JEE Mains Physics · Magnetism and Matter

The Earth's Field: Dip and Oscillating Magnets

The earth's field B splits into a horizontal part B cos δ and a vertical part B sin δ, where δ is the angle of dip; the horizontal part is the one that cancels a magnet's field at a neutral point and sets the period of a magnet swinging in a horizontal plane.

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Why this matters

Nine PYQs, one of them asking for a number, and none since 2023. Five resolve the earth's field into components or compare the dip read in some other plane with the true dip; four use the horizontal component, either to cancel a magnet's field or to time an oscillating needle. Each needs one relation, used with the right component.

Concept 1 of 2: Components of the earth's field and the angle of dip

The earth's field at a place points down into the ground at the angle of dip δ. Its horizontal part, B cos δ, lies along the magnetic meridian; its vertical part is B sin δ. A dip circle set in some other vertical plane still feels the whole vertical part but only a share of the horizontal part, so it reads a steeper dip than the true one.

Definition

$B_H = B\cos\delta$ , $B_V = B\sin\delta$ , $\tan\delta = B_V/B_H$ , $B = \sqrt{B_H^{2} + B_V^{2}}$ .
Declination is the angle between the geographic and the magnetic meridian. The dip formulas use the MAGNETIC meridian.
A dip circle in a vertical plane at α to the magnetic meridian reads $\delta'$ : $\tan\delta' = \dfrac{\tan\delta}{\cos\alpha}$ . So $\delta' \geq \delta$ , equal only in the meridian itself.
In the plane perpendicular to the meridian (α = 90°) the needle stands vertical: $\delta' = 90^{\circ}$ .
Apparent dips $\delta_1$ and $\delta_2$ in two perpendicular vertical planes: $\cot^{2}\delta = \cot^{2}\delta_1 + \cot^{2}\delta_2$ .
At the magnetic equator δ = 0 and $B_V = 0$ ; at a magnetic pole δ = 90° and $B_H = 0$ .

Components, and dip in a plane at α to the meridian

B_H = B\cos\delta \qquad B_V = B\sin\delta \qquad \tan\delta' = \frac{\tan\delta}{\cos\alpha} \qquad \cot^{2}\delta = \cot^{2}\delta_1 + \cot^{2}\delta_2

Worked example

At a place the earth's total field is 0.5 G and the angle of dip is 53°

(\tan 53^{\circ} = 4/3)

. Find

B_H

and

B_V

. A dip circle is then set in a vertical plane at 60° to the magnetic meridian. What dip does it show?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q108Moderate

Example 1 · Magnetism and Matter · The Earth's Field: Dip and Oscillating Magnets

The vertical component of the earth's magnetic field is

6 \times10^{- 5}\text{ }T

at any place where the angle of dip is

37^{\circ}

. The earth's resultant magnetic field at that place will be (Given

\tan37^{\circ}=\frac{3}{4}

)

Swapping sine and cosine

The horizontal component is B cos δ and the vertical one is B sin δ. At a large dip the field is nearly vertical, so the vertical part is the larger one.

Geographic meridian in place of magnetic

The dip relations use the angle from the magnetic meridian. They give the same answer for the geographic meridian only when the declination is zero.

Drill 4 more on components of the earth's field and the angle of dip

Concept 2 of 2: Horizontal component: neutral points and oscillating magnets

A magnet that can turn only in a horizontal plane feels only the horizontal part of the earth's field. Pull it aside and it swings about the meridian like a pendulum, faster when B_H is larger. Near a fixed magnet there are points where its field exactly cancels B_H; a compass there points nowhere in particular. These are the neutral points.

Definition

Period of a magnet swinging in a horizontal plane: $T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ . The frequency n (oscillations per minute) goes as $\sqrt{MB_H/I}$ .
Two magnets in the same field: $\dfrac{T_1^{2}}{T_2^{2}} = \dfrac{I_1}{I_2}\cdot\dfrac{M_2}{M_1}$ .
One needle at two places: $n^{2} \propto B_H = B\cos\delta$ , so $\dfrac{n_1^{2}}{n_2^{2}} = \dfrac{B_1\cos\delta_1}{B_2\cos\delta_2}$ .
Neutral point: the magnet's field equals $B_H$ and points the other way.
N pole pointing north: the neutral points lie on the magnet's equatorial line, $\dfrac{\mu_0}{4\pi}\dfrac{M}{(r^{2} + l^{2})^{3/2}} = B_H$ .
N pole pointing south: they lie on its axis, $\dfrac{\mu_0}{4\pi}\dfrac{2Mr}{(r^{2} - l^{2})^{2}} = B_H$ .

Oscillation period and the frequency at two places

T = 2\pi\sqrt{\frac{I}{MB_H}} \qquad \frac{n_1^{2}}{n_2^{2}} = \frac{B_1\cos\delta_1}{B_2\cos\delta_2}

Worked example

A magnet swinging in a horizontal plane has a period of 2 s where

B_H = 0.36\ G

. What is its period where

B_H = 0.25\ G

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q102Moderate

Example 2 · Magnetism and Matter · The Earth's Field: Dip and Oscillating Magnets

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place

P

of dip

30^{\circ}

. The number of oscillations per minute become 10 at another place

Q

60^{\circ}dip

. The ratio of the total magnetic field at the two places

\left( B_{Q}:B_{P} \right)

is:

Using the total field for the period

A needle swinging in a horizontal plane responds to B cos δ only. When the dip changes from place to place, the cos δ factor must go into the comparison.

Oscillations per minute are a frequency

The square of the number of oscillations per minute is proportional to B_H. The square of the period is inversely proportional to it. Mixing the two inverts the ratio.

Drill 3 more on horizontal component: neutral points and oscillating magnets

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Components of the earth's field and the angle of dip
Components, and dip in a plane at α to the meridian
$B_H = B\cos\delta \qquad B_V = B\sin\delta \qquad \tan\delta' = \frac{\tan\delta}{\cos\alpha} \qquad \cot^{2}\delta = \cot^{2}\delta_1 + \cot^{2}\delta_2$
Horizontal component: neutral points and oscillating magnets
Oscillation period and the frequency at two places
$T = 2\pi\sqrt{\frac{I}{MB_H}} \qquad \frac{n_1^{2}}{n_2^{2}} = \frac{B_1\cos\delta_1}{B_2\cos\delta_2}$

Watch out for (4)

Swapping sine and cosine
→ Components of the earth's field and the angle of dip
Geographic meridian in place of magnetic
→ Components of the earth's field and the angle of dip
Using the total field for the period
→ Horizontal component: neutral points and oscillating magnets
Oscillations per minute are a frequency
→ Horizontal component: neutral points and oscillating magnets

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