Question

Three quantities are needed to specify the magnetic field of the Earth on its
Surface-the horizontal component, the magnetic declination and ....​

Answer 1

EXPLANATION.

Three quantities are needed to specify the magnetic field of the earth on it's surface.

1. The horizontal component.
2. The magnetic declination.
3. The angle of dip.

Earth's magnetic field.

The earth magnetic field $B_{e}$ in the magnetic meridian may be resolved into a horizontal component $B_{H}$ and the vertical component $B_{V}$ at any place.

θ = dip Or inclination (α) = declination.

Horizontal component of earth's magnetic field.

$B_{H} = B_{e} cos \theta. - - - - - (1).$

Vertical component of earth's magnetic field.

$B_{V} = B_{e} sin \theta. - - - - - (2).$

$B_{e} = \sqrt{(B_{H}^{2} + B_{V}^{2}) }$

Dividing equation (2) by equation (1), we get.

$\displaystyle \frac{B_{V}}{B_{H}} = \frac{B_{e} sin \theta}{B_{e} cos \theta} = tan \theta$

$\displaystyle \frac{B_{V}}{B_{H}} = tan \theta$

Apparent Dip.

If the dip circle is not kept in the magnetic meridian, the needle will not show the correct direction of earth magnetic field. The angle made by the needle with the horizontal is called the apparent dip for this place. if the dip circle is at an angle θ to the meridian.

The effective horizontal component in this place is,

$\displaystyle B_{H}' = B_{H} cos \theta$

The vertical component is still : $B_{V}$

If $\delta_{1}$ is the apparent dip and $\delta$ is the true dip then,

$\displaystyle tan \delta_{1} = \frac{B_{V}}{B_{H}'} = \frac{B_{V}}{B_{H} cos \theta}$

$\displaystyle tan \delta_{1} = \frac{tan \delta}{cos \theta}. - - - - - (1).$

Now, supposed the dip circle is rotated through an angle of 90° from this position, it will now make an angle (90° - θ) with the meridian.

The effective horizontal component in this place is,

$B_{H}'' = B_{H} sin \theta$

$\delta_{2}$ be the apparent dip, we shall have.

$\displaystyle tan \delta_{2} = \frac{B_{V}}{B_{H}'} = \frac{B_{V}}{B_{H} sin \theta}$

$\displaystyle tan \delta_{2} = \frac{tan \delta}{sin \theta} . - - - - - (2).$

From equation (1) and equation (2), we get.

$cot^{2} \delta_{1} + cot^{2} \delta_{2} = cot^{2} \delta$

Thus, one can get the true dip $\delta$ without locating the magnetic meridian.

More about angle of dip (δ).

(1) = At a place on poles earth magnetic field is perpendicular to the surface of earth.

δ = 90°.

$B_{V} = B sin (90^{\circ}) = B$

$B_{H} = B cos (90^{\circ} ) = 0$

So except at poles the earth has a horizontal component of magnetic induction field.

(2) At a place on equator, earth's magnetic field is parallel to the surface of earth.

δ = 0°.

$B_{H} = B cos (0^{\circ}) = B$

$B_{V} = B_{H} sin (0^{\circ} )= 0$

So, except at equator the earth's has vertical component of magnetic induction field.

(3) In a vertical plane at an angle θ to magnetic meridian.

$B_{H}' = B_{H} cos \theta \ \ and \ \ B_{V}' = B_{V}$

So the angle of dip $\delta '$ in a vertical plane making an angle θ to magnetic meridian is given by,

$\displaystyle tan \delta ' = \frac{B_{V}'}{B_{H}'} = \frac{B_{V}}{B_{H} cos \theta}$

$\displaystyle tan \delta ' = \frac{tan \delta}{cos \theta}$

(a) for a vertical plane other than magnetic meridian,

θ > 0°  and  cosθ < 1.

$\delta ' > \delta \ \ [ angle \ of \ dip \ increase]$

(b) For a plane perpendicular to magnetic meridian.

θ = 90°.

$\displaystyle tan \delta' = \frac{tan \delta}{cos (90^{\circ} )} = \infty$

$\delta ' = 90^{\circ}$

This shows that in a plane perpendicular to magnetic meridian the dip needle will become vertical.