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Earth's magnetism · geomagnetic elements · deep dive

The Earth Is a Magnet
Installed Upside Down

Three numbers are enough to describe the Earth's magnetic field anywhere on the planet: which way it points on a map, how steeply it dives into the ground, and how much of it is left lying flat. This page builds all three from scratch, derives the triangle that connects them, and then handles the one genuinely tricky case — apparent dip.

Spin the compassTilt the fieldTravel the globeRotate the dip circle

Before you start — where this topic counts

Earth's magnetism has been removed from the rationalised NCERT Chapter 5, and "magnetic elements" is on the deleted list for the JEE Main syllabus too. For NEET 2027 and JEE Main 2027 specifically, declination, dip and the tangent galvanometer are out of scope. Older papers do contain them — NEET 2019 and 2017, AIPMT 2012, several JEE Main 2021 and 2022 questions — which is why they still float around in practice sets.

It is still examined in KCET, MHT-CET, WBJEE, AP and TS EAMCET, BITSAT and several state boards, and it remains genuinely useful physics. So this page is built in full. Just be deliberate about how much time you give it relative to torque, energy and the dipole fields, which are worth several times as many marks.

PART 01

The riddle in the name

Why the Earth's north is magnetically a south
Story

A giant magnet buried in the planet

Picture an enormous bar magnet buried inside the Earth, running roughly top to bottom. That is not literally what is down there — the real cause is molten iron churning in the outer core — but as a picture for calculating, it works beautifully.

Now here is the riddle. A compass needle's north end swings towards the Earth's geographic north. But north attracts south, never north. So whatever is buried up there must be a magnetic south pole.

The Earth's magnet is installed upside down compared with its name. Nobody made a mistake — sailors named the compass ends thousands of years before anyone knew why the needle turned. The name came first and the explanation came later, and by then it was too late to change.

Numbers worth carrying

QuantityTypical valueNote
Total field B≈ 0.3 to 0.6 GThat is 3 to 6 × 10−5 T — a very weak field
Tilt of the magnetic axis≈ 11°Why magnetic north and geographic north disagree
Dip at the magnetic equatorField is entirely horizontal there
Dip at the magnetic poles90°Field is entirely vertical there
Dip in India (roughly)10° to 45°Rises as you go north from the tip of the peninsula

Remember 1 gauss = 10−4 tesla, so a value quoted as "0.5 G" is 5 × 10−5 T. Questions switch between the two units freely.

Two meridians, and the difference between them

A meridian is just a vertical plane — imagine a huge sheet of glass standing on its edge.

Geographic meridian: the vertical plane through your position and the Earth's geographic north–south axis. This is what a map means by north.

Magnetic meridian: the vertical plane through your position containing the Earth's magnetic field. This is where a freely hanging compass needle lies.

These two sheets of glass do not coincide, and the angle between them is the first of our three elements.

PART 02

Three numbers, and why exactly three

Declination D, dip δ, horizontal component BH
Story

Describing a thrown stick

Suppose a stick is lying on the ground, tilted into the earth at one end, and you have to describe its position over the phone. What must you say?

One: which compass direction it points along — "north-east-ish". Two: how steeply it dives into the ground — "about thirty degrees down". Three: how long it is.

Three numbers, and the person on the phone can reproduce it exactly. The Earth's field is an arrow in space, so it needs those same three numbers. We call them declination (which way), dip (how steeply), and horizontal component (how much).

Lab 1 · Declination, seen from above

drag the needle

This is a bird's-eye view, looking straight down at the ground. The grey arrow is where a map says north is. The red needle is where your compass actually points. The angle between them is the declination.

Declination D = Direction: Compass error:

Declination is the angle between the geographic meridian and the magnetic meridian, measured in the horizontal plane. It is called east declination when the compass points east of true north, west declination otherwise.

Why a pilot cares

If declination is 15° and you fly by compass alone for 1000 km believing you are heading true north, you end up roughly 260 km off target. This is why every aeronautical chart prints the local declination, and why it is the practical reason the quantity has a name at all.

PART 03

The triangle that runs everything

BH = B cos δ · BV = B sin δ · tan δ = BV/BH
Story

A ladder against a wall

Lean a ladder against a wall. The ladder itself is the total field B. How far its foot sits from the wall is the horizontal part. How high it reaches up the wall is the vertical part.

Stand the ladder almost upright and it barely sticks out along the ground — big vertical part, small horizontal part. Lay it almost flat and the opposite happens.

The dip angle δ is simply how steep the ladder is. And because the ladder, the ground and the wall make a right-angled triangle, everything follows from sines and cosines you already know.

Maths

Resolving B into two components

The total field B lies in the magnetic meridian, tilted below the horizontal by the dip angle δ. Resolve it along the horizontal and the vertical:

BH = B cos δ    BV = B sin δ

Divide the second by the first and B cancels:

tan δ = BV / BH

Square and add them, and use sin2δ + cos2δ = 1:

B = √(BH2 + BV2)

Four relations, but only one triangle. Given any two of B, BH, BV, δ you can always find the other two.

Lab 2 · The dip triangle

drag the field arrow

Side view, looking along the magnetic meridian. Drag the blue arrow to tilt the field, or use the sliders. The triangle and all four quantities update together.

δ = B = BH = B cosδ = BV = B sinδ = tanδ =

Watch the two extremes. At δ = 0° the whole field is horizontal and BV vanishes. At δ = 90° the whole field is vertical and BH vanishes — and a compass, which can only respond to the horizontal part, becomes completely useless.

The one thing a compass can and cannot do

An ordinary compass needle is pivoted so it can only swing horizontally. It therefore responds to BH alone and is completely blind to BV.

That is why AIPMT 2012 asks what happens to a compass taken to a geomagnetic pole: there BH = 0, so there is no horizontal field to align with, and the needle can point anywhere at all. To measure the dip you need a different instrument — a dip circle, whose needle swings in a vertical plane.

Quick check

At a certain place the dip is 60°. The ratio BV : BH is

PART 04

Travelling from equator to pole

How dip changes with latitude · tan δ = 2 tan λ
Story

Field lines leaving and arriving

Picture the field lines of the buried magnet. Near the magnetic equator they run flat, sliding along parallel to the ground. Stand there and the field has no downward part at all — the dip is zero.

Walk towards a magnetic pole and the lines begin to tilt, then steepen, until right at the pole they plunge straight down into the earth. The dip has climbed to 90°.

Everywhere in between the field is slanting, and the dip angle tells you exactly how slanted.

Lab 3 · Around the globe

drag your position

Drag the marker anywhere on the surface. The small needle shows how a dip circle would sit at that spot, and the readouts give the dip and the components there.

Latitude λ = Dip δ = B / Bequator = Needle:

For a simple dipole model the two relations are tan δ = 2 tan λ and B = B0√(1 + 3sin²λ). Note the factor of 2 — the dip angle is always steeper than the latitude, which is why India sees dips of 40°+ at latitudes of only about 25°.

Where the factor of 2 comes from

It is the same 2 you already know. At magnetic latitude λ, the vertical part of a dipole's field is the axial-type component and carries the factor 2, while the horizontal part is the equatorial-type component and does not:

BV = 2B0 sin λ    BH = B0 cos λ

Divide: tan δ = BV/BH = 2 sinλ/cosλ = 2 tan λ. The famous 2 : 1 axial-to-equatorial ratio from the bar magnet reappears here wearing geography's clothes.

Quick check

At a place A the angle of dip is +25°; at a place B it is −25°. We can conclude that

PART 05

Apparent dip — the only tricky bit

tan δ′ = tan δ / cos θ · and the cot² relation
Story

Reading a hill at an angle

Imagine a hillside sloping down at 30°. Walk straight down it and you feel the full 30°. But walk diagonally across the face and the path under your feet feels gentler — you are not taking the steepest line.

The dip circle works the opposite way round, and this is the bit that catches people. When you rotate the instrument away from the magnetic meridian, it sees less of the horizontal field but the same vertical field. Less horizontal with the same vertical means the needle appears steeper.

So apparent dip is always greater than true dip. Never smaller. That single fact catches most of the wrong answers.

Maths

Deriving the apparent dip formula

Let the dip circle's vertical plane be rotated by an angle θ away from the magnetic meridian.

Step 1. The vertical component lies along the rotation axis itself, so it is unaffected: still BV.

Step 2. The horizontal component BH lies in the magnetic meridian. Only its projection into the new plane acts on the needle, and that projection is BH cos θ.

Step 3. The needle settles along the resultant of what it can feel, so the apparent dip δ′ satisfies

tan δ′ = BV / (BH cos θ) = tan δ / cos θ

Check the limits. At θ = 0 the plane is the magnetic meridian, cosθ = 1, and δ′ = δ — the true dip. At θ = 90° the plane is perpendicular to the meridian, cosθ = 0, and tanδ′ → ∞, so δ′ = 90°. The needle stands bolt upright, because no horizontal field reaches it at all.

Lab 4 · Rotate the dip circle

two views, one instrument

Left is the view from above, showing how far the instrument has been swung from the magnetic meridian. Right is what the needle inside it actually does. The true dip stays fixed throughout — only the reading changes.

θ = True dip δ = Effective horizontal = BHcosθ = Apparent dip δ′ =

Slide θ up and watch the apparent dip climb while the true dip sits still. The gap between them is entirely an artefact of how the instrument is aimed — the Earth has not changed.

Two perpendicular planes — the cot² relation

Suppose you take two readings in vertical planes at right angles to each other. Call them δ1 and δ2. If the first plane is at angle θ to the meridian, the second is at (90° − θ), so

tan δ1 = tanδ/cosθ    tan δ2 = tanδ/sinθ

Take cotangents and square them:

cot²δ1 + cot²δ2 = (cos²θ + sin²θ)/tan²δ = cot²δ

The unknown rotation θ vanishes completely. That is the point of the trick: two readings taken at right angles give you the true dip without ever knowing which way you were pointing. This exact relation was asked as NEET 2017.

Quick check

A dip circle is rotated away from the magnetic meridian. The apparent dip it reads is

PART 06

Worked problems

Real past questions, in the standard solution format
Find B and BV from BH and dip · JEE Main 2022 (29 June)
Given
Dip δ = 30°; horizontal component BH = 0.5 G.
Asked
The Earth's total magnetic field at that place, in gauss.
Concept
BH is the projection of B onto the horizontal, so B is recovered by dividing by cos δ.
Formula
BH = B cos δ ⇒ B = BH/cos δ
Solution
cos 30° = √3/2 ≈ 0.866
B = 0.5 / 0.866 = 0.577 G, i.e. 1/√3 G ≈ 5.77 × 10−5 T
Sense check
B must always exceed BH, since the horizontal part is only a piece of the whole. 0.577 > 0.5 ✔
Find B from BV and dip · JEE Main 2022 (29 July)
Given
Vertical component BV = 6 × 10−5 T; dip δ = 37°, with tan 37° = 3/4.
Asked
The resultant field B.
Concept
Get BH from the tangent relation, then combine the two components by Pythagoras. The 3-4-5 triangle is hidden in the given tangent.
Formula
tan δ = BV/BH;   B = √(BH² + BV²)
Solution
BH = BV/tan δ = 6 × 10−5 / (3/4) = 8 × 10−5 T
B = √(8² + 6²) × 10−5 = √100 × 10−5
B = 10−4 T
Shortcut
tan 37° = 3/4 is the signal for a 3-4-5 triangle. Once BV = 6 pairs with BH = 8, the hypotenuse is 10 without any square roots.
True dip from apparent dip · JEE Main 2021 (20 July)
Given
The dip circle is at θ = 30° to the magnetic meridian; the apparent dip read is δ′ = 45°.
Asked
The true dip δ.
Concept
Rotating the plane reduces the effective horizontal field to BHcosθ while leaving BV untouched, which inflates the reading. Undo that inflation.
Formula
tan δ′ = tan δ / cos θ ⇒ tan δ = tan δ′ × cos θ
Solution
tan δ = tan 45° × cos 30° = 1 × 0.866 = 0.866
δ = tan−1(0.866) ≈ 40.9°, i.e. tan−1(√3/2)
Sense check
True dip must be less than the apparent dip. 40.9° < 45° ✔ If your answer had come out above 45°, you multiplied where you should have divided.
Magnet hung at an angle · JEE Main 2022 (27 July)
Given
A magnet hung in a vertical plane at 45° to the magnetic meridian settles at 60° to the horizontal.
Asked
The true angle of dip.
Concept
Identical structure to the previous problem. The 60° reading is an apparent dip, taken in a plane rotated by 45°.
Formula
tan δ = tan δ′ × cos θ
Solution
tan δ = tan 60° × cos 45° = √3 × (1/√2) = √3/√2 = √1.5 ≈ 1.225
δ = tan−1(√(3/2)) ≈ 50.8°
Watch out
The phrase "hung at 45° with the magnetic meridian" describes the plane, and "makes 60° with the horizontal" describes the needle. Mixing up which angle is θ and which is δ′ inverts the whole calculation.
Two perpendicular planes · NEET 2017
Given
Apparent dips δ1 and δ2 observed in two vertical planes at right angles to each other.
Asked
The true angle of dip δ, in terms of δ1 and δ2.
Concept
Write both apparent dips in terms of the same unknown rotation θ, then combine so that θ cancels. Cotangents are the natural form because they put cosθ and sinθ in the numerators.
Formula
tan δ1 = tanδ/cosθ and tan δ2 = tanδ/sinθ
Solution
cot δ1 = cosθ/tanδ and cot δ2 = sinθ/tanδ
Square and add: cot²δ1 + cot²δ2 = (cos²θ + sin²θ)/tan²δ = 1/tan²δ
cot²δ = cot²δ1 + cot²δ2
Why it matters
The unknown θ has disappeared. Two readings taken at right angles give the true dip without you ever having to find the magnetic meridian — which is exactly how the measurement is made in practice.
PART 07

Everything on one page

The whole topic, compressed
BH = B cos δ · BV = B sin δ · tan δ = BV/BH · B = √(BH²+BV²) Apparent dip: tan δ′ = tan δ / cos θ  (δ′ ≥ δ always) Two perpendicular planes: cot²δ = cot²δ1 + cot²δ2 Dipole model: tan δ = 2 tan λ · B = B0√(1 + 3 sin²λ)
ElementSymbolWhat it measuresMeasured in which plane
DeclinationDAngle between geographic and magnetic meridiansHorizontal
Dip / inclinationδAngle of B below the horizontalVertical (the magnetic meridian)
Horizontal componentBHThe part of B lying flat along the groundHorizontal

Five sentences that carry the whole topic

1. A magnetic south pole sits near the geographic north, which is why compass norths point that way.
2. Three numbers fix the field: which way (D), how steep (δ), how much lies flat (BH).
3. B, BH and BV form a right-angled triangle with δ as the angle — every formula is just trigonometry on it.
4. Dip runs from 0° at the magnetic equator to 90° at the magnetic poles, following tanδ = 2tanλ.
5. Rotating the dip circle keeps BV but shrinks the horizontal to BHcosθ, so the apparent dip is always steeper than the truth.

Four traps this topic sets

1. Thinking a magnetic north pole is buried under geographic north. It is a south.
2. Multiplying instead of dividing in the apparent dip formula. Check the direction: apparent must be the bigger angle.
3. Mixing up which given angle is the plane's rotation θ and which is the needle's reading δ′.
4. Forgetting that an ordinary compass only feels BH — which is why it fails completely at a magnetic pole.

A word on time allocation

This topic is not in the current NEET or JEE Main syllabus. If those are your target exams, treat this page as background reading and pattern recognition — enough that an old practice paper cannot ambush you — rather than as material to drill.

If you are also sitting KCET, MHT-CET, WBJEE, EAMCET or BITSAT, it is live and worth real attention. In that case the highest-value items are the dip triangle in Part 3 and the apparent dip formula in Part 5, in that order. The five worked problems in Part 6 cover essentially every way those two have been asked.