CBSE Class 12 Physics Magnetism Theory and Examples

CBSE Class 12 Physics Magnetism Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

1. INTRODUCTION

Magnets and the magnetic field of the Earth are the subject of this Class 12 Physics chapter on magnetism. These notes cover the properties of magnets, the magnetic dipole and its moment, magnetic lines of force, the field of a bar magnet on its axis and on the equatorial line, the potential energy and torque of a dipole, a current loop as a magnetic dipole, the elements of the Earth's magnetic field, neutral points, tangent law, the tangent galvanometer, the oscillation magnetometer, magnetic permeability, susceptibility, intensity of magnetization, para-, dia- and ferromagnetism, hysteresis and Curie law, with solved examples.

INTRODUCTION

In nature, first time iron ore was found which had a property to attack small pieces of iron, nickel and cobalt etc. And this property of attracting small pieces of material such as iron, nickel and cobalt etc. by a piece of material is called magnetism.
There are two types of magnets. One which are found in nature are called natural magnets. These are of distorted shape.
Second which are artificially manufactured are called artificial magnets. These are of defined shape and size i.e. bar magnets.

BASIC PROPERTIES OF MAGNETS

1. A magnet attracts magnetic substances such as nickel, cobalt, iron etc.
2. When a magnet is suspended with the help of unspun thread, its two end, which are called poles come to rest in north south direction.
3. Poles exists always in pair and having equal strength i.e. magnetic monopoles do not exist.
4. Like poles repel each other and unlike poles attract each other.
5. North and south pole of magnet are a little inwards from geometrical end. The magnetic length of magnet is about 0.84 times the geometrical length.
6. The force of attraction or repulsion between two magnetic poles is directly proportional to product of pole strengths (m₁, m₂) and inversely proportional to square of distance (r) between their centres
F ∝ \( \frac{m_1 m_2}{r^2} \)
F = \( \frac{\mu_0}{4\pi}\,\frac{m_1 m_2}{r^2} \)
\( \frac{\mu_0}{4\pi} \) = 1 in CGS units = 10⁻⁷ wbA⁻¹m⁻¹
where μ₀ is absolute permeability of free space.

MAGNETIC DIPOLE AND It's MOMENT

It has two unlike poles of equal strength separated by a definite small distance.
Unit of pole strength is ampere meter or Newton/Tesla. Magnetic dipole moment (M⃗) is defined as the product of pole strength denoted by m and distance between the two poles, called magnetic length, represented by 2ℓ⃗

[Figure: A bar magnet with its south pole S at the left end and north pole N at the right end, with the dipole moment M drawn as an arrow from S to N]

M⃗ = m (2ℓ⃗)
It is a vector quantity and its direction is from south to north pole.
Its unit is ampere meter² or Joule/Tesla

MAGNETIC FIELD

It is the space around magnet or a current carrying conductor, in which magnetic effects can be felt.
Magnetic field strength at a point is defined as the force experience by a unit positive charge moving with unit velocity in the direction perpendicular to magnetic field.
When the magnitude and direction of magnetic field in a region is same throughout, field is uniform magnetic field. i.e. Earth's magnetic field. Otherwise field is called non uniform field i.e. magnetic field of a bar magnet.

[Figure: Uniform magnetic field: five equally spaced parallel straight lines with arrows pointing to the right. Non uniform: the field lines of a bar magnet with N at the left and S at the right, curving from N round to S outside the magnet]

MAGNETIC LINE OF FORCES AND THEIR PROPERTIES

As electric field lines of forces we also have imaginary magnetic field lines to represent magnetic field in any region. These are closed lines when electric field lines are open lines.

[Figure: Two curved lines of force with arrows; the direction of the magnetic field at the point A is towards AB]

To draw magnetic lines we use a magnetic dipole i.e. small compose needle.
The direction of magnetic field at a point is the same as direction of tangent at the point.

PROPERTIES

1. No two magnetic lines can cut each other.
2. These are closed lines whose direction outside magnet is from north to south pole and inside from south to north pole.
3. Tangent at a point of magnetic line give direction of field at that point.
4. Crowding of magnetic lines of force represent stronger magnetic field.
5. There is always tension in magnetic field line and repulsion between two lines.

Ex.1 The main difference between electriclines of force and magnetic lines of force is
(A) Electric lines of force are closed curves whereas magnetic lines of force are open curves.
(B) Electric lines of force are open curves whereas magnetic lines of force are closed curves.
(C) Magnetic lines of force cut each other whereas electric lines of force do not cut.
(D) Electric lines of force cut each other whereas magnetic lines of force do not cut.

Answer: (B) Electric lines of force are open curves whereas magnetic lines of force are closed curves.

1. The magnetic lines of force are in the form of closed curves whereas electric lines of force are open curves.
2. Hence the correct answer will be (B).

Teacher's Note:
a) Neither kind of line can cut another, so options (C) and (D) are wrong.
b) Electric lines begin on positive charges and end on negative charges, whereas magnetic lines have no beginning or end.

MAGNETIC FIELD STRENGTH DUE TO BAR MAGNET

(i) End-on position : In this position point lies on the axis of magnet.
Magnetic field strength due to south pole
B_S = \( \frac{\mu_0}{4\pi}\,\frac{m}{(r+\ell)^2} \) along PS
due to north pole. B_N = \( \frac{\mu_0}{4\pi}\,\frac{m}{(r-\ell)^2} \) along NP
Net magnetic field strength at P
B⃗₁ = B_N - B_S along NP
= \( \frac{\mu_0}{4\pi}\,\frac{m\,4\ell r}{(r^2-\ell^2)^2} \) = \( \frac{\mu_0}{4\pi}\,\frac{2Mr}{(r^2-\ell^2)^2} \) along NP.
For small magnet ℓ << r.
So B_E = \( \frac{\mu_0}{4\pi}\,\frac{2M}{r^3} \) along NP.
(ii) Broad - side - on - position : When point lies on equatorial line of magnet.
Magnetic field strength at point P.
Due to north pole
B_N = \( \frac{\mu_0}{4\pi} \) × \( \frac{m}{(r^2+\ell^2)} \) along NP
Due to south pole
B_S = \( \frac{\mu_0}{4\pi} \) × \( \frac{m}{(\ell^2+r^2)} \) along PS

[Figure: A magnet with its north pole N above and south pole S below, separated by 2ℓ; the point P on the equatorial line at distance r from the centre; the lines PN and PS each make the angle α with the perpendicular PO]

Net magnetic field
B⃗_B = 2 \( \frac{\mu_0}{4\pi} \) × \( \frac{m}{(\ell^2+r^2)} \) cos α along PS
= \( \frac{\mu_0}{4\pi}\,\frac{m \times 2\ell}{(r^2+\ell^2)^{3/2}} \) = \( \frac{\mu_0}{4\pi}\,\frac{M}{(r^2+\ell^2)^{3/2}} \)
For small magnet ℓ < < r
B⃗_B = \( \frac{\mu_0}{4\pi}\,\frac{M}{r^3} \) along PQ
Note \( \frac{B_E}{B_B} \) = \( \frac{2}{1} \)

Ex.2 A bar magnet with its poles 25 cm apart and of pole strength 24.0 A-m rests with its centre on a frictionless pivot. A force F is applied on the magnet at a distance of 12 cm from the pivot. So that it is held in equilibrium at an angle of 30º with respect to a magnetic field of induction 0.25 T. The value of force F is
(A) 65.62 N       (B) 2.56 N
(C) 6.52            (D) 6.25 N

Answer: (D) 6.25 N

1. In equilibrium, the torque due to the external magnetic field equals the torque due to the applied force: MB sin 30º = F × (perpendicular distance).
2. M = m × (pole separation) = 24 × 0.25 = 6 A-m².
3. 0.25 × 24 × 0.25 × (1/2) = F × 0.12
4. F = 0.75/0.12 = 6.25 N.
5. Hence, the correct option is (D).

Teacher's Note:
a) The torque on a dipole of moment M in a field B at the angle θ is MB sin θ.
b) The force is applied perpendicular to the magnet at 12 cm from the pivot, so the lever arm is 0.12 m.

Ex.3 A small magnet of magnetic moment 4A-m² is placed on a deflection magnetometer in tan-B position at a distance of 20 cm from the compass needle. At what distance from compass needle should another small magnet of moment 0.5 A-m² be placed such that the deflection of the needle remains zero ?

Answer: 10 cm

1. The deflection of the needle remains zero only if the field of the second magnet, placed on the other side of the needle, cancels the field of the first magnet. Both magnets are small and are used in the same position, so each field is B = \( \frac{\mu_0}{4\pi}\,\frac{2M}{x^3} \).
2. For equal fields, M/x³ must be the same for both magnets, so B = \( \frac{\mu_0}{4\pi}\,\frac{2M}{x^3} \) = constant.
3. ∴ x₂ = x₁ \( \left( \frac{M_2}{M_1} \right)^{1/3} \) = 20 × \( \left( \frac{1}{2 \times 4} \right)^{1/3} \) = 20 × 0.5 = 10 cm.
4. Hence, the second magnet should be placed at 10 cm from the compass needle.

Teacher's Note:
a) Cube roots: (1/8)^{1/3} = 1/2.
b) The same dependence B ∝ M/x³ holds on the axial and equatorial lines of a short magnet, so the result does not depend on the position used.

Ex.4 The ratio of intensities of magnetic field, at distance x and 2x from the centre of magnet of length 2cm on its axis, will be
(A) 4 : 1           (B) 4 : 1 approx
(C) 8 : 1           (D) 8 : 1 approx

Answer: (D) 8 : 1 approx

1. On the axis of a short magnet, B = \( \frac{\mu_0}{4\pi}\,\frac{2Mx}{(x^2-\ell^2)^2} \), which for x much larger than the half-length ℓ becomes B = \( \frac{\mu_0}{4\pi}\,\frac{2M}{x^3} \).
2. ∴ \( \frac{B_1}{B_2} \) = \( \left( \frac{x_2}{x_1} \right)^3 \) = 8 : 1 approximately.
3. Hence, the correct option is (D).

Teacher's Note:
a) The ratio is only approximate because the magnet has a finite length of 2 cm.
b) The printed solution uses (ℓ² + x²)^{3/2} in the denominator, which is the equatorial form; for the axial line the exact form is shown above, and both reduce to 1/x³ for a short magnet.

[Figure: Magnet A lying horizontally at the left (hatched) and magnet B standing vertically at the right (hatched); the needle at P lies between them at the distance d₁ from A and d₂ from B; the needle is deflected by the angle θ]

Ex.5 Two magnets A and B are identical and these are arranged as shown in the figure. Their length is negligible in comparison to the separation between them. A magnetic needle is placed between the magnets at point P which gets deflected through an angle θ under the influence of magnets. The ratio of distances d₁ and d₂ will be
(A) (2 tan θ)^1/3      (B) (2 tan θ)^-1/3
(C) (2 cot θ)^1/3      (D) (2 cot θ)^-1/3

Answer: (C) (2 cot θ)^1/3

1. The point P is on the axis of magnet A (distance d₁) and on the equatorial line of magnet B (distance d₂). The two fields are perpendicular to each other.
2. According to tangent law, B_A = B_B tan θ.
3. or \( \frac{\mu_0}{4\pi}\,\frac{2M}{d_1^3} \) = \( \frac{\mu_0}{4\pi}\,\frac{M}{d_2^3} \) tan θ
4. ∴ \( \frac{d_1}{d_2} \) = (2 cot θ)^1/3.
5. Hence, the correct option is (C).

Teacher's Note:
a) Axial field of a short magnet is twice its equatorial field at the same distance.
b) The figure shows magnet A horizontal at the left and magnet B vertical at the right, with the needle between them.

Ex.6 The force experienced by a pole of strength 100 A-m at a distance of 0.2 m from a short magnet of length 5cm and pole strength of 200A-m on its axial line will be
(A) 2.5 × 10⁻² N     (B) 2.5 × 10⁻³ N
(C) 5.0 × 10⁻² N     (D) 5.0 × 10⁻³ N

Answer: (A) 2.5 × 10⁻² N

1. The force on the pole is F = mB, where the field on the axial line of a short magnet is B = \( \frac{\mu_0}{4\pi}\,\frac{2M}{x^3} \) and the magnetic moment is M = 200 × 0.05 = 10 A-m².
2. F = \( \frac{10^{-7} \times 2 \times 200 \times 0.05 \times 100}{8 \times 10^{-3}} \) = 2.5 × 10⁻² N.
3. Hence, the correct option is (A).

Teacher's Note:
a) Here 2ℓ = 0.05 m is the magnetic length, so M = 200 × 0.05 = 10 A-m², and x³ = (0.2)³ = 8 × 10⁻³ m³.
b) Check: B = 10⁻⁷ × 2 × 10/(8 × 10⁻³) = 2.5 × 10⁻⁴ T and F = 100 × 2.5 × 10⁻⁴ = 2.5 × 10⁻² N.

Ex.7 A magnet 10 cm long has a pole strength of 12 A.m. Find the magnitude of magnetic field strength B at a point on its axis at a distance of 20 cm from it. What would be the value of B, if the point were to lie at the same distance on equatorial of magnet ?
(A) 3.4 × 10⁻⁵ T     (B) 1.4 × 10⁻⁵ T
(C) 1.7 × 10⁻⁵ T     (D) 0.85 × 10⁻⁵ T

Answer: (A) 3.4 × 10⁻⁵ T on the axis and (B) 1.4 × 10⁻⁵ T on the equatorial line

1. The magnetic moment is M = m × 2ℓ = 12 × 0.10 = 1.2 A-m², with ℓ = 0.05 m and r = 0.20 m.
2. On the axis: B = \( \frac{\mu_0}{4\pi}\,\frac{2Mr}{(r^2-\ell^2)^2} \) = \( \frac{10^{-7} \times 2 \times 1.2 \times 0.20}{(0.04 - 0.0025)^2} \) = 3.4 × 10⁻⁵ T.
3. On the equatorial line: B = \( \frac{\mu_0}{4\pi}\,\frac{M}{(r^2+\ell^2)^{3/2}} \) = \( \frac{10^{-7} \times 1.2}{(0.0425)^{3/2}} \) = 1.4 × 10⁻⁵ T.
4. Hence, the field is 3.4 × 10⁻⁵ T on the axis, option (A), and 1.4 × 10⁻⁵ T on the equatorial line, option (B).

Teacher's Note:
a) The printed answer is (C) 1.7 × 10⁻⁵ T, which is half the axial value; that holds only for a short magnet (B_equatorial = B_axial/2). Here the magnet is 10 cm long and the distance is only 20 cm, so the exact formulas give 1.4 × 10⁻⁵ T.
b) Use the exact formulas when the length of the magnet is not small compared with the distance.

POTENTIAL ENERGY OF A MAGNETIC DIPOLE

Potential energy of a magnetic dipole in magnetic field is the energy possessed by the magnet due to its particular position.
If magnetic dipole is kept at an angle θ with magnetic field then
The workdone by an external agent in changing angle θ to θ + dθ is dw = (MBsinθ) dθ.
The workdone in rotating the magnet from an angle θ₀ to θ is
W = \( \int_{\theta_0}^{\theta} MB \sin\theta\, d\theta \) = MB (cosθ₀ - cosθ)
This work is stored as the potential energy of the field magnet system. Thus
U (θ) - U (θ₀) = MB (cosθ₀ - cosθ)
If U (90º) = 0
U (θ) - 0 = MB (cos90 - cosθ) = - MBcosθ
U(θ) = - M⃗ · B⃗

Ex.8 A magnet of moment M is lying in a magnetic field of induction B. W₁ is the work done in turning it from 0º to 60º and W₂ is the work done in turning it from 30º to 90º. Then -
(A) W₂ = W₁        (B) W₂ = \( \frac{W_1}{2} \)
(C) W₂ = 2W₁       (D) W₂ = √3 W₁

Answer: (D) W₂ = √3 W₁

1. W = MB (cos θ₁ - cos θ₂).
2. W₁ = MB (cos 0º - cos 60º) = \( \frac{MB}{2} \).
3. W₂ = MB (cos 30º - cos 90º) = \( \frac{\sqrt{3}\,MB}{2} \).
4. ∴ W₂ = √3 W₁.
5. Hence, the correct option is (D).

Teacher's Note:
a) cos 0° = 1, cos 60° = 1/2, cos 30° = √3/2 and cos 90° = 0.
b) The work done depends only on the initial and final angles, not on the path.

Ex.9 A bar magnet of magnetic moment 4.0 A-m² is free to rotate about a vertical axis through its centre. The magnet is released from rest from east-west position. Kinetic energy of the magnet in north-south position will be (H = 25μT)
(A) 10⁻² J         (B) 10⁻⁴ J
(C) 10⁻⁶ J         (D) 0

Answer: (B) 10⁻⁴ J

1. Loss in P.E. = gain in K.E.
2. E_k = U_i - U_f = - MB cos 90º - (- MB cos 0º) = MB.
3. = 4 × 25 × 10⁻⁶ = 10⁻⁴ J.
4. Hence, the correct option is (B).

Teacher's Note:
a) In the east-west position the magnet is at 90° to the horizontal field H, and in the north-south position it is parallel to H.
b) The potential energy at 90° is taken as zero, and it is -MB at 0°.

Ex.10 A magnet having a magnetic moment of 1.0 × 10⁴ J/T is free to rotate in a horizontal plane where a magnetic field 4 × 10⁻⁵ T exists. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction 60º from the field
(A) 0.4 J        (B) 2J
(C) 0.2J        (D) 1J

Answer: (C) 0.2 J

1. W = MB (cos θ₁ - cos θ₂) = 1.0 × 10⁴ × 4 × 10⁻⁵ × (cos 0º - cos 60º).
2. = 0.4 × (1 - 0.5) = 0.2 J.
3. Hence, the correct option is (C).

Teacher's Note:
a) MB = 1.0 × 10⁴ × 4 × 10⁻⁵ = 0.4 J.
b) Rotating slowly means the work done by the external agent equals the increase in potential energy.

TORQUE ON A MAGNET IN MAGNETIC FIELD

When a bar magnet of pole strength m and magnetic length 2ℓ is placed at an angle θ with a uniform magnetic field B⃗ .

[Figure: A bar magnet SN of length 2ℓ in a uniform field B (horizontal parallel lines) at the angle θ; the force mB acts on the N pole along B and mB on the S pole opposite to B; the perpendicular distance between the forces is marked]

Force on N-pole = mB along B⃗
Force on S-pole = mB opposite B⃗
These forces are equal in magnitude and opposite in since so they will form couple.
Torque τ = mB × ON = MB 2ℓ sin θ = MB sinθ
τ⃗ = M⃗ × B⃗
The direction τ⃗ is perpendicular to plane containing M⃗ and B⃗ , and is given by right hand screw rule

Ex.11 The length of a bar magnet is 10 cm and its pole strength is 10⁻³ Weber. It is placed in a magnetic field of induction 4 π x 10⁻³ Tesla in a direction making an angle of 30º with the field direction. The value of torque acting on the magnet will be
(A) 2π × 10⁻⁷ N-m    (B) 2π × 10⁻⁵ N-m
(C) 0.5 × 10² N-m    (D) None of these

Answer: (A) 2π × 10⁻⁷ N-m

1. The torque on the magnet is τ = MB sin θ = m ℓ B sin θ, where ℓ here is the length of the magnet (0.1 m).
2. τ = 10⁻³ × 0.1 × 4π × 10⁻³ × 0.5
3. = 2π × 10⁻⁷ N-m.
4. Hence, the correct option is (A).

Teacher's Note:
a) As in the printed solution, the pole strength is used numerically as 10⁻³ in the formula τ = mℓB sin θ.
b) Use sin 30° = 0.5.

CURRENT LOOP AS MAGNETIC DIPOLE

Consider a current loop in adjacent figure. Looking from left we find anticlockwise current so it has north polarity.
Similarly, looking from right, current is clockwise hence it has south polarity. So a current loop work as magnetic dipole.
The magnetic dipole moment (M) of current loop is directly proportional to current (I) through the loop and area (A) enclosed by the loop.
∴ M = I A

[Figure: A circular current loop with N on its left face and S on its right face, labelled "equivalent magnetic dipole" with poles +m and -m, and "Current loop"]

For N turns
M = NIA
In vector from we can write M⃗ = NIA n̂
Where n̂ in a unit vector perpendicular to the plane of loop in the direction give by right hand screw rule.

Ex.12 A current of 1 ampere is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be
(A) 0.314 A-m²       (B) 3140 A-m²
(C) 100 A-m²         (D) μ₀ A-m²

Answer: (A) 0.314 A-m²

1. M = iA = πR²Ni
2. = 3.14 × 0.01 × 10 × 1 = 0.314 A-m².
3. Hence, the correct option is (A).

Teacher's Note:
a) R = 10 cm = 0.1 m, so R² = 0.01 m².
b) The moment is the product of the number of turns, the current and the area.

MAGNETIC FIELD DUE TO EARTH

Due to earth's magnetic dipole, magnetic field is present everywhere on earth's surface. The axis of dipole makes an angle of 11.5º with the earth's axis of rotation. The dipole axis cuts the earth's surface at two points, one near the geographical north pole and the other near the geographical south pole. These are called geomagnetic northpole and geomagnetic south pole.
Earth's is magnetic field change in magnitude and direction both. But this change is appreciable for longtime interval say 20 years.
The exact cause of earth's magnetism is not yet known. At present it seems that the field results mainly due to circulating electric currents induced in the molten liquid and other conducting material inside earth.

ELEMENTS OF THE EARTH'S MAGNETIC FIELD

Our earth behaves in such a way that we can assume the existence of a powerful magnet inside it. This magnet is so oriented that its south pole is towards earth's north pole while the north pole of the magnet is towards earth's south pole.

[Figure: Fig.3: The magnetic field of the Earth drawn as the field of a bar magnet inside it. Labels: axis of rotation of earth, magnetic axis, geographical north pole, magnetic south pole, geographical equator, magnetic equator, magnetic north pole, geographical south pole; NORTH at the top and SOUTH at the bottom]

TERMS RELATED TO EARTH'S MAGNETIC FIELD

(a) Geographical Meridian : Planes passing through geographical north and south pole is called geographical meridian and they are infinite in numbers.
(b) Magnetic Meridian : Planes passing through magnetic south and north pole is called magnetic meridian.
(c) Angle of Declination (α) : The angle between geographical meridian and magnetic meridian is called the angle of declination and its value lies between 15º to 18º As the north-south axis of earth's magnet and earth's geographical poles do not coincide, a freely suspended magnetic needle makes an angle with geographical axis. This angle between the axis of magnetic needle and geographical axis is called the angle of declination.

[Figure: Fig.4: A rectangular plane showing the geographical meridian and the magnetic meridian with the angle of declination φ between them; the earth's field B at the point A is resolved into the horizontal component B_H and the vertical component B_v, with the angle of dip θ between B and B_H; the points P, C, K, E, L, D, Q, S, H are marked]

(d) Angle of Dip (θ) : If a magnetic needle is pivoted on a horizontal axis through its centre of gravity so that it is free to rotate in a vertical plane only, then such a needle is known as dip needle. If dip needle is so adjusted that its vertical plane (in which it is free to rotate) coincides with the magnetic meridian (which is also a vertical plane), the needle then aligns itself along the direction of B⃗ (total magnetic field intensity of earth).

[Figure: Fig.5: A dip needle pivoted at its centre on a stand in front of a vertical circular scale; the needle lies along the total field B making the angle θ with the horizontal component B_H; the vertical component B_v is marked; N and S are the ends of the needle]

The angle which the axis of the needle makes with horizontal is called angle of dip. Thus, angle of dip at a place is defined as the angle between the direction of the earth's magnetic field and the horizontal in the magnetic meridian at that place. Angle of dip at magnetic poles is 90º as the needle rests vertically at that places and at magnetic equator angle of dip is 0º.
(e) Horizontal component of Earth's Magnetic field : Earth's magnetic field is not horizontal except at magnetic equator, i.e., at any place, the earth's magnetic field 'B' in the magnetic meridian can be resolved into a horizontal component, B_H and a vertical component, B_v. In fig. (4) , AE represents the total intensity of earth magnetic field & ∠PAE = θ. The resultant intensity B is resolved into two rectangular components. Horizontal component, B_H = AK = B cosθ and Vertical component, B_v = AL = B sin θ.(It is vertically upward in southern hemisphere and vertically downward in northern hemisphere)
∴ B = \( \sqrt{B_H^2 + B_v^2} \) and tanθ = B_v/B_H
(i) At poles, B_H = 0
(ii) At equator, B_v = 0

NEUTRAL POINT

The point at which the horizontal component of magnetic field due to a magnet is equal and opposite to the earth's horizontal component of magnetic field is called neutral point because net horizontal component of magnetic field is zero at that point.
Generally we consider two cases.

[Figure: Earth's field lines (GEOG. NORTH at top, GEOG. SOUTH at bottom) with a bar magnet whose north pole N is towards the north; the two neutral points P and Q lie on the equatorial line of the magnet, where the magnet's field B₂ and the horizontal component H of the Earth's field are marked]

(A) When magnet is placed with its N-pole towards north of earth. We get two neutral points P and Q lying on the equatorial line of the magnet. If 2ℓ = magnetic length, M = magnetic dipole moment of the magnet. d = distance of each neutral point from the centre of magnet.
Due to magnet, magnetic field at neutral point B = \( \frac{\mu_0}{4\pi}\,\frac{M}{(d^2+\ell^2)^{3/2}} \)
Let horizontal component of earth's field = H
At neutral point B = H
⇒ M = \( \frac{4\pi}{\mu_0} \) H (d² + ℓ²)^3/2
(B) When magnet is placed its south pole towards north of earth. We get two neutral points P and Q lying on the axis of magnet.

[Figure: Earth's field lines with a bar magnet whose south pole S is towards the north; the two neutral points P and Q lie on the axis of the magnet, above and below it]

At neutral point B = H
\( \frac{\mu_0}{4\pi}\,\frac{2Md}{(d^2-\ell^2)^2} \) = H
M = \( \frac{4\pi H (d^2-\ell^2)^2}{2\mu_0 d} \)
As we see when we rotate magnet by 180º, neutral point is rotated by 90º. Similarly if we rotate magnet by angle θ, neutral point will rotate by an angle θ/2.

Ex.13 A short bar magnet is placed with its north pole pointing south. The neutral point is 10 cm away from the centre of the magnet. If H = 0.4 gauss, calculate magnetic moment of the magnet.
(A) 2 Am²         (B) 1A m²
(C) 0.1 A m²      (D) 0.2 Am²

Answer: (D) 0.2 Am²

1. Here, d = 10 cm = 0.1 m, H = 0.4 gauss = 0.4 × 10⁻⁴ T and M = ?
2. Neutral points in this case lie on axial line, such that \( \frac{\mu_0}{4\pi}\,\frac{2M}{d^3} \) = H.
3. ∴ 10⁻⁷ × \( \frac{2M}{(0.1)^3} \) = 0.4 × 10⁻⁴
4. M = 0.2 A m².
5. Hence, the correct option is (D).

Teacher's Note:
a) With the north pole pointing south, the neutral points lie on the axis of the magnet.
b) The short-magnet formula is used because the magnet is short.

TANGENT LAW

When a magnet (e.g. a compass needle) is placed under the combination of two perpendicular magnetic fields of intensities V and H. Let the magnet make an angle θ with H at equilibrium position then
V = H tan θ

[Figure: A needle SN at the angle θ to the horizontal axis H, in a vertical field V; the forces mV and mH act on the poles, forming opposing couples]

Ex.14 At any place on earth, the horizontal component of earth's magnetic field is √3 times the vertical component. The angle of dip at that place will be
(A) 60º          (B) 45º
(C) 90º          (D) 30º

Answer: (D) 30º

1. tan θ = \( \frac{B_v}{B_H} \) = \( \frac{B_v}{\sqrt{3}\,B_v} \) = \( \frac{1}{\sqrt{3}} \) = tan 30º
2. ∴ θ = 30º.
3. Hence, the correct option is (D).

Teacher's Note:
a) The angle of dip is the angle between the total field and the horizontal.
b) Since B_H is larger than B_v, the dip is less than 45°.

TANGENT GALVANOMETER (Moving Magnet Type Galvanometer)

This is based on tangent law. It is used for detection and measurement of low electric currents.
If galvanometer has coil in which, n is the number of turns, r is radius of soild and μ₀ magnetic permittivity of free space.

[Figure: A tangent galvanometer: a vertical circular coil C on a stand with a small magnetic needle at its centre above a horizontal scale, a pointer A, the terminals T₁ and T₂ and levelling screws]

Then magnetic field intensity due to coil
B = \( \frac{\mu_0}{4\pi}\,\frac{2\pi n I}{r} \)
This field will be perpendicular to the horizontal component of earth's magnetic field so B = H tan θ

[Figure: A right triangle of fields: the horizontal component H along the base, the field B of the coil vertical, and the resultant at the angle θ to H]

\( \frac{\mu_0}{4\pi}\,\frac{2n\pi I}{r} \) = H tan θ
I = K tan θ
K = \( \frac{2rH}{\mu_0 n} \) is a constant for a galvanometer.
It is called reduction factor of galvanometer.

Ex.15 The radius of the coil of a tangent galvanometer is 16 cm. How many turns of the wire should be used if a current of 40 m A is to produce a deflection of 45º. given, horizontal component of earth's field is 0.36 × 10⁻⁴ T.
(A) 458         (B) 229
(C) 200         (D) 115

Answer: (B) 229

1. Here, r = 16 cm = 0.16 m, I = 40 mA = 40 × 10⁻³ A, θ = 45º and H = 0.36 × 10⁻⁴ T.
2. As \( \frac{\mu_0 n I}{2r} \) = H tan θ
3. ∴ n = \( \frac{2rH\tan\theta}{\mu_0 I} \) = \( \frac{2 \times 0.16 \times 0.36 \times 10^{-4} \times \tan 45^\circ}{4\pi \times 10^{-7} \times (40 \times 10^{-3})} \)
4. n = 229.
5. Hence, the correct option is (B).

Teacher's Note:
a) tan 45° = 1, so the deflection is 45° when the field of the coil equals H.
b) Check: 1.152 × 10⁻⁵/(5.03 × 10⁻⁸) = 229.

OSCILLATION MAGNETOMETER

When a magnet is suspended freely in a uniform magnetic field, using an unspun thread, it comes to rest in a direction NS parallel to the direction of the field. If this magnet is slightly deflected from its equilibrium position and left, magnet begins to vibrate simple harmonically about the direction of field and its mean position.
The time period of T of vibration is given by

[Figure: An oscillation magnetometer: a magnet suspended by a thread T from a torsion head inside a glass-fronted box, with a mirror and a reference line below; the field H is marked]

T = 2π \( \sqrt{\frac{I}{MH}} \)
where,  I = moment of inertia of magnetic about the suspension fibre as the axis of rotation,
M = magnetic moment of the magnet,
H = intensity of uniform magnetic field

Uses :
(A) To compare of horizontal component of earth's magnetic field at two places, horizontal component of earth's magnetic field be H₁ and H₂ and time perior of oscillation T₁ and T₂ respectively.
Then \( \frac{H_2}{H_1} \) = \( \frac{T_1^2}{T_2^2} \)
(B) To compare magnetic moments of two magnets of same size and same mass

\( \frac{M_2}{M_1} \) = \( \frac{T_1^2}{T_2^2} \)
(C) To determine the magnetic moment of a magnet
T = 2π \( \sqrt{\frac{I}{MH}} \)
If mass, length and breadth of magnet are m, ℓ and b respectively then I = \( \frac{m(\ell^2 + b^2)}{12} \).
If H is known, M can be calculated
(D) To compare the magnetic moments of two magnets of unequal size and masses. When north pole of two magnets are in same direction, Then
Time period T₁ = 2π \( \sqrt{\frac{I_1 + I_2}{(M_1 + M_2)H}} \) ................(1)

[Figure: Two bar magnets placed one above the other with their north poles together, giving the combined moment (M₁ + M₂) and combined moment of inertia (I₁ + I₂); the points A and B are marked on the suspension]

and when they are in opposite direction.
Time period T₂ = 2π \( \sqrt{\frac{I_1 + I_2}{(M_1 - M_2)H}} \) ................(2)

[Figure: Two bar magnets placed one above the other with their poles opposite, giving the net moment (M₁ - M₂) and combined moment of inertia (I₁ + I₂); the points A and B are marked on the suspension]

By equation (1) and (2) \( \frac{M_1}{M_2} \) = \( \frac{T_2^2 + T_1^2}{T_2^2 - T_1^2} \)

Ex.16 The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts length wise then the time period of each part will be
(A) 4 sec.        (B) 2 sec.
(C) 0.5 sec      (D) 0.25 sec

Answer: (A) 4 sec.

1. T = 2π \( \sqrt{\frac{I}{MB}} \) = 2π \( \sqrt{\frac{m\ell^2}{12 \times m_p \ell B}} \) = 4 sec, where m is the mass, ℓ the length and m_p the pole strength of the magnet.
2. Cutting it into two equal parts length wise halves the mass and the pole strength and leaves the length unchanged.
3. T' = 2π \( \sqrt{\frac{\frac{m}{2}\ell^2}{12 \times \frac{m_p}{2}\,\ell B}} \) = 4 sec.
4. Hence, the correct option is (A).

Teacher's Note:
a) Cutting lengthwise means cutting along the length, so the length stays the same and only the width changes.
b) Both the moment of inertia and the magnetic moment are halved, so their ratio and the time period do not change.

Ex.17 The length , breadth and mass of two bar magnets are same but their magnetic moments are 3M and 2M respectively. These are joined pole to pole and are suspended by a string. When oscillated in a magnetic field of strength B, the time period obtained is 5s. If the poles of either of the magnets are reverse then time period of the combination in the same magnetic field will be -
(A) 3 √3 s        (B) 2 √2 s
(C) 5 √5 s        (D) 1s

Answer: (C) 5 √5 s

1. T = 2π \( \sqrt{\frac{I}{MB}} \) or T ∝ \( \frac{1}{\sqrt{M}} \), since the moment of inertia I of the combination is the same in both cases.
2. When the like poles are together the net moment is 3M + 2M = 5M, and when one magnet is reversed the net moment is 3M - 2M = M.
3. ∴ \( \frac{T_1}{T_2} \) = \( \sqrt{\frac{3M - 2M}{3M + 2M}} \) = \( \frac{1}{\sqrt{5}} \), so T₂ = 5√5 s.
4. Hence, the correct option is (C).

Teacher's Note:
a) The moment of inertia is the same because the two magnets have the same size and mass.
b) Reversing one magnet reduces the net moment, so the time period increases.

Ex.18 A thin magnetic needle oscillates in a horizontal plane with a period T. It is broken into n equals parts. The time period of each part will be
(A) T            (B) \( \frac{T}{n^2} \)
(C) Tn²        (D) \( \frac{T}{n} \)

Answer: (D) \( \frac{T}{n} \)

1. T = 2π \( \sqrt{\frac{I}{MB}} \) = 2π \( \sqrt{\frac{m\ell^2/12}{m_p \ell B}} \), so T ∝ \( \sqrt{\frac{m\ell}{m_p}} \).
2. Each part has mass m/n, length ℓ/n and the same pole strength m_p.
3. ∴ \( \frac{T\'}{T} \) = \( \left( \frac{\frac{m}{n}\cdot\frac{\ell}{n}}{m\ell} \right)^{1/2} \) = \( \frac{1}{n} \)
4. T' = \( \frac{T}{n} \).
5. Hence, the correct option is (D).

Teacher's Note:
a) The pole strength of a piece is unchanged when a magnet is broken across its length.
b) The magnetic moment becomes M/n and the moment of inertia I/n³, so T becomes T/n.

SOME TERMS RELATED TO MAGNETISM

(A) Magnetic permeability : It is the ability of material to permit the passage of magnetic lines through it. If it denoted by μ. Its unit is weber/ammeter.
Relative magnetic permeability (μ_r), is a factor by which the magnetic field B is increased when a material is brought in the field. μ_r is unitless.
μ_r = \( \frac{\mu}{\mu_0} \) = 1 + χ
where μ₀ is permeability of vaccum and χ is susceptibily of material.
(B) Intensity of magnetization (I) : It is defined as magnetic moment per unit volume.
I⃗ = \( \frac{\text{magnetic moment}}{\text{volume}} \) = \( \frac{\vec{m}}{V} \)
I = \( \frac{m}{V} \) = \( \frac{2m\ell}{A(2\ell)} \) = \( \frac{m}{A} \)
For bar magnet of pole strength m, length x and area of cross-section A.
(C) Magnetic susceptibility : For paramagnetic and diamagnetic substances, the intensity of magnetization (I⃗) of a material is directly proportional to magnetic in intensity (H⃗)
Thus I⃗ = χ H⃗
The proportionality content χ is called susceptibility. It is a dimensionless quantity.
For vaccum χ = 0.
Note : μ_r = 1 + χ (in M.K.S. system)
μ_r = 1 + 4πχ (in C.G.S. system)
(D) Magnetic Intensity : The actual magnetic field inside the material is the sum of the applied magnetic field and magnetic field due to magnetization.
Hence H⃗ = \( \frac{\vec{B}}{\mu_0} \) - I⃗ = \( \frac{\vec{B}}{\mu} \)
where B⃗ is the resultant magnetic field, and μ permeability of the material and I⃗ intensity of magnetization. The quantity H⃗ is called magnetic field intensity or magnetizing field intensity. Unit of it is ampere/meter.

Ex.19 The magnetic susceptibility of a paramagnetic substance is 3 × 10⁻⁴. It is placed in a magnetising field of 4 × 10⁴ amp/m. The intensity of magnetisation will be
(A) 3 × 10⁸ A/m     (B) 12 × 10⁸ A/m
(C) 12 A/m         (D) 24 A/m

Answer: (C) 12 A/m

1. I = χH = 3 × 10⁻⁴ × 4 × 10⁴ = 12 A/m.
2. Hence, the correct option is (C).

Teacher's Note:
a) The printed solution shows 4 × 10³, which is a misprint; with the field 4 × 10⁴ A/m given in the question the product is 12 A/m.
b) The magnetisation is small because χ for a paramagnetic substance is very small.

Ex.20 The horizontal component of flux density of earth's magnetic field is 1.7 × 10⁻⁵ tesla. The value of horizontal component of intensity of earth's magnetic field will be ?
(A) 24.5 A/m     (B) 13.5 A/m
(C) 0.135 a/m    (D) 1.35 A/m

Answer: (B) 13.5 A/m

1. H = \( \frac{B}{\mu_0} \) = \( \frac{1.7 \times 10^{-5}\ \text{Wb/m}^2}{4\pi \times 10^{-7}\ \text{Wb/A-m}} \) = 13.5 A/m.
2. Hence, the correct answer will be (B).

Teacher's Note:
a) The magnetic intensity of free space is H = B/μ₀.
b) Check: 1.7 × 10⁻⁵/(1.2566 × 10⁻⁶) = 13.53 A/m.

Ex.21 A magnetising field of 2 × 10³ amp/m produces a magnetic flux density of 8π Tesla in an iron rod. The relative permeability of the rod will be
(A) 10²          (B) 10⁰
(C) 10⁴          (D) 10¹

Answer: (C) 10⁴

1. μ_r = \( \frac{\mu}{\mu_0} \) = \( \frac{B}{H\mu_0} \)
2. or μ_r = \( \frac{8\pi}{2 \times 10^3 \times 4\pi \times 10^{-7}} \) = 10⁴.
3. Hence, the correct option is (C).

Teacher's Note:
a) The relative permeability is a ratio, so it has no unit.
b) A value as high as 10⁴ is typical of a ferromagnetic material such as iron.

Ex.22 A bar magnet made of steel has a magnetic moment of 2.5 A-m² and a mass of 6.6 × 10³ kg. If the density of steel is 7.9 × 10⁹ kg/m³, find the intensity of magnetization of the magnet.

Answer: 3.0 × 10⁶ A/m

1. The volume of the bar magnet is V = \( \frac{\text{mass}}{\text{density}} \) = \( \frac{6.6 \times 10^{-3}\ \text{kg}}{7.9 \times 10^{3}\ \text{kg/m}^3} \) = 8.3 × 10⁻⁷ m³.
2. The intensity of magnetization is I = \( \frac{M}{V} \) = \( \frac{2.5\ \text{A-m}^2}{8.3 \times 10^{-7}\ \text{m}^3} \) = 3.0 × 10⁶ A/m.
3. Hence, the intensity of magnetization is 3.0 × 10⁶ A/m.

Teacher's Note:
a) The ratio of mass to density is the same for 6.6 × 10³ kg / 7.9 × 10⁹ kg/m³ as printed in the question and for 6.6 × 10⁻³ kg / 7.9 × 10³ kg/m³ used in the printed solution, so V = 8.3 × 10⁻⁷ m³ either way.
b) Magnetization is the magnetic moment per unit volume.

MAGNETIC BEHAVIOUR OF ATOM

If an electron having charge e is revoling in an orbit of radius r having uniform angular velocity ω. Then
Magnetic moment of atom contributed by the electron M = \( \frac{1}{2} \) e ω r² and magnetic field
B = \( \frac{eh}{4\pi m} \) = 9.27 x 10⁻²⁴ ampere meter².
If source as the unit of atomic magnetic dipole moment. Where h is plank's constant.

MOLECULAR / ATOMIC THEORY OF MAGNETIZATION

According to this theory
(i) Every molecule/atom of a magnetic substance has north pole as well as south pole of equal strength, so it is a magnetic dipole.
(ii) In unmagnetized condition, atomic dipoles are randomly oriented and they cancel dipole moment of one another.
(iii) After applying magnetic field, atomic dipoles align in its direction.
(iv) Extent of magnetization depends on extent of realignment of the molecular dipoles.
(v) On heating magnetization reduces or vanish completely.
On the basis of this theory magnetic materials are classified into three categories
(a) Paramagnetic substances : Every molecules / atoms of these substances has its own dipole moment. When we place such substances in magnetic field tries to align every individual molecule / atom in its direction e.g. chromium, CuSO₄ , crown glass, aluminimium, platinum, manganese, solution of salts of iron and nickel are paramagnetic substances.
Permeability of paramagnetic substance is greater than unity. Susceptibility of paramagnetic substances varies inversely to the temperature of substance.
(b) Diamagnetic substances : Molecules / atoms of this type of substances do not have dipole moment individually. When they are kept in any magnetic field, magnetic moment is induces but it is in opposite direction to the applied field. e.g. copper, gold, mercury , quart, antimony and Bismoth etc.
Permeability of diamagnetic substances is always less than unity.
Susceptibility of diamagnetic substances does not change with temperature.
(c) Ferromagnetic substances : In these substance each individual atom/molecule has magnetic moment. They have strong tendency to align themselves so they make domain. In a domain, a large no. of atoms/ molecules align in same direction.
Different domains have different directions of magnetic moment hence, the material remains un magnetized.
When a magnetic field is applied, the domains which are aligned along the direction of the field grown in size and those opposite to it get reduced. iron nickel and cobalt etc. are examples of such materials.

[Figure: Three squares divided into four triangular domains by their diagonals: "Unmagnetized" with the domain arrows pointing in different directions (right, down, left, up) so that they cancel; "Domain-growing" in a field B (upwards) where the domains along B have grown; "Domain-alignment" in a field B where the arrows of the domains are all turned towards the direction of B]

HYSTERESIS

As shown in adjacent curve, as we change field intensity, magnetization changes.
In the beginning , the field is zero and sample has no magnetization, this is corresponding to point O.

[Figure: The hysteresis loop with the magnetization I on the vertical axis and the field H on the horizontal axis, with the points marked C, D, E, F, G, A and the field values -H₀ and H₀ on the H axis; the curve is traced along the path ACDEFGA]

As we increase field intensity, magnetization point A corresponds to saturation magnetization. As value of and direction of field intensity varies, magnetization along path ACDEFGA is followed.
Hence as field intensity increased or decreased, magnetization does not return to its initial value. This fact is called hysteresis. The curve ACDEFGA is called hysteresis loop. This are of the hysteresis loop is proportional to the thermal energy developed per unit volume of the material as it goes through the hysteresis cycle.
Hysteresis loop for soft iron is narrow and large, whereas hysteresis loop for steel is wide and short.

Ex.23 The mass of a speciment of a ferromagnetic material is 0.6 kg. and its density is 7.8 x 10³ kg/m³. If the area of hysteresis loop of alternating magnetising field of frequency 50Hz is 0.722 MKS units then the hysteresis loss per second will be -
(A) 277.7 × 10⁻⁵ Joule
(B) 277.7 × 10⁻⁶ Joule
(C) 277.7 × 10⁻⁴ Joule
(D) 277.7 × 10⁻⁴ Joule

Answer: (A) 277.7 × 10⁻⁵ Joule

1. The hysteresis loss in time t is W_H = VAft = \( \frac{m}{d} \) Aft, where V = m/d is the volume of the specimen, A is the area of the loop and f is the frequency.
2. W_H = \( \frac{0.6}{7.8 \times 10^3} \) × 0.722 × 50 (for t = 1 s)
3. = 277.7 × 10⁻⁵ Joule.
4. Hence, the correct option is (A).

Teacher's Note:
a) Options (C) and (D) are printed identically in the source.
b) Check: 7.69 × 10⁻⁵ m³ × 0.722 × 50 = 2.777 × 10⁻³ J = 277.7 × 10⁻⁵ J.

[Figure: The hysteresis loops of soft iron (narrow, tall loop) and of steel (wide loop) drawn on the same I - H axes (X' - X and Y' - Y)]

CURIE LAW

According to this law, intensity of magnetization (I) of a magnetic material is directly proportional to magnetic induction (B) and inversely proportional to the temperature of magnetic material.
I ∝ \( \frac{B}{T} \)
we know B ∝ H
So \( \frac{I}{H} \) ∝ \( \frac{1}{T} \)
\( \frac{I}{H} \) = χ = \( \frac{C}{T} \)
where χ is susceptibility of material and C is a constant called curie constant.

[Figure: A graph of I against H/T: a curve that rises steeply from the origin and bends over to reach a flat saturation level]

The variation of I with graph it is clear that saturation is reached when H/T exceeds a certain limit. Magnetic thermometers are based on curie law and filled with paramagnetism substance (e).