theta }}{{\tan \theta '}} = \cos x \cr & \cr}

Therefore, the correct option is A.

Note:
Don’t be confused - Magnetic dip, dip, and magnetic inclination are interchangeably used for dip angle. Dip angles play a prominent role in the field of aviation. The lines plotted along the places where the dip angle or magnetic dip is the same are known as isoclinic lines. The line drawn joining the points where the dip angle is zero is called the magnetic equator.
Please observe that the magnetic meridian and geographic meridian of earth are completely different. The angle between the magnetic meridian and the geographic median is called the magnetic declination.