Terrestrial Magnetism (Earth's Field)

PHYTerrestrial Magnetism

The Earth's magnetic field is described by three elements: the angle of declination, the angle of dip $\delta$ , and the horizontal component $B_H$ . The total field $B$ resolves into a horizontal part $B_H=B\cos\delta$ and a vertical part $B_V=B\sin\delta$ , so $\tan\delta=B_V/B_H$ .

ISC numericals ask you to resolve the field into components, find the dip and total field from the components, and (a linked skill) compute the field of a current loop that might be compared with the Earth's field.

Horizontal and vertical components

B_H = B\cos\delta, \qquad B_V = B\sin\delta

B

total field,

\delta

angle of dip;

B_H

is the horizontal component,

B_V

the vertical.

Angle of dip from components

\tan\delta = \dfrac{B_V}{B_H}

\delta

dip angle, the angle the resultant field makes with the horizontal.

Total field from components

B = \sqrt{B_H^2 + B_V^2}

Resultant of the perpendicular horizontal and vertical components.

Field at centre of a square loop (side

a

)

B = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi a}

a

side length,

I

current; sum of the fields from the four finite straight sides at the centre.

The three magnetic elements of the Earth are declination, dip (inclination) $\delta$ , and the horizontal component $B_H$ .
Angle of dip $\delta=0^{\circ}$ at the magnetic equator (field horizontal) and $\delta=90^{\circ}$ at the magnetic poles (field vertical).
$1\ \mathrm{gauss}\ (\mathrm{G}) = 10^{-4}\ \mathrm{T}$ ; the Earth's field is of order a few $\times10^{-5}\ \mathrm{T}$ , i.e. a few tenths of a gauss.
A compass needle aligns with $B_H$ (horizontal), while a dip needle aligns with the full field $B$ at angle $\delta$ to the horizontal.
$B_H$ , $B_V$ and $B$ form a right-angled triangle, so any two elements determine the rest.
For the square loop, each side is a finite straight wire; the four equal contributions add to give $B=2\sqrt2\,\mu_0 I/\pi a$ .
Always convert the side length to metres and keep the field in tesla (convert gauss if given) before substituting.

Where the marks go

Swapping $\cos\delta$ and $\sin\delta$ : the horizontal component uses $\cos\delta$ , the vertical uses $\sin\delta$ .
Writing $\tan\delta = B_H/B_V$ upside down; it is $\tan\delta = B_V/B_H$ .
Forgetting the gauss-to-tesla conversion ( $1\ \mathrm{G}=10^{-4}\ \mathrm{T}$ ) when components are given in gauss.
Using the circular-loop formula $\mu_0 I/2R$ for a square loop instead of $2\sqrt2\,\mu_0 I/\pi a$ .

How the board asks it

Numericaltan(delta)=B_V/B_H and B=sqrt(B_H^2+B_V^2)
At a place the horizontal component of the Earth's magnetic field is $B_H = 0.26\ \mathrm{G}$ and the vertical component is $B_V = 0.45\ \mathrm{G}$ . Calculate the angle of dip $\delta$ and the total magnetic field $B$ of the Earth at that place.
Define / statethe three magnetic elements of the earth
Name the three magnetic elements of the Earth. Define the angle of dip (inclination) $\delta$ at a place and state its value at the magnetic equator and at the magnetic poles.
Derive / prove $B_H = B\cos\delta$ , $B_V = B\sin\delta$
The Earth's total magnetic field $B$ acts at an angle of dip $\delta$ to the horizontal. Obtain expressions for its horizontal component $B_H$ and vertical component $B_V$ , and hence show that $\tan\delta = B_V/B_H$ .
Numericalfield at centre of a square loop, $B=2\sqrt2\,\mu_0 I/\pi a$
A square loop of side $a = 10\ \mathrm{cm}$ carries a current $I = 5\ \mathrm{A}$ . Calculate the magnetic field at the centre of the loop and compare its magnitude with the Earth's horizontal field $B_H = 0.4\ \mathrm{G}$ .
Give reasonsdip needle aligns with full field B at angle delta
Give reasons: a freely suspended dip needle remains horizontal at the magnetic equator but stands vertical at the magnetic poles.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.