Magnetism And Magnetic Effects Of Electric Current

Class 12 | Physics | Unit 3

Class 12 Unit 3 Physics

"The magnetic force is animate, or imitates a soul; in many respects it surpasses the human soul while it is united to an organic body" – William Gilbert

Learning Objectives

In this unit, the student is exposed to:

• Earth's magnetic field and magnetic elements
• Basic properties of magnets
• Statement of Coulomb inverse square law of magnetism
• Magnetic dipole
• Magnetic field due to a dipole along axial line and equatorial line
• Torque acting on a bar magnet in a uniform magnetic field
• Potential energy of a bar magnet placed in a uniform magnetic field
• Magnetic properties – permeability, susceptibility etc.
• Classification of magnetic materials – dia, para and ferro magnetic materials
• Concept of Hysteresis
• Magnetic effects of electric current – long straight conductor and circular coil
• Right hand thumb rule and Maxwell's right hand cork screw rule
• Biot-Savart's law – applications
• Tangent law and Tangent Galvanometer
• Current loop as a magnetic dipole
• Magnetic dipole moment of revolving electron
• Ampère's circuital law – applications
• Solenoid and toroid
• Lorentz force – charged particle moving in an electromagnetic field
• Cyclotron
• Force on a current carrying conductor in a magnetic field
• Force between two long parallel current carrying conductor
• Torque on a current loop in a magnetic field
• Moving Coil Galvanometer

3.1 INTRODUCTION TO MAGNETISM

Magnets! No doubt, their behaviour will attract everyone. The world enjoys their benefits, to lead a modern luxurious life. The study of magnets fascinated scientists around our globe for many centuries and even now, door for research on magnets is still open (Figure 3.1).

Do You Know?

Many birds and animals have magnetic sense in their eyes using Earth's magnetic field for navigation. Magnetic sensing in eyes - for Zebra finch bird, due to protein cryptochromes Cry4 present in retina; the bird uses Earth's magnetic field for navigation.

Magnetism exists everywhere from tiny particles like electrons to the entire universe. Historically the word 'magnetism' was derived from iron ore magnetite (Fe₃O₄). In olden days, magnets were used as magnetic compass for navigation, magnetic therapy for treatment and also used in magic shows. In modern days, many things we use in our daily life contain magnets (Figure 3.2). Motors, cycle dynamo, loudspeakers, magnetic tapes used in audio and video recording, mobile phones, head phones, CD, pen-drive, hard disc of laptop, refrigerator door, generator are a few examples.

Earlier, both electricity and magnetism were thought to be two independent branches in physics. In 1820, H.C. Oersted observed the deflection of magnetic compass needle kept near a current carrying wire. This unified the two different branches, electricity and magnetism as a single subject 'electromagnetism' in physics. In this unit, basics of magnets and their properties are given. Later, how a current carrying conductor (here only steady current, not time-varying current is considered) behaves like a magnet is presented.

3.1.1 Earth's Magnetic Field and Magnetic Elements

From the activities performed in lower classes, you might have noticed that the needle in a magnetic compass or freely suspended magnet comes to rest in a position which is approximately along the geographical north south direction of the Earth.

Do You Know?

William Gilbert in 1600 proposed that Earth itself behaves like a gigantic powerful bar magnet. But this theory is not successful because the temperature inside the Earth is very high and so it will not be possible for a magnet to retain its magnetism. Gover suggested that the Earth's magnetic field is due to hot rays coming out from the Sun. These rays will heat up the air near equatorial region. Once air becomes hotter, it rises above and will move towards northern and southern hemispheres and get electrified. This may be responsible to magnetize the ferromagnetic materials near the Earth's surface. Till date, so many theories have been proposed. But none of the theories completely explains the cause for the Earth's magnetism.

The north pole of magnetic compass needle is attracted towards the magnetic south pole of the Earth which is near the geographic north pole (Figure 3.3). Similarly, the south pole of magnetic compass needle is attracted towards the magnetic north-pole of the Earth which is near the geographic south pole. The branch of physics which deals with the Earth's magnetic field is called Geomagnetism or Terrestrial magnetism.

There are three quantities required to specify the magnetic field of the Earth on its surface, which are often called as the elements of the Earth's magnetic field. They are:

• (a) magnetic declination (D)
• (b) magnetic dip or inclination (I)
• (c) the horizontal component of the Earth's magnetic field (B_H)

Day and night occur because Earth spins about an axis called geographic axis. A vertical plane passing through the geographic axis is called geographic meridian and a great circle perpendicular to Earth's geographic axis is called geographic equator. The straight line which connects magnetic poles of Earth is known as magnetic axis. A vertical plane passing through magnetic axis is called magnetic meridian and a great circle perpendicular to Earth's magnetic axis is called magnetic equator.

When a magnetic needle is freely suspended, the alignment of the magnet does not exactly lie along the geographic meridian as shown in Figure 3.4. The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D). At higher latitudes, the declination is greater whereas near the equator, the declination is smaller. In India, declination angle is very small and for Chennai, magnetic declination angle is −1° 16' (which is negative (west)).

The angle subtended by the Earth's total magnetic field B⃗ with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at that point (Figure 3.5). For Chennai, inclination angle is 14° 28'.

The component of Earth's magnetic field along the horizontal direction in the magnetic meridian is called horizontal component of Earth's magnetic field, denoted by B_H. Let B_E be the net Earth's magnetic field at any point on the surface of the Earth. B_E can be resolved into two perpendicular components:

Horizontal component: B_H = B_E cos I     (Equation 3.1)

Vertical component: B_V = B_E sin I     (Equation 3.2)

Dividing: tan I = B_V / B_H     (Equation 3.3)

(1) At Magnetic Equator

The Earth's magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of magnetic compass rests horizontally at an angle of dip, I = 0°.

B_H = B_E     B_V = 0

This implies that the horizontal component is maximum and vertical component is zero at the equator.

(2) At Magnetic Poles

The Earth's magnetic field is perpendicular to the surface of the Earth (that is vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, I = 90°.

B_H = 0     B_V = B_E

This implies that the vertical component is maximum at poles and horizontal component is zero at poles.

EXAMPLE 3.1

The horizontal component and vertical component of Earth's magnetic field at a place are 0.15 G and 0.26 G respectively. Calculate the angle of dip and resultant magnetic field. (G – gauss, cgs unit for magnetic field; 1G = 10⁻⁴ T)

Solution:

B_H = 0.15 G and B_V = 0.26 G

The resultant magnetic field of the Earth is obtained using the components.

Do You Know? – Aurora Borealis and Aurora Australis

People living at high latitude regions (near Arctic or Antarctic) might experience dazzling coloured natural lights across the night sky. This ethereal display on the sky is known as aurora borealis (northern lights) or aurora australis (southern lights). These lights are often called as polar lights. The lights are seen above the magnetic poles of the northern and southern hemispheres. They are called as "Aurora borealis" in the north and "Aurora australis" in the south. This occurs as a result of interaction between the gaseous particles in the Earth's atmosphere with highly charged particles released from the Sun's atmosphere through solar wind. These particles emit light due to collision and variations in colour are due to the type of the gas particles that take part in the collisions. A pale yellowish–green colour is produced when the ionized oxygen takes part in the collision and a blue or purplish–red aurora is produced due to ionized nitrogen molecules.

3.1.2 Basic Properties of Magnets

Some basic terminologies and properties used in describing bar magnet.

(a) Magnetic Dipole Moment

Consider a bar magnet as shown in Figure 3.6. Let q_m be the pole strength of the magnetic pole and let l be the distance between the geometrical centre of bar magnet O and one end of the pole. The magnetic dipole moment is defined as the product of its pole strength and magnetic length. It is a vector quantity, denoted by p⃗_m.

Where d⃗ is the vector drawn from south pole to north pole and its magnitude |d⃗| = 2l.

The magnitude of magnetic dipole moment is: p_m = 2q_ml

The SI unit of magnetic moment is A m². The direction of magnetic moment is from south pole to north pole.

(b) Magnetic Field

Magnetic field is the region or space around every magnet within which its influence will be felt by keeping another magnet in that region. The magnetic field B⃗ at a point is defined as a force experienced by the bar magnet of unit pole strength. Its unit is N A⁻¹ m⁻¹.

(c) Types of Magnets

Magnets are classified into natural magnets and artificial magnets. For example, iron, cobalt, nickel, etc. are natural magnets. Strengths of natural magnets are very weak and the shapes of the magnet are irregular. Artificial magnets are made in order to have desired shape and strength. If the magnet is in the form of rectangular shape or cylindrical shape, then it is known as bar magnet.

Properties of Magnet

The following are the properties of bar magnet (Figure 3.7):

1. A freely suspended bar magnet will always point along the north-south direction.
2. A magnet attracts or repels another magnet or magnetic substances towards itself. The attractive or repulsive force is maximum near the end of the bar magnet. When a bar magnet is dipped into iron filings, they cling to the ends of the magnet.
3. When a magnet is broken into pieces, each piece behaves like a magnet with poles at its ends.
4. Two poles of a magnet have pole strength equal to one another.
5. The length of the bar magnet is called geometrical length and the length between two magnetic poles in a bar magnet is called magnetic length. Magnetic length is always slightly smaller than geometrical length. The ratio of magnetic length and geometrical length = 5/6 = 0.833

EXAMPLE 3.2

Let the magnetic moment of a bar magnet be p⃗_m whose magnetic length is d = 2l and pole strength is q_m. Compute the magnetic moment of the bar magnet when it is cut into two pieces (a) along its length (b) perpendicular to its length.

Solution:

(a) Bar magnet cut into two pieces along its length:
When the bar magnet is cut along the axis into two pieces, new magnetic pole strength is q'_m = q_m/2, but magnetic length does not change. So, the magnetic moment is reduced to half.

(b) Bar magnet cut into two pieces perpendicular to the axis:
When the bar magnet is cut perpendicular to the axis into two pieces, magnetic pole strength will not change but magnetic length will be halved. So the magnetic moment is reduced to half.

EXAMPLE 3.3

Compute the magnetic length of a uniform bar magnet if the geometrical length of the magnet is 12 cm. Mark the positions of magnetic pole points.

Solution:

Geometrical length of the bar magnet is 12 cm.

Magnetic length = (5/6) × geometrical length = (5/6) × 12 = 10 cm.

Note

1. Pole strength is a scalar quantity with dimension [M⁰ L T⁰ A]. Its SI unit is N T⁻¹ (newton per tesla) or A m (ampere-metre).
2. Like positive and negative charges in electrostatics, north pole of a magnet experiences a force in the direction of magnetic field while south pole of a magnet experiences force opposite to the magnetic field.
3. Pole strength depends on the nature of materials of the magnet, area of cross section and the state of magnetization.
4. If a magnet is cut into two equal halves along the length, then pole strength is reduced to half.
5. If a magnet is cut into two equal halves perpendicular to the length, then pole strength remains same.
6. If a magnet is cut into two pieces, we will not get separate north and south poles. Instead, we get two magnets. In other words, isolated monopole does not exist in nature.

Magnetic Field Lines

1. Magnetic field lines are continuous closed curves. The direction of magnetic field lines is from North pole to South pole outside the magnet and from South pole to North pole inside the magnet.
2. The direction of magnetic field at any point on the curve is known by drawing tangent to the magnetic field lines at that point.
3. Magnetic field lines never intersect each other. Otherwise, the magnetic compass needle would point towards two different directions, which is not possible.
4. The degree of closeness of the field lines determines the relative strength of the magnetic field. The magnetic field is strong where magnetic field lines crowd and weak where magnetic field lines are well separated.

(d) Magnetic Flux

The number of magnetic field lines crossing any area normally is defined as magnetic flux Φ_B through the area. Mathematically, the magnetic flux through the surface of area A⃗ in a uniform magnetic field B⃗ is defined as:

Φ_B = B⃗ · A⃗ = BA cos θ

where θ is the angle between B⃗ and A⃗.

Special cases:

• (a) When B⃗ is normal to the surface (θ = 0°): Φ_B = BA (maximum).
• (b) When B⃗ is parallel to the surface (θ = 90°): Φ_B = 0.

Magnetic flux is a scalar quantity. The SI unit for magnetic flux is weber, symbol Wb. Dimensional formula: M L² T⁻² A⁻¹. The CGS unit of magnetic flux is maxwell. 1 weber = 10⁸ maxwell.

The magnetic flux density is defined as the number of magnetic field lines crossing per unit area kept normal to the direction of lines of force. Its unit is Wb m⁻² or tesla (T).

(e) Uniform and Non-Uniform Magnetic Field

Uniform magnetic field: Magnetic field is said to be uniform if it has same magnitude and direction at all the points in a given region. Example: locally Earth's magnetic field is uniform.

Non-uniform magnetic field: Magnetic field is said to be non-uniform if the magnitude or direction or both vary at different points in a region. Example: magnetic field of a bar magnet.

EXAMPLE 3.4

Calculate the magnetic flux coming out from closed surface containing magnetic dipole (say, a bar magnet) as shown in figure.

Solution:

The total flux emanating from the closed surface S enclosing the dipole is zero. So, ∮ B⃗ · dA⃗ = 0. Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero. This is similar to Gauss's law in electrostatics.

3.2 COULOMB'S INVERSE SQUARE LAW OF MAGNETISM

Consider two bar magnets A and B as shown in Figure 3.11. When the north pole of magnet A and the north pole of magnet B or the south pole of magnet A and the south pole of magnet B are brought closer, they repel each other. On the other hand, when the north pole of magnet A and the south pole of magnet B or the south pole of magnet A and the north pole of magnet B are brought closer, their poles attract each other.

So analogous to Coulomb's law in electrostatics, we can state Coulomb's law for magnetism (Figure 3.12) as follows:

Coulomb's Law for Magnetism

The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them.

Mathematically: F = k · (q_mA × q_mB) / r²

where q_mA and q_mB are pole strengths of two poles and r is the distance between two magnetic poles.

where k is a proportionality constant whose value depends on the surrounding medium. In SI unit, the value of k for free space is: k = μ₀/4π = 10⁻⁷ H m⁻¹

where μ₀ is the absolute permeability of free space (air or vacuum) and H stands for henry.

EXAMPLE 3.5

The repulsive force between two magnetic poles in air is 9 × 10⁻³ N. If the two poles are equal in strength and are separated by a distance of 10 cm, calculate the pole strength of each pole.

Solution: The magnitude of the force between two poles is given by F = μ₀/(4π) × q_m²/r²

3.2.1 Magnetic Field at a Point Along the Axial Line of the Magnetic Dipole (Bar Magnet)

Consider a bar magnet NS as shown in Figure 3.13. Let N be the north pole and S be the south pole of the bar magnet, each of pole strength q_m and are separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet) at a distance r from the geometrical centre O of the bar magnet can be computed by keeping unit north pole (q_mc = 1 Am) at C.

The magnetic field at C due to the north pole is in the direction away from N, where (r − l) is the distance between north pole of the bar magnet and unit north pole at C.

The magnetic field at C due to the south pole is in the direction towards S, where (r + l) is the distance between south pole of the bar magnet and unit north pole at C.

The net magnetic field due to the magnetic dipole at point C is along the axial direction.

Since the magnitude of magnetic dipole moment |p⃗_m| = p_m = q_m · 2l, the magnetic field at a point C (axial line) is:

B_axial = (μ₀/4π) · (2p_mr) / (r² − l²)²

If r ≫ l (short magnet): B_axial = (μ₀/4π) · (2p_m) / r³    (Equation 3.13)

3.2.2 Magnetic Field at a Point Along the Equatorial Line Due to a Magnetic Dipole (Bar Magnet)

Consider a bar magnet NS as shown in Figure 3.14. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength q_m and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical centre O of the bar magnet can be computed by keeping unit north pole (q_mc = 1 Am) at C.

The net magnetic field at point C at the equatorial point is:

B_equatorial = (μ₀/4π) · p_m / (r² + l²)^(3/2)

If r ≫ l (short magnet): B_equatorial = (μ₀/4π) · p_m / r³

Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.

EXAMPLE 3.6

A short bar magnet has a magnetic moment of 0.5 J T⁻¹. Calculate magnitude and direction of the magnetic field produced by the bar magnet which is kept at a distance of 0.1 m from the centre of the bar magnet along (a) axial line of the bar magnet and (b) normal bisector of the bar magnet.

Solution:

Given: magnetic moment = 0.5 J T⁻¹ and distance r = 0.1 m

(a) B_axial = (μ₀/4π) · (2p_m) / r³ = 1 × 10⁻⁴ T; direction is towards South to North.

(b) B_equatorial = (μ₀/4π) · p_m / r³ = 0.5 × 10⁻⁴ T; direction is towards North to South.

Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.

3.3 TORQUE ACTING ON A BAR MAGNET IN UNIFORM MAGNETIC FIELD

Consider a magnet of length 2l and pole strength q_m kept in a uniform magnetic field B⃗ as shown in Figure 3.16. Each pole experiences a force of magnitude q_mB but acting in opposite directions. Therefore, the net force exerted on the magnet is zero and hence, there is no translatory motion. These two equal and opposite forces constitute a couple (about midpoint of bar magnet) tend to align the magnet in the direction of the magnetic field B⃗.

The magnitude of total torque about point O:

τ = p_m B sin θ

In vector notation: τ⃗ = p⃗_m × B⃗    (Equation 3.25)

Do You Know?

(a) Why a freely suspended bar magnet in your laboratory experiences only torque (rotational motion) but not any translator motion even though Earth has non-uniform magnetic field? It is because Earth's magnetic field is locally (physics laboratory) uniform.

(b) Suppose we keep a freely suspended bar magnet in a non-uniform magnetic field. What will happen? It will undergo translatory motion (net force) and rotational motion (torque).

3.3.1 Potential Energy of a Bar Magnet in a Uniform Magnetic Field

When a bar magnet (magnetic dipole) of dipole moment p⃗_m is held at an angle θ with the direction of a uniform magnetic field B⃗, as shown in Figure 3.17, the magnitude of the torque acting on the dipole is τ = p_mB sin θ.

If the dipole is rotated through a very small angular displacement dθ against the torque τ_b at constant angular velocity, then the work done by external torque for this small angular displacement is dW = τ dθ.

Total work done in rotating the dipole from θ' to θ:

W = p_mB (cos θ' − cos θ)

If we choose the reference point as θ' = 90°, so that second term in the equation becomes zero:

U = −p_mB cos θ = −p⃗_m · B⃗    (Equation 3.27)

The potential energy stored in a bar magnet in a uniform magnetic field is given by U = −p⃗_m · B⃗.

Case 1

The potential energy of the bar magnet is minimum when it is aligned along the external magnetic field (θ = 0°, U = −p_mB) and maximum when the bar magnet is aligned anti-parallel to external magnetic field (θ = 180°, U = +p_mB).

EXAMPLE 3.7

Consider a magnetic dipole which on switching ON external magnetic field orient only in two possible ways i.e., one along the direction of the magnetic field (parallel to the field) and another anti-parallel to magnetic field. Compute the energy for the possible orientation.

Solution:

Let p⃗_m be the dipole and before switching ON the external magnetic field, there is no orientation. Therefore, the energy U = 0.

As soon as external magnetic field is switched ON:

Parallel (θ = 0°): U = −p_mB cos 0° = −p_mB

Anti-parallel (θ = 180°): U = −p_mB cos 180° = +p_mB

3.4 MAGNETIC PROPERTIES

All materials are not magnetic in nature. Further, all the magnetic materials will not behave identically. So, in order to differentiate one magnetic material from another, some basic parameters are used.

(a) Magnetising Field

The magnetic field which is used to magnetize a sample or specimen is called the magnetising field. Magnetising field is a vector quantity and is denoted by H⃗ and its unit is A m⁻¹.

(b) Magnetic Permeability

The magnetic permeability is the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance. In free space, the permeability (or absolute permeability) is denoted by μ₀ and for any other medium it is denoted by μ. The relative permeability μ_r is defined as:

μ_r = μ / μ₀    (Equation 3.29)

Relative permeability is a dimensionless number and has no units. For free space (air or vacuum), the relative permeability is unity i.e., μ_r = 1.

(c) Intensity of Magnetisation

Any bulk material (any object of finite size) contains a large number of atoms. Each atom consists of electrons which undergo orbital motion. Due to orbital motion, electron has magnetic moment which is a vector quantity. In general, these magnetic moments orient randomly, therefore, the net magnetic moment is zero per unit volume of the material. When such a material is kept in an external magnetic field, atomic dipoles are induced and hence, they will try to align partially or fully along the direction of external field. The net magnetic moment per unit volume of the material is known as intensity of Magnetisation. It is a vector quantity.

The SI unit of intensity of magnetisation is ampere metre⁻¹ (A m⁻¹).

For a bar magnet, the intensity of magnetisation can be defined as the pole strength per unit area (face area).    (Equation 3.31)

(d) Magnetic Induction or Total Magnetic Field

When a substance like soft iron bar is placed in a uniform magnetising field H⃗, the substance gets magnetised. The magnetic induction (total magnetic field) inside the specimen B⃗ is equal to the sum of the magnetic field B⃗₀ produced in vacuum due to the magnetising field and the magnetic field B⃗_m due to the induced magnetism of the substance:

B⃗ = μ₀(H⃗ + M⃗)

(e) Magnetic Susceptibility

When a substance is kept in a magnetising field H⃗, magnetic susceptibility gives information about how a material responds to the external (applied) magnetic field. In other words, the magnetic susceptibility measures how easily and how strongly a material can be magnetised. It is defined as the ratio of the intensity of magnetization M⃗ induced in the material to the magnetising field H⃗:

χ_m = M / H

It is a dimensionless quantity.

Table 3.1 – Magnetic Susceptibility for Various Materials

Material	Magnetic Susceptibility (χm)
Aluminium	2.3 × 10⁻⁵
Copper	−0.98 × 10⁻⁵
Diamond	−2.2 × 10⁻⁵
Gold	−3.6 × 10⁻⁵
Mercury	−3.2 × 10⁻⁵
Silver	−2.6 × 10⁻⁵
Titanium	7.06 × 10⁻⁵
Tungsten	6.8 × 10⁻⁵
Carbon dioxide (1 atm)	−2.3 × 10⁻⁵
Oxygen (1 atm)	2090 × 10⁻⁹

EXAMPLE 3.8

Compute the intensity of magnetisation of the bar magnet whose mass, magnetic moment and density are 200 g, 2 A m² and 8 g cm⁻³, respectively.

Solution:

Density = mass / volume ⟹ Volume = mass / density

Magnitude of magnetic moment p_m = 2 A m²

Intensity of Magnetisation = p_m / Volume

EXAMPLE 3.9

Using the relation B⃗ = μ₀(H⃗ + M⃗), show that χ_m = μ_r − 1.

Solution: B = μ₀(H + M) = μ₀H(1 + M/H) = μ₀H(1 + χ_m) = μH. So μ = μ₀(1 + χ_m), therefore μ_r = 1 + χ_m, hence χ_m = μ_r − 1.

EXAMPLE 3.10

Two materials X and Y are magnetised whose values of intensity of magnetisation are 500 A m⁻¹ and 2000 A m⁻¹ respectively. If the magnetising field is 1000 A m⁻¹ then which one among these materials can be easily magnetized?

Solution:

Susceptibility of X: χ_mX = 500/1000 = 0.5

Susceptibility of Y: χ_mY = 2000/1000 = 2

Since susceptibility of material Y is greater than that of material X, material Y can be easily magnetized.

3.5 CLASSIFICATION OF MAGNETIC MATERIALS

The magnetic materials are generally classified into three types based on their behaviour in a magnetising field. They are diamagnetic, paramagnetic and ferromagnetic materials.

(a) Diamagnetic Materials

The orbital motion of electrons around the nucleus produces a magnetic field perpendicular to the plane of the orbit. Thus each electron orbit has finite orbital magnetic dipole moment. Since the orbital planes of the other electrons are oriented in random manner, the vector sum of magnetic moments is zero and there is no resultant magnetic moment for each atom.

In the presence of a uniform external magnetic field, some electrons are speeded up and some are slowed down. The electrons whose moments were anti-parallel are speeded up according to Lenz's law and this produces an induced magnetic moment in a direction opposite to the field. The induced moment disappears as soon as the external field is removed.

When placed in a non-uniform magnetic field, the interaction between induced magnetic moment and the external field creates a force which tends to move the material from stronger part to weaker part of the external field. It means that diamagnetic material is repelled by the field.

Examples: Bismuth, Copper and Water etc.

Properties of diamagnetic materials:

1. Magnetic susceptibility is negative.
2. Relative permeability is slightly less than unity.
3. The magnetic field lines are repelled or expelled by diamagnetic materials when placed in a magnetic field.
4. Susceptibility is nearly temperature independent.

Note: Superconductors are perfect diamagnetic materials. The expulsion of magnetic flux from a superconductor during its transition to the superconducting state is known as Meissner effect.

(b) Paramagnetic Materials

In some magnetic materials, each atom or molecule has net magnetic dipole moment which is the vector sum of orbital and spin magnetic moments of electrons. Due to the random orientation of these magnetic moments, the net magnetic moment of the materials is zero.

Do You Know? – Magnetic Levitated Train

Magnetic levitated train is also called Maglev train. This train floats few centimetres above the guideway because of electromagnet used. Maglev train does not need wheels and also achieve greater speed. The basic mechanism of working of Maglev train involves two sets of magnets. One set is used to repel which makes train to float above the track and another set is used to move the floating train ahead at very great speed. These trains are quieter, smoother and environmental friendly compared conventional trains and have potential for moving with much higher speeds with technology in future.

In the presence of an external magnetic field, the torque acting on the atomic dipoles will align them in the field direction. As a result, there is net magnetic dipole moment induced in the direction of the applied field. The induced dipole moment is present as long as the external field exists. When placed in a non-uniform magnetic field, the paramagnetic materials will tend to move from weaker to stronger part of the field. Materials which exhibit weak magnetism in the direction of the applied field are known as paramagnetic materials.

Examples: Aluminium, Platinum, Chromium and Oxygen etc.

Properties of Paramagnetic Materials

1. Magnetic susceptibility is positive and small.
2. Relative permeability is greater than unity.
3. The magnetic field lines are attracted into the paramagnetic materials when placed in a magnetic field.
4. Susceptibility is inversely proportional to temperature.

Curie's Law

When temperature is increased, thermal vibration will upset the alignment of magnetic dipole moments. Therefore, the magnetic susceptibility decreases with increase in temperature. In many cases, the susceptibility of the materials is χ_m = C/T. This relation is called Curie's law. Here C is called Curie constant and temperature T is in kelvin. The graph drawn between magnetic susceptibility and temperature is a rectangular hyperbola.

(c) Ferromagnetic Materials

An atom or a molecule in a ferromagnetic material possesses net magnetic dipole moment as in a paramagnetic material. A ferromagnetic material is made up of smaller regions, called ferromagnetic domains (Figure 3.20). Within each domain, the magnetic moments are spontaneously aligned in a direction. This alignment is caused by strong interaction arising from electron spin which depends on the inter-atomic distance. Each domain has net magnetisation in a direction. However the direction of magnetisation varies from domain to domain and thus net magnetisation of the specimen is zero.

In the presence of external magnetic field, two processes take place:

1. The domains having magnetic moments parallel to the field grow bigger in size.
2. The other domains (not parallel to field) are rotated so that they are aligned with the field.

As a result of these mechanisms, there is a strong net magnetisation of the material in the direction of the applied field.

When placed in a non-uniform magnetic field, the ferromagnetic materials will have a strong tendency to move from weaker to stronger part of the field. Materials which exhibit strong magnetism in the direction of applied field are called ferromagnetic materials.

Examples: Iron, Nickel and Cobalt.

Properties of Ferromagnetic Materials

1. Magnetic susceptibility is positive and large.
2. Relative permeability is large.
3. The magnetic field lines are strongly attracted into the ferromagnetic materials when placed in a magnetic field.
4. Susceptibility is inversely proportional to temperature.

Curie-Weiss Law

As temperature increases, the ferromagnetism decreases due to the increased thermal agitation of the atomic dipoles. At a particular temperature, ferromagnetic material becomes paramagnetic. This temperature is known as Curie temperature T_C. The susceptibility of the material above the Curie temperature is given by χ_m = C/(T − T_C). This relation is called Curie-Weiss law. The constant C is called Curie constant and temperature T is in kelvin scale.

Do You Know? – Keezhadi and Magnetism

Magnetism plays interesting role in various aspects of life. It has connection with archaeological place Keezhadi too. To find whether any archaeological structure exists under the surface of a given place, well established technique called 'magnetometer surveying' is used. In this technique, the variation of the magnetic field in comparison with the neighbouring place is studied. The magnetic field variation is due to the presence of magnetic mineral magnetite and its related minerals present in the archaeological structures like buried wall, pottery, bricks, buried tombs, monuments and inhabited sites. Those minerals are either diamagnetic or paramagnetic or ferromagnetic in nature and each type has different range of magnetic susceptibilities. Indian Institute of Geomagnetism (IIG), Mumbai conducted magnetometer survey on Keezhadi site and found out that there were archaeological structures like wall, pottery etc. There was magnetic field variation in the range of 10 to 100 nT over the particular area. In fact, the existence of massive brick structures at Keezhadi has been revealed through magnetism.

Do You Know? – Spin

Like mass and charge for particles, spin is also another important attribute for an elementary particle. Spin is a quantum mechanical phenomenon which is responsible for magnetic properties of the material. Spin in quantum mechanics is entirely different from spin we encounter in classical mechanics. Spin in quantum mechanics does not mean rotation; it is intrinsic angular momentum which does not have classical analogue. For historical reason, the name spin is retained. Spin of a particle takes only positive values, but the orientation of the spin vector takes plus or minus values in an external magnetic field. For example, electron has spin s = 1/2. In the presence of magnetic field, the spin will orient either parallel or anti-parallel to the direction of magnetic field. Spin is parallel to the magnetic field direction (Spin up) or anti-parallel (Spin down). This implies that the magnetic spin m_s takes two values for an electron: m_s = +1/2 (spin up) and m_s = −1/2 (spin down). Spin for proton and neutron is s = 1/2. For photon, spin s = 1.

Comparison of Types of Magnetism (NOT FOR EXAMINATION)

Type of magnetism	Magnetising field absent (H = 0)	Magnetising field present (H ≠ 0)	Susceptibility	Relative permeability
Diamagnetism	Zero magnetic moment	Aligned opposite to the field	Negative	Less than unity
Paramagnetism	Net magnetic moment but random alignment	Aligned with the field	Positive and small	Greater than unity
Ferromagnetism	Net magnetic moment in a domain but they are randomly aligned	Aligned with the field	Positive and large	Very large

3.6 HYSTERESIS

When a ferromagnetic material is kept in a magnetising field, the material gets magnetised by induction. An important characteristic of ferromagnetic material is that the variation of magnetic induction B⃗ with magnetising field H⃗ is not linear. It means that the ratio B/H = μ is not a constant.

A ferromagnetic material (example, Iron) is magnetised slowly by a magnetising field H⃗. The magnetic induction B⃗ of the material increases from point A with the magnitude of the magnetising field and then attains a saturation level. This response of the material is depicted by the path AC as shown in Figure 3.23. Saturation magnetization is defined as the maximum point up to which the material can be magnetised by applying the magnetising field.

If the magnetising field is now reduced, the magnetic induction also decreases but does not retrace the original path CA. It takes different path CD. When the magnetising field is zero, the magnetic induction is not zero and it has positive value. This implies that some magnetism is left in the specimen even when H = 0. The residual magnetism AD present in the specimen is called remanence or retentivity. Remanence is defined as the ability of the materials to retain the magnetism in them even after the magnetising field disappears.

In order to demagnetise the material, the magnetising field is gradually increased in the reverse direction. Now the magnetic induction decreases along DE and becomes zero at E. The magnetising field AE in the reverse direction is required to bring residual magnetism to zero. The magnitude of the reverse magnetising field for which the residual magnetism of the material vanishes is called its coercivity.

Further increase of H⃗ in the reverse direction causes the magnetic induction to increase along EF until it reaches saturation at F in the reverse direction. If magnetising field is decreased and then increased with direction reversed, the magnetic induction traces the path F, G, K, C. This closed curve A, C, D, E, F, G, K, C is called hysteresis loop and it corresponds to one cycle of magnetisation. In the entire cycle, the magnetic induction B lags behind the magnetising field H. This phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means 'lagging behind'.

Hysteresis Loss

During the magnetisation of the specimen through a cycle, there is loss of energy in the form of heat. This loss is attributed to the rotation and orientation of molecular magnets in various directions. It is found that the energy lost (or dissipated) per unit volume of the material when it is carried through one cycle of magnetisation is equal to the area of the hysteresis loop.

Hard and Soft Magnetic Materials

Based on the shape and size of the hysteresis loop, ferromagnetic materials are classified as soft magnetic materials with smaller area and hard magnetic materials with larger area.

Table 3.2 – Differences between Soft and Hard Ferromagnetic Materials

S.No.	Properties	Soft Ferromagnetic Materials	Hard Ferromagnetic Materials
1	When external field is removed	Magnetisation disappears	Magnetisation persists
2	Area of the loop	Small	Large
3	Retentivity	Low	High
4	Coercivity	Low	High
5	Susceptibility and magnetic permeability	High	Low
6	Hysteresis loss	Less	More
7	Uses	Solenoid core, transformer core and electromagnets	Permanent magnets
8	Examples	Soft iron, Mumetal, Stalloy etc.	Carbon steel, Alnico, Lodestone etc.

Applications of Hysteresis Loop

The significance of hysteresis loop is that it provides information such as retentivity, coercivity, permeability, susceptibility and energy loss during one cycle of magnetisation for each ferromagnetic material. Some examples:

1. Permanent magnets: The materials with high retentivity, high coercivity and low permeability are suitable for making permanent magnets. Examples: Carbon steel and Alnico.
2. Electromagnets: The materials with high initial permeability, low retentivity, low coercivity and thin hysteresis loop with smaller area are preferred to make electromagnets. Examples: Soft iron and Mumetal (Nickel Iron alloy).
3. Core of the transformer: The materials with high initial permeability, large magnetic induction and thin hysteresis loop with smaller area are needed to design transformer cores. Examples: Soft iron.

EXAMPLE 3.11

The following figure shows the variation of intensity of magnetisation with the applied magnetic field intensity for three magnetic materials X, Y and Z. Identify the materials X, Y and Z.

Solution:

The slope of M-H graph is a measure of the magnetic susceptibility, χ_m = M/H.

Material X: Slope is positive and larger value. So, it is a ferromagnetic material.

Material Y: Slope is positive and lesser value than X. So, it could be a paramagnetic material.

Material Z: Slope is negative and hence, it is a diamagnetic material.

3.7 MAGNETIC EFFECTS OF CURRENT

3.7.1 Oersted Experiment

In 1820 Hans Christian Oersted, while preparing for his lecture in physics, noticed that electric current passing through a wire deflects the nearby magnetic needle in the compass. By proper investigation, he observed that the deflection of magnetic needle is due to the change in magnetic field produced around current carrying conductor (Figure 3.25). When the direction of current is reversed, the magnetic needle is deflected in the opposite direction. This lead to the development of the theory 'electromagnetism' which unifies two branches in physics namely, electricity and magnetism.

3.7.2 Magnetic Field Around a Straight Current-Carrying Conductor and Circular Loop

(a) Current Carrying Straight Conductor

Suppose we keep a magnetic compass near a current-carrying straight conductor, then the needle of the magnetic compass experiences a torque and deflects to align in the direction of the magnetic field at that point. For a straight current-carrying conductor, the nature of magnetic field is like concentric circles having their common centre on the axis of the conductor as shown in Figure 3.26(a). The direction of circular magnetic field lines will be clockwise or anticlockwise depending on the direction of current in the conductor. If the strength (or magnitude) of the current is increased then the density of the magnetic field will also increase. The strength of the magnetic field (B) decreases as the distance (r) from the conductor increases.

(b) Circular Coil Carrying Current

Suppose we keep a magnetic compass near a current carrying circular conductor, then the needle of the magnetic compass experiences a torque and deflects to align in the direction of the magnetic field at that point. We can notice that at the points A and B in the vicinity of the coil, the magnetic field lines are circular. The magnetic field lines are nearly parallel to each other near the centre of the loop, indicating that the field present near the centre of the coil is almost uniform (Figure 3.27). The strength of the magnetic field is increased if either the current in the coil or the number of turns or both are increased. The polarity (north pole or south pole) depends on the direction of current in the loop.

3.7.3 Right Hand Thumb Rule

The right hand rule is used to find the direction of magnetic field when the direction of current in a conductor is known. Assume that we hold the current carrying conductor in our right hand such that the thumb points in the direction of current flow, then the fingers encircling the conductor point in the direction of the magnetic field lines produced. Figure 3.28 shows the right hand rule for current carrying straight conductor and circular coil.

3.7.4 Maxwell's Right Hand Cork Screw Rule

This rule can also be used to find the direction of the magnetic field around the current-carrying conductor. If we rotate a right-handed screw using a screw driver, then the direction of current is same as the direction in which screw advances and the direction of rotation of the screw gives the direction of the magnetic field. (Figure 3.29)

EXAMPLE 3.12

The magnetic field shown in the figure is due to the current carrying wire. In which direction does the current flow in the wire?

Solution: Using right hand rule, current flows upwards.

3.8 BIOT-SAVART LAW

Soon after Oersted's discovery, both Jean-Baptiste Biot and Felix Savart in 1819 did quantitative experiments on the force experienced by a magnet kept near current carrying wire and arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the magnetic field. This is true for any shape of the conductor.

3.8.1 Definition and Explanation of Biot-Savart Law

Biot and Savart experimentally observed that the magnitude of magnetic field dB⃗ at a point P (Figure 3.30) at a distance r from the small elemental length taken on a conductor carrying current varies:

• directly as the strength of the current I
• directly as the magnitude of the length element dl⃗
• directly as the sine of the angle θ between dl⃗ and r̂
• inversely as the square of the distance r between the point P and length element dl⃗

This is expressed as: dB⃗ = (μ₀/4π) · I dl⃗ × r̂ / r²

where k = μ₀/4π in SI units.

Here dB⃗ is perpendicular to both I dl⃗ (pointing the direction of current flow) and the unit vector r̂ directed from dl⃗ toward point P (Figure 3.31).

The net magnetic field at P due to the conductor is: B⃗ = ∫ (μ₀/4π) · I dl⃗ × r̂ / r²    (Equation 3.34)

Cases

1. If the point P lies on the conductor, then θ = 0°. Therefore |dB⃗| is zero.
2. If the point lies perpendicular to the conductor, then θ = 90°. Therefore dB⃗ is maximum.

Note – Electric Current as a Vector Quantity

Electric current is not a vector quantity. It is a scalar quantity. But electric current in a conductor has direction of flow. Therefore, the electric current flowing in a small elemental conductor can be taken as vector quantity, i.e., I dl⃗.

Similarities between Electric Field (Coulomb's Law) and Magnetic Field (Biot-Savart's Law)

• Electric and magnetic fields obey inverse square law, so they are long range fields.
• Obey the principle of superposition and are linear with respect to source.

Differences between Electric Field and Magnetic Field

S.No.	Electric Field	Magnetic Field
1	Produced by a scalar source (electric charge q)	Produced by a vector source (current element I dl⃗)
2	It is directed along the position vector joining the source and the point at which the field is calculated	It is directed perpendicular to the position vector r̂ and the current element I dl⃗
3	Does not depend on angle	Depends on the angle between the position vector r̂ and the current element I dl⃗

3.8.2 Magnetic Field Due to Long Straight Conductor Carrying Current

Let YY' be an infinitely long straight conductor and I be the steady current through the conductor as shown in Figure 3.32. In order to calculate magnetic field at a point P which is at a distance a from the wire, let us consider a small line element dl (segment AB). The magnetic field at a point P due to current element I dl can be calculated from Biot-Savart's law.

For infinitely long conductor, the total magnetic field at P is:

B = μ₀I / (2πa)

3.8.3 Magnetic Field Produced Along the Axis of the Current-Carrying Circular Coil

Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction as shown in Figure 3.33. The magnetic field at a point P on the axis of the circular coil at a distance z from the centre of the coil O is computed by taking two diametrically opposite line elements of the coil each of length dl⃗ at C and D.

The magnetic field B⃗ at point P due to the current-carrying circular loop is:

B = μ₀IR² / (2(R² + z²)^(3/2))

If the circular coil contains N turns: B = μ₀NIR² / (2(R² + z²)^(3/2))

The magnetic field at the centre of the coil (z = 0): B = μ₀NI / (2R)    (Equation 3.38)

EXAMPLE 3.13

What is the magnetic field at the centre of the loop shown in figure?

Solution: The magnetic field due to current in the upper semicircle and lower semicircle of the circular coil are equal in magnitude but opposite in direction. Hence, the net magnetic field at the centre of the loop (at point O) is zero: B⃗ = 0⃗.

3.8.4 Tangent Law and Tangent Galvanometer

Tangent galvanometer is a device used to detect very small currents. It is a moving magnet type galvanometer. Its working is based on tangent law (Figure 3.34).

Tangent Law

When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields. Let B be the magnetic field produced by passing current through the coil of the tangent galvanometer and B_H be the horizontal component of Earth's magnetic field. Under the action of two magnetic fields, the needle comes to rest making angle θ with B_H, such that:

B = B_H tan θ    (Equation 3.39)

Construction

Tangent Galvanometer (TG) consists of copper coil of several turns wound on a non-magnetic circular frame. The frame is made up of brass or wood which is mounted vertically on a horizontal base table (turn table) with three levelling screws. The TG is provided with two or more coils of different number of turns. Most of the equipments we use in laboratory, contains coils of 2 turns, 5 turns and 50 turns which are of different thickness and are used for measuring currents of different strengths. At the centre of turn table, there is a small upright projection on which a compass box is placed. Compass box consists of a small magnetic needle which is pivoted at its centre, such that the centres of both magnetic needle and circular coil exactly coincide. A thin aluminium pointer attached perpendicular to the magnetic needle moves over a graduated circular scale. The circular scale is divided into four quadrants and they are graduated in degrees, each quadrant being numbered from 0° to 90°. In order to avoid parallax error in measurement, a mirror is placed below the aluminium pointer.

Precautions

1. All the nearby magnets and magnetic materials are kept away from the instrument.
2. Using spirit level, the levelling screws at the base are adjusted so that the small magnetic needle is exactly horizontal and also coil (mounted on the frame) is exactly vertical.
3. The plane of the coil is kept parallel to the small magnetic needle by rotating the coil about its vertical axis. So that, the coil remains in magnetic meridian.
4. The compass box alone is rotated such that the aluminium pointer reads 0° to 0°.

Theory

In the tangent galvanometer experiment, when no current is passed through the coil, the small magnetic needle lies along horizontal component of Earth's magnetic field. When the circuit is closed, the electric current will pass through the circular coil and produce magnetic field at the centre of the coil. Now there are two fields which are acting mutually perpendicular to each other:

1. the magnetic field (B) due to the electric current in the coil acting normal to the plane of the coil.
2. the horizontal component of Earth's magnetic field (B_H).

Because of these crossed fields, the pivoted magnetic needle deflects through an angle θ. From tangent law: B = B_H tan θ.

The horizontal component of Earth's magnetic field: B_H = μ₀NI / (2Rtan θ)

EXAMPLE 3.14

A coil of a tangent galvanometer of diameter 0.24 m has 100 turns. If the horizontal component of Earth's magnetic field is 25 × 10⁻⁶ T, then calculate the current which gives a deflection of 60°.

Solution:

Diameter = 0.24 m, so radius R = 0.12 m. N = 100. B_H = 25 × 10⁻⁶ T. θ = 60°.

B = B_H tan 60° and B = μ₀NI/(2R). Solving for I gives the required current.

3.8.5 Current Loop as a Magnetic Dipole

The magnetic field at a point on the axis of the current-carrying circular loop of radius R at a distance z from its centre is given by the standard formula. At larger distance z ≫ R:

B = μ₀IA / (2πz³)

where A = πR² is the area of the loop.

Comparing with axial field of a magnetic dipole: p_m = IA

In vector notation: p⃗_m = IA n̂    (Equation 3.44)

This implies that a current carrying circular loop behaves as a magnetic dipole of magnetic moment p_m. So, the magnetic dipole moment of any current loop is equal to the product of the current and area of the loop.

Right Hand Thumb Rule for Current Loop

If we curl the fingers of right hand in the direction of current in the loop, then the stretched thumb gives the direction of the magnetic moment associated with the loop.

Table 3.3 – End Rule: Polarity with Direction of Current in Circular Loop

Current in circular loop	Polarity
Anti-clockwise current	North Pole
Clockwise current	South Pole

3.8.6 Magnetic Dipole Moment of Revolving Electron

Suppose an electron undergoes circular motion around the nucleus as shown in Figure 3.36. The circulating electron in a loop is like current in a circular loop (since flow of charge constitutes current). The magnetic dipole moment due to current carrying circular loop is μ_l = IA.

If T is the time period of revolution of an electron, the current due to circular motion of the electron is I = −e/T, where −e is the charge of an electron.

If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit, then T = 2πR/v.

μ_l = −evR/2

The angular momentum of the electron about O: L = mvR

Therefore: μ_l = −(e/2m) × L

The negative sign indicates that the magnetic moment and angular momentum are in opposite direction.

The ratio μ_l/L = e/(2m) is a constant known as gyro-magnetic ratio.

According to Bohr's quantization rule: L = nh/(2π)

The minimum value of magnetic moment (n = 1): μ_B = eh/(4πm) = 9.27 × 10⁻²⁴ A m²

μ_B is called Bohr magneton which is used to measure atomic magnetic moments.

3.9 AMPÈRE'S CIRCUITAL LAW

Ampère's circuital law is used to calculate magnetic field at a point whenever there is a symmetry in the problem. This is similar to Gauss's law in electrostatics.

3.9.1 Ampère's Circuital Law

Ampère's Law

The line integral of magnetic field over a closed loop is μ₀ times net current enclosed by the loop:

∮ B⃗ · dl⃗ = μ₀ I_enclosed

where I_enclosed is the net current linked by the closed loop C. Note that the line integral does not depend on the shape of the path or the position of the conductor with the magnetic field.

3.9.2 Magnetic Field Due to the Current Carrying Wire of Infinite Length Using Ampère's Law

Consider a straight conductor of infinite length carrying current I and the direction of magnetic field lines is shown in Figure 3.37. Since the wire is geometrically cylindrical in shape and symmetrical about its axis, we construct an Ampèrian loop in the form of a circular shape at a distance r from the centre of the conductor as shown in Figure 3.37.

From Ampère's law: ∮ B⃗ · dl⃗ = μ₀I

Due to the symmetry, the magnitude of the magnetic field is uniform over the Amperian loop.

B × 2πr = μ₀I

B = μ₀I / (2πr)

In vector form: B⃗ = (μ₀I / 2πr) n̂

EXAMPLE 3.15

Compute the magnitude of the magnetic field of a long, straight wire carrying a current of 1 A at distance of 1 m from it. Compare it with Earth's magnetic field.

Solution:

Given: I = 1 A and radius r = 1 m

B = μ₀I/(2πr) = (4π × 10⁻⁷ × 1)/(2π × 1) = 2 × 10⁻⁷ T

Earth's magnetic field is approximately 10⁻⁵ T. So B_{straight wire} is one hundred times smaller than B_Earth.

Solenoid

A solenoid is a long coil of wire closely wound in the form of helix as shown in Figure 3.38. When electric current is passed through the solenoid, the magnetic field is produced. The magnetic field of the solenoid is due to the superposition of magnetic fields of each turn of the solenoid. The direction of magnetic field due to solenoid is given by right hand palm-rule. Inside the solenoid, the magnetic field is nearly uniform and parallel to its axis whereas, outside the solenoid the field is negligibly small. Based on the direction of the current, one end of the solenoid behaves like North Pole and the other end behaves like South Pole. The magnetic field of a solenoid looks like the magnetic field of a bar magnet.

3.9.3 Magnetic Field Due to a Long Current Carrying Solenoid

Consider a solenoid of length L having N turns. The diameter of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely. In order to calculate the magnetic field at any point inside the solenoid, we use Ampere's circuital law.

From Ampère's circuital law with rectangular Amperian loop abcd:

∮ B⃗ · dl⃗ = μ₀ × (total current enclosed by Amperian loop)

B × h = μ₀ × n × h × I

B = μ₀nI

where n = N/L is the number of turns per unit length.

Note

Solenoid can be used as electromagnet. It produces strong magnetic field that can be turned ON or OFF. This is not possible in case of permanent magnet. Further the strength of the magnetic field can be increased by keeping iron bar inside the solenoid. This is because the magnetic field of the solenoid magnetizes the iron bar and hence the net magnetic field is the sum of magnetic field of the solenoid and magnetic field of magnetised iron.

EXAMPLE 3.16

Calculate the magnetic field inside a solenoid, when (a) the length of the solenoid becomes twice with fixed number of turns (b) both the length of the solenoid and number of turns are doubled (c) the number of turns becomes twice for the fixed length of the solenoid. Compare the results.

Solution: The magnetic field of a solenoid (inside) is B = μ₀nI = μ₀(N/L)I

(a) L → 2L, N → N: n = N/(2L) → B becomes B/2 (halved)

(b) L → 2L, N → 2N: n = 2N/(2L) = N/L → B remains same

(c) L → L, N → 2N: n = 2N/L → B becomes 2B (doubled)

Thus, strength of the magnetic field is increased when we pack more loops into the same length for a given current.

Do You Know? – MRI

MRI is Magnetic Resonance Imaging which helps the physicians to diagnose or monitor treatment for a variety of abnormal conditions happening within the head, chest, abdomen and pelvis. It is a non invasive medical test. The patient is placed in a circular opening (actually interior of a solenoid which is made up of superconducting wire) and large current is sent through the superconducting wire to produce a strong magnetic field. So, it uses more powerful magnet, radio frequency pulses and a computer to produce pictures of organs which helps the physicians to examine various parts of the body.

3.9.4 Toroid

A solenoid is bent in such a way its ends are joined together to form a closed ring shape, is called a toroid which is shown in Figure 3.42. The magnetic field has constant magnitude inside the toroid whereas in the interior region (say, at point P) and exterior region (say, at point Q), the magnetic field is zero.

(a) Open Space Interior to the Toroid

We construct an Amperian loop 1 of radius r₁ around the point P. Since the loop 1 encloses no current, ∮ B⃗ · dl⃗ = 0. This is possible only if the magnetic field at point P vanishes: B_P = 0.

(b) Open Space Exterior to the Toroid

We construct an Amperian loop 3 of radius r₃ around the point Q. Since in each turn of the toroid loop, current coming out of the plane of paper is cancelled by the current going into the plane of paper, I_enclosed = 0. This is possible only if the magnetic field at point Q vanishes: B_Q = 0.

(c) Inside the Toroid

We calculate the magnetic field B_S at point S by constructing an Amperian loop 2 of radius r₂. The length of the loop is L₂ = 2πr₂.

From Ampere's circuital law: B × 2πr₂ = μ₀NI

If n = N/(2πr₂) is the number of turns per unit length, then: B = μ₀nI

3.10 LORENTZ FORCE

When an electric charge q is kept at rest in a magnetic field, no force acts on it. At the same time, if the charge moves in the magnetic field, it experiences a force. This force is different from Coulomb force. This force is known as magnetic force.

In general, if the charge is moving in both the electric and magnetic fields, the total force experienced by the charge is given by:

F⃗ = q(E⃗ + v⃗ × B⃗)

It is known as Lorentz force.

3.10.1 Force on a Moving Charge in a Magnetic Field

When an electric charge q is moving with velocity v⃗ in the magnetic field B⃗, it experiences a force, called magnetic force F⃗_m. After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field:

F⃗_m = q(v⃗ × B⃗)

Implications of Lorentz Force

1. F⃗_m is directly proportional to the magnetic field B⃗.
2. F⃗_m is directly proportional to the velocity v⃗ of the moving charge.
3. F⃗_m is directly proportional to sine of the angle between the velocity and magnetic field.
4. F⃗_m is directly proportional to the magnitude of the charge q.
5. The direction of F⃗_m is always perpendicular to v⃗ and B⃗.
6. The direction of F⃗_m on negative charge is opposite to the direction of F⃗_m on positive charge provided other factors are identical.
7. If velocity v⃗ of the charge q is along magnetic field B⃗ then F⃗_m is zero.

Definition of Tesla

The strength of the magnetic field is one tesla if a unit charge moving in it with unit velocity experiences unit force.

EXAMPLE 3.17

A particle of charge q moves with velocity v⃗ along positive y-direction in a magnetic field B⃗. Compute the Lorentz force experienced by the particle (a) when magnetic field is along positive y-direction (b) when magnetic field points in positive z-direction (c) when magnetic field is in z-y plane and making an angle θ with velocity of the particle. Mark the direction of magnetic force in each case.

Solution:

Velocity of the particle: v⃗ = v ĵ

(a) B⃗ = B ĵ → F⃗_m = q(v ĵ × B ĵ) = 0. No force acts on the particle when it moves along the direction of magnetic field.

(b) B⃗ = B k̂ → F⃗_m = q(v ĵ × B k̂) = qvB î. The magnitude is qvB and direction is along positive x-direction.

(c) B⃗ = B cos θ ĵ + B sin θ k̂ → F⃗_m = q(v ĵ × (B cos θ ĵ + B sin θ k̂)) = qvB sin θ î.

EXAMPLE 3.18

Compute the work done and power delivered by the Lorentz force on the particle of charge q moving with velocity v⃗. Calculate the angle between Lorentz force and velocity of the charged particle and also interpret the result.

Solution:

For a charged particle moving in a magnetic field: F⃗ = q(v⃗ × B⃗)

The work done by the magnetic field: dW = F⃗ · v⃗ dt = q(v⃗ × B⃗) · v⃗ dt = 0

Since (v⃗ × B⃗) is perpendicular to v⃗, the dot product is zero. This means that Lorentz force does no work on the particle. The angle between Lorentz force and velocity of the charged particle is 90°. Thus Lorentz force changes the direction of the velocity but not the magnitude of the velocity. Hence Lorentz force does no work and also does not alter kinetic energy of the particle.

3.10.2 Motion of a Charged Particle in a Uniform Magnetic Field

Consider a charged particle of charge q having mass m entering into a region of uniform magnetic field B⃗ with velocity v⃗ such that velocity is perpendicular to the magnetic field. As soon as the particle enters into the field, Lorentz force acts on it in a direction perpendicular to both magnetic field B⃗ and velocity v⃗. As a result, the charged particle moves in a circular orbit as shown in Figure 3.45.

The Lorentz force acts as centripetal force:

qvB = mv²/r → r = mv/(qB) = p/(qB)    (Equation 3.57)

Time period: T = 2πm/(qB)    (Equation 3.59) — called cyclotron period.

Frequency: f = qB/(2πm)    (Equation 3.60) — called cyclotron frequency or gyro-frequency.

Angular frequency: ω = qB/m    (Equation 3.61)

From equations (3.59), (3.60) and (3.61), we infer that time period and frequency depend only on charge-to-mass ratio (specific charge) but not on velocity or the radius of the circular path.

If a charged particle moves in a region of uniform magnetic field such that its velocity is not perpendicular to the magnetic field, then the velocity of the particle is split up into two components; one component is parallel to the field while the other component perpendicular to the field. The component of velocity parallel to field remains unchanged and the component perpendicular to the field keeps changing due to Lorentz force. Hence the path of the particle is not a circle; it is a helical path around the field lines as shown in Figure 3.46.

EXAMPLE 3.19

An electron moving perpendicular to a uniform magnetic field 0.500 T undergoes circular motion of radius 2.50 mm. What is the speed of electron?

Solution:

|q| = 1.60 × 10⁻¹⁹ C, B = 0.500 T, m = 9.11 × 10⁻³¹ kg, r = 2.50 × 10⁻³ m

v = |q|Br/m = (1.60 × 10⁻¹⁹ × 0.500 × 2.50 × 10⁻³) / (9.11 × 10⁻³¹) = 2.195 × 10⁸ m s⁻¹

EXAMPLE 3.20

A proton moves in a uniform magnetic field of strength 0.500 T magnetic field is directed along the x-axis. At initial time, t = 0 s, the proton has velocity v⃗ = (1.95 × 10⁵ î + 2.00 × 10⁵ k̂) m s⁻¹. (a) At initial time, what is the acceleration of the proton? (b) Is the path circular or helical? If helical, calculate the radius of helical trajectory and also calculate the pitch of the helix.

Solution:

B⃗ = 0.500 î T, q = 1.60 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg

(a) F⃗ = q(v⃗ × B⃗); acceleration is obtained from F⃗ = ma⃗.

(b) Trajectory is helical. Radius of helical path and pitch of the helix (distance travelled along x-axis per revolution P = v_xT) can be calculated. The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.

EXAMPLE 3.21

Two singly ionized isotopes of uranium ₉₂U²³⁵ and ₉₂U²³⁸ are sent with velocity 1.00 × 10⁵ m s⁻¹ into a magnetic field of strength 0.500 T normally. Compute the distance between the two isotopes after they complete a semi-circle. Also compute the time taken by each isotope to complete one semi-circular path. (Given: m₂₃₅ = 3.90 × 10⁻²⁵ kg and m₂₃₈ = 3.95 × 10⁻²⁵ kg)

Solution:

q = 1.6 × 10⁻¹⁹ C, B = 0.500 T, v = 1.00 × 10⁵ m s⁻¹

r₂₃₅ = m₂₃₅v/(qB) → diameter d₂₃₅ = 2r₂₃₅ = 97.6 cm

r₂₃₈ = m₂₃₈v/(qB) → diameter d₂₃₈ = 2r₂₃₈ = 98.8 cm

Separation distance: Δd = d₂₃₈ − d₂₃₅ = 1.2 cm

This arrangement is known as mass spectrometer. A mass spectrometer is used in many areas in sciences, especially in medicine, in space science, in geology etc.

3.10.3 Motion of a Charged Particle Under Crossed Electric and Magnetic Field (Velocity Selector)

Let us consider an experimental arrangement to illustrate velocity selector as shown in Figure 3.48. In the region of space between the parallel plates of a capacitor which produce uniform electric field E⃗, a uniform magnetic field B⃗ is maintained perpendicular to the direction of electric field. Suppose a charged particle with charge q enters the space from left side with a velocity v⃗, the net force on the particle is:

For a positive charge, the electric force acts in downward direction whereas the Lorentz force acts upwards. When these two forces balance each other:

qE = qvB → v_n = E/B

This speed is independent of mass and charge. By proper choice of electric and magnetic fields, the particle with particular speed can be selected. Such an arrangement of fields is called a velocity selector.

EXAMPLE 3.22

Let E be the electric field of magnitude 6.0 × 10⁶ N C⁻¹ and B be the magnetic field magnitude 0.83 T. Suppose an electron is accelerated with a potential of 200 V, will it show zero deflection? If not, at what potential will it show zero deflection?

Solution:

E = 6.0 × 10⁶ N C⁻¹, B = 0.83 T

v_n = E/B = 6.0 × 10⁶ / 0.83 ≈ 7.23 × 10⁶ m s⁻¹

m = 9.1 × 10⁻³¹ kg, |q| = e = 1.6 × 10⁻¹⁹ C

Velocity at 200 V: v₂₀₀ = √(2|q|V/m). Since v₂₀₀ ≠ v_n, the electron is deflected. The correct potential for null deflection can then be calculated.

3.10.4 Cyclotron

Cyclotron (Figure 3.49) is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle

When a charged particle moves perpendicular to the magnetic field, it experiences magnetic Lorentz force.

Construction

The schematic diagram of a cyclotron is shown in Figure 3.50. The particles are allowed to move in between two semicircular metal containers called Dees (hollow D-shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the centre in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working

Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion moves in a circular path. After one semi-circular path inside Dee 1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force on the charged particle q is provided by Lorentz force. The increase in velocity increases the radius of circular path. This process continues and hence the particle moves in spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target T.

The important condition in cyclotron operation is that when the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_oscillation. This is called resonance condition.

Cyclotron frequency: f = qB/(2πm)

Time period: T = 2πm/(qB)

Kinetic energy of the charged particle: KE = q²B²r²/(2m)

Limitations of Cyclotron

1. The speed of the ion is limited.
2. Electron cannot be accelerated.
3. Uncharged particles cannot be accelerated.

Note: Deutrons (bundles of one proton and one neutron) can be accelerated because it has same charge as that of proton. But neutron (electrically neutral particle) cannot be accelerated by the cyclotron. When a deutron is bombarded with a beryllium target, a beam of high energy neutrons are produced. These high-energy neutrons are sent into the patient's cancerous region to break the bonds in the DNA of the cancer cells (killing the cells). This is used in treatment of fast-neutron cancer therapy.

EXAMPLE 3.23

Suppose a cyclotron is operated to accelerate protons with a magnetic field of strength 1 T. Calculate the frequency in which the electric field between two Dees could be reversed.

Solution:

B = 1 T, m_p = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C

f = qB/(2πm_p) = (1.60 × 10⁻¹⁹ × 1) / (2π × 1.67 × 10⁻²⁷)

3.10.5 Force on a Current Carrying Conductor Placed in a Magnetic Field

When a current carrying conductor is placed in a magnetic field, the force experienced by the conductor is equal to the sum of Lorentz forces on the individual charge carriers in the conductor. Consider a small segment of conductor of length dl, with cross-sectional area A and current I as shown in Figure 3.51.

The force on the small elemental section of the current-carrying conductor is:

dF⃗ = I dl⃗ × B⃗

The force on a straight current carrying conductor of length l placed in a uniform magnetic field is:

F⃗ = I l⃗ × B⃗

In magnitude: F = BIl sin θ

• (a) If θ = 0°: F = 0 (conductor along field direction).
• (b) If θ = 90°: F_total = BIl (maximum, conductor perpendicular to field).

Fleming's Left Hand Rule (FLHR)

When a current carrying conductor is placed in a magnetic field, the direction of the force experienced by it is given by Fleming's Left Hand Rule as shown in Figure 3.52. Stretch out forefinger, the middle finger and the thumb of the left hand such that they are in three mutually perpendicular directions. If the forefinger points in the direction of magnetic field, the middle finger in the direction of the electric current, then thumb will point in the direction of the force experienced by the conductor.

EXAMPLE 3.24

A metallic rod of linear density is 0.25 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 45° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of strength 0.25 T is acting on it in the vertical direction. Calculate the electric current flowing in the rod to keep it stationary.

Solution:

Linear density (mass per unit length) = 0.25 kg m⁻¹, B = 0.25 T, angle = 45°.

For equilibrium: IBl cos 45° = (m/l)gl sin 45°

I = (m/l)g / B = (0.25 × 10) / 0.25 = 9.8 A

So, we need to supply current of 9.8 A to keep the metallic rod stationary.

3.10.6 Force Between Two Long Parallel Current Carrying Conductors

Let two long straight parallel current carrying conductors separated by a distance r be kept in air medium as shown in Figure 3.53. Let I₁ and I₂ be the electric currents passing through the conductors A and B in same direction (i.e. along z-direction) respectively.

The force per unit length of the conductor B due to current in the conductor A:

F/l = μ₀I₁I₂ / (2πr)

The force between two parallel current carrying conductors is attractive if they carry current in the same direction. The force between two parallel current carrying conductors is repulsive if they carry current in opposite directions.

Definition of Ampere

One ampere is defined as that constant current which when passed through each of the two infinitely long parallel straight conductors kept side by side parallelly at a distance of one metre apart in air or vacuum causes each conductor to experience a force of 2 × 10⁻⁷ newton per meter length of conductor.

Do You Know? – Redefinition of Ampere

In November 2018, the redefinition of the ampere along with three other SI base units: the kilogram (mass), kelvin (temperature) and mole (amount of substance) was approved. Starting on May 20, 2019, the ampere is based on a fundamental physical constant: the elementary charge (e). Thus, one ampere is defined as the electric current corresponding to the flow of 1/(1.602 176 634 × 10⁻¹⁹) elementary charges per second.

3.11 TORQUE ON A CURRENT LOOP

The force on a current carrying wire in a magnetic field is responsible for the motor operation.

3.11.1 Torque on a Current Loop Placed in a Magnetic Field

Consider a rectangular loop PQRS carrying current I is placed in a uniform magnetic field B⃗. Let a and b be the length and breadth of rectangular loop respectively. The unit vector n̂ normal to the plane of the loop makes an angle θ with the magnetic field as shown in Figure 3.57.

F_PQ = IaB sin(π/2) = IaB (upwards)

F_RS = IaB (downwards)

F_QR = IbB cos θ (direction as shown)

F_SP = IbB cos θ (direction as shown)

Since F_QR and F_SP are equal, opposite and collinear, they cancel each other. But F_PQ and F_RS constitute a couple which exerts a torque:

τ = IAB sin θ (along direction AB)

In vector form: τ⃗ = p⃗_m × B⃗ = NIAB sin θ

If there are N turns: τ = NIAB sin θ

Special cases:

• (a) θ = 90° (plane of loop parallel to field): τ_max = IAB
• (b) θ = 0° or 180° (plane perpendicular to field): τ = 0

3.11.2 Moving Coil Galvanometer

Moving coil galvanometer is a device which is used to detect the flow of current in an electrical circuit.

Principle

When a current carrying loop is placed in a uniform magnetic field, it experiences a torque.

Construction

A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil as shown in Figure 3.59. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet. The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. In a fine suspension strip, a small plane mirror is attached in order to measure the deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head. In order to pass electric current through the galvanometer, the suspension strip and the spring S are connected to terminals.

Working

Consider a single turn of the rectangular coil PQRS whose length is l and breadth b. PQ = RS = l and QR = SP = b. Let I be the electric current flowing through the rectangular coil PQRS as shown in Figure 3.60. The horse-shoe magnet has hemi-spherical magnetic poles which produces a radial magnetic field. Due to this radial field, the sides QR and SP are always parallel to the magnetic field B and experience no force. The sides PQ and RS are always perpendicular to the magnetic field and experience equal forces in opposite directions. Due to this, torque is produced.

For single turn, the deflecting torque is: τ = bF = bBIl = (lb)BI = ABI (since area A = lb)

For coil with N turns: τ = NABI    (Equation 3.69)

Due to this deflecting torque, the coil gets twisted and restoring torque is developed: τ = Kθ    (Equation 3.70)

where K is the restoring couple per unit twist.

At equilibrium: NABI = Kθ → I = (K/NAB)θ = Gθ

where G = K/(NAB) is called galvanometer constant or current reduction factor.

Figure of Merit of a Galvanometer

It is defined as the current required to produce a deflection of one scale division in the galvanometer.

Sensitivity of a Galvanometer

The galvanometer is said to be sensitive if it shows large scale deflection even for a small current passed through it or a small voltage applied across it.

Current Sensitivity

It is defined as the deflection produced per unit current flowing through galvanometer: I_s = θ/I = NAB/K = 1/G

The current sensitivity of a galvanometer can be increased by:

1. Increasing the number of turns, N
2. Increasing the magnetic induction, B
3. Increasing the area of the coil, A
4. Decreasing the couple per unit twist of the suspension wire, K

Phosphor-bronze wire is used as the suspension wire because the couple per unit twist is very small.

Voltage Sensitivity

It is defined as the deflection produced per unit voltage applied across galvanometer: V_s = θ/V = I_s/R_g

where R_g is the resistance of galvanometer.

EXAMPLE 3.25

The coil of a moving coil galvanometer has 5 turns and each turn has an effective area of 2 × 10⁻² m². It is suspended in a magnetic field whose strength is 4 × 10⁻² Wb m⁻². If the torsional constant K of the suspension fibre is 4 × 10⁻⁹ N m deg⁻¹. (a) Find its current sensitivity in division per micro-ampere. (b) Calculate the voltage sensitivity of the galvanometer for it to have full scale deflection of 50 divisions for 25 mV. (c) Compute the resistance of the galvanometer.

Solution:

N = 5, A = 2 × 10⁻² m², B = 4 × 10⁻² Wb m⁻², K = 4 × 10⁻⁹ N m deg⁻¹

(a) Current sensitivity = NAB/K = (5 × 2 × 10⁻² × 4 × 10⁻²)/(4 × 10⁻⁹) = 10⁵ div A⁻¹ = 0.1 div μA⁻¹ (1 μA = 10⁻⁶ A)

(b) Voltage sensitivity = 50 divisions / 25 mV = 2 div mV⁻¹

(c) Resistance of the galvanometer: R_g = I_s/V_s

EXAMPLE 3.26

The resistance of a moving coil galvanometer is made twice its original value in order to increase current sensitivity by 50%. Find the percentage change in voltage sensitivity.

Solution:

Voltage sensitivity: V_s = I_s/R_g

When resistance is doubled: R' = 2R_g

Increase in current sensitivity by 50%: new I_s' = 1.5 I_s

New voltage sensitivity: V_s' = 1.5 I_s/(2R_g) = 0.75 V_s

Hence the voltage sensitivity decreases. The percentage decrease in voltage sensitivity = 25%.

Conversion of Galvanometer into Ammeter and Voltmeter

Galvanometer to an Ammeter

Ammeter is an instrument used to measure current flowing in the electrical circuit. The ammeter must offer low resistance such that it will not change the current passing through it. So ammeter is connected in series to measure the circuit current. A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance S.

The potential difference across galvanometer is same as the potential difference across shunt resistance:

I_g R_g = (I − I_g) S

Resistance of ammeter: R_a = R_gS / (R_g + S)

Since the shunt resistance is very low, R_a is also small. For an ideal ammeter, the resistance must be equal to zero.

Key Points

1. An ammeter is a low resistance instrument and it is always connected in series to the circuit.
2. An ideal ammeter has zero resistance.
3. In order to increase the range of an ammeter n times, the value of shunt resistance to be connected in parallel is S = R_g/(n − 1).

Galvanometer to a Voltmeter

A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. It should not draw any current from the circuit otherwise the value of potential difference to be measured will change. Voltmeter must have high resistance and when it is connected in parallel, it will not draw appreciable current so that it will indicate the true potential difference.

A galvanometer is converted into a voltmeter by connecting high resistance R_h in series with galvanometer as shown in Figure 3.62.

I_g = V / (R_g + R_h)

Resistance of voltmeter: R_v = R_g + R_h

Key Points

1. Voltmeter is a high resistance instrument and it is always connected in parallel with the circuit element across which the potential difference is to be measured.
2. An ideal voltmeter has infinite resistance.
3. In order to increase the range of voltmeter n times the value of resistance to be connected in series with galvanometer is R_h = (n − 1)R_g.

SUMMARY

Key Points

1. A vertical plane passing through geographic axis is called geographic meridian.
2. A vertical plane passing through magnetic axis is called magnetic meridian.
3. The angle between magnetic meridian at a point with the geographic meridian is called the declination or magnetic declination.
4. The angle subtended by the Earth's total magnetic field B⃗ with the horizontal direction in the magnetic meridian is called dip or magnetic inclination at that point.
5. The magnetic moment is defined as the product of its pole strength and magnetic length. It is a vector quantity, denoted by p⃗_m.
6. The region surrounding magnet where magnetic pole of strength unity experiences a force is known as magnetic field. It is a vector quantity and denoted by B⃗. Its unit is N A⁻¹ m⁻¹.
7. The number of magnetic field lines crossing normally through a given area is called magnetic flux Φ_B. It is a scalar quantity. In SI unit, magnetic flux Φ_B is weber, symbol Wb.
8. Statement of Coulomb's law in magnetism: "The force of attraction or repulsion between two magnetic poles is proportional to the product of their pole strengths and inversely proportional to the square of distance between them".
9. Magnetic dipole kept in a uniform magnetic field experiences torque.
10. Tangent galvanometer is a device used to measure very small currents. It is a moving magnet type galvanometer. Its working is based on tangent law.
11. Tangent law is B = B_H tan θ.
12. The magnetic field which is used to magnetize a sample or specimen is called the magnetising field. It is a vector quantity and denoted by H⃗ and its unit is A m⁻¹.
13. The measure of ability of the material to allow the passage of magnetic lines of force through it is known as magnetic permeability.
14. The net magnetic moment per unit volume of material is known as intensity of magnetisation.
15. Magnetic susceptibility is defined as the ratio of the intensity of magnetisation M⃗ induced in the material due to the magnetising field H⃗.
16. Magnetic materials are classified into three categories: diamagnetic, paramagnetic and ferromagnetic materials.
17. The lagging of magnetic induction B⃗ behind the cyclic variation in magnetising field H⃗ is defined as "Hysteresis", which means "lagging behind".
18. The right hand thumb rule: "Assume that we hold the current carrying conductor in our right hand such that the thumb points in the direction of current flow, then the rest of the fingers encircling the wire point in the direction of the magnetic field lines produced".
19. Maxwell right hand cork screw rule: "If we rotate a screw by a screw driver, then the direction of current is same as the direction in which screw advances, and the direction of rotation of the screw will determine the direction of the magnetic field".
20. Ampère's circuital law: ∮ B⃗ · dl⃗ = μ₀ I_enclosed.
21. Magnetic field inside the solenoid is B = μ₀nI, where n is the number of turns per unit length.
22. Magnetic field interior to the toroid is B = μ₀nI, where n is the number of turns per unit length.
23. Lorentz force is F⃗ = q(E⃗ + v⃗ × B⃗).
24. Charged particle moving in a uniform magnetic field will undergo circular motion.
25. Fleming's Left Hand Rule: Stretch forefinger, the middle finger and the thumb of the left hand such that they are in mutually perpendicular directions. If we keep the forefinger in the direction of magnetic field, the middle finger in the direction of the electric current, then the thumb points in the direction the force experienced by the conductors.
26. One ampere is defined as that constant current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 × 10⁻⁷ newton per meter length of the conductor.
27. When a current carrying coil is placed in a uniform magnetic field, the net force on it is always zero but net torque is not zero. The magnitude of net torque is τ = NABI sin θ.
28. Moving coil galvanometer is an instrument used for the detection and measurement of small currents.
29. In moving coil galvanometer, current passing through the galvanometer is directly proportional to the deflection: I = Gθ, where G = K/(NAB) is called galvanometer constant or current reduction factor.
30. Current sensitivity: I_s = θ/I = NAB/K = 1/G.
31. Voltage sensitivity: V_s = θ/V = 1/(GR_g) = I_s/R_g.
32. Ammeter is an instrument used to measure current in an electrical circuit.

Book-back Exercises — Questions & Answers

Answers generated strictly from the Samacheer Kalvi textbook content. Attempt each question first, then click "Show Answer" to check.

I. Multiple Choice Questions

1. The magnetic field at the centre O of the following current loop is

a) μI/4r b) μI/4r c) μI/2r d) μI 2r

Answer: a) μI/4r

Explanation: The magnetic field at the centre of a circular current loop is derived from Biot-Savart law; for a semicircular loop the field is half that of a full loop, giving μI/4r.

2. An electron moves in a straight line inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor undeflected when the plates of the capacitor are kept under constant magnetic field of induction B is

a) ε₀ elB/σ b) ε₀ lB/σl c) ε₀ lB/eσ d) ε₀ lB/σ

Answer: d) ε₀ lB/σ

Explanation: For the electron to move undeflected, the electric and magnetic forces balance. Using the velocity selector condition and the relation for electric field from charge density, the time comes out as ε₀lB/σ.

3. A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field B.

a) √(2q³BV/m) b) √(2q³B²V/2m) c) √(2q³B²V/m) d) √(2q³BV/m³)

Answer: c) √(2q³B²V/m)

Explanation: A particle accelerated through potential V gains kinetic energy qV = ½mv², giving velocity v = √(2qV/m). The magnetic force is F = qvB, substituting v gives F = qB√(2qV/m) = √(2q³B²V/m).

4. A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is nearly

a) 1.0 A m² b) 1.2 A m² c) 0.5 A m² d) 0.8 A m²

Answer: b) 1.2 A m²

Explanation: Magnetic dipole moment = NIA = 50 × 3 × π × (0.05)² ≈ 50 × 3 × 7.854 × 10⁻³ ≈ 1.178 ≈ 1.2 A m².

5. A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current I = 8 mA. The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the centre of the spiral is

a) 5 μT b) 7 μT c) 8 μT d) 10 μT

Answer: b) 7 μT

Explanation: The magnetic field at the centre of a spiral coil is calculated by integrating contributions from each turn; using the given values the result is approximately 7 μT.

6. Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque?

a) Circle b) Semi-circle c) Square d) All of them

Answer: a) Circle

Explanation: Torque τ = NIAB sinθ. For a given perimeter (wire length), the circle encloses the maximum area among all shapes, so the circular loop experiences the greatest torque.

7. Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P at a distance of R/2 from the centre of each coil is

a) 8Nμ₀I / √5 R b) 8Nμ₀I / (√5)^(3/2) R c) 8Nμ₀I / 5R d) 4Nμ₀I / √5 R

Answer: b) 8Nμ₀I / (√5)^(3/2) R

Explanation: Using the formula for magnetic field on the axis of a circular coil, B = μ₀NIR²/2(R²+z²)^(3/2), and adding contributions from both coils at point P, the result is 8Nμ₀I/(√5)^(3/2) R.

8. A wire of length l carrying a current I along the Y direction is kept in a magnetic field given by B⃗ = β/√3 (î + ĵ + k̂) T. The magnitude of Lorentz force acting on the wire is

a) √(2/3) βIl b) √(1/3) βIl c) √2 βIl d) √(1/2) βIl

Answer: a) √(2/3) βIl

Explanation: The force on a current-carrying wire is F = Il × B. With current along ĵ and B = β/√3(î+ĵ+k̂), computing the cross product Il ĵ × B and taking magnitude gives √(2/3) βIl.

9. A bar magnet of length l and magnetic moment p_m is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be (NEET 2013)

a) P_m b) 3/π × P_m c) 2/π × P_m d) 1/2 × P_m

Answer: b) 3/π × P_m

Explanation: When the bar magnet is bent into an arc subtending 60° at the centre, the effective magnetic length (distance between poles) changes; using geometry the new dipole moment becomes 3/π × P_m.

10. A non-conducting charged ring carrying a charge of q, mass m and radius r is rotated about its axis with constant angular speed ω. Find the ratio of its magnetic moment with angular momentum.

a) q/m b) 2q/m c) q/2m d) q/4m

Answer: c) q/2m

Explanation: Magnetic moment μ = IA = (qω/2π)(πr²) = qωr²/2. Angular momentum L = mr²ω. Therefore μ/L = q/2m, which is the gyromagnetic ratio as discussed in the chapter.

11. The B-H curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is

a) 1.00 mA b) 1.25 mA c) 1.50 mA d) 1.75 mA

Answer: c) 1.50 mA

Explanation: To demagnetize the ferromagnet, the coercive field H_c must be applied. Reading from the B-H curve and using H = nI (n = 1000 turns/cm = 10⁵ turns/m), the required current is 1.50 mA.

12. Two short bar magnets have magnetic moments 1.20 A m² and 1.00 A m² respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is (Horizontal component of Earth's magnetic induction is 3.6 × 10⁻⁵ Wb m⁻²) (NSEP 2000-2001)

a) 3.60 × 10⁻⁵ Wb m⁻² b) 3.5 × 10⁻⁵ Wb m⁻² c) 2.56 × 10⁻⁴ Wb m⁻² d) 2.2 × 10⁻⁴ Wb m⁻²

Answer: c) 2.56 × 10⁻⁴ Wb m⁻²

Explanation: At the midpoint on the common equatorial line, the equatorial fields of both magnets add together (both pointing North to South) and the resultant is added to Earth's horizontal component, giving 2.56 × 10⁻⁴ Wb m⁻².

13. The vertical component of Earth's magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?

a) 30° b) 45° c) 60° d) 90°

Answer: b) 45°

Explanation: From the chapter, tan I = B_V/B_H. When B_V = B_H, tan I = 1, therefore I = 45°.

14. A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the axis of rotation.

a) (1/4)σωπBR b) (1/2)σωπBR² c) (1/4)σωπBR³ d) (1/4)σωπBR⁴

Answer: d) (1/4)σωπBR⁴

Explanation: By integrating the magnetic moment contributions from each ring element of the rotating disc and multiplying by the magnetic field B, the torque is found to be (1/4)σωπBR⁴.

15. The potential energy of magnetic dipole whose dipole moment is p_m = (−0.5 î + 0.4 ĵ) A m² kept in uniform magnetic field B⃗ = 0.2 î T

a) −0.1 J b) −0.8 J c) 0.1 J d) 0.8 J

Answer: c) 0.1 J

Explanation: Potential energy U = −p_m · B = −(−0.5 î + 0.4 ĵ) · (0.2 î) = −(−0.5 × 0.2) = −(−0.1) = +0.1 J.

II. Short Answer Questions

1. What is meant by magnetic induction?

Answer: When a substance like soft iron bar is placed in a uniform magnetising field H, the substance gets magnetised. The magnetic induction (total magnetic field) inside the specimen B is equal to the sum of the magnetic field B₀ produced in vacuum due to the magnetising field and the magnetic field B_m due to the induced magnetism of the substance. Thus B = B₀ + B_m.

2. Define magnetic flux.

Answer: The number of magnetic field lines crossing any area normally is defined as magnetic flux Φ_B through the area. Mathematically, Φ_B = B·A cosθ, where θ is the angle between B and the area vector A. It is a scalar quantity and its SI unit is weber (Wb).

3. Define magnetic dipole moment.

Answer: The magnetic dipole moment is defined as the product of its pole strength and magnetic length. It is a vector quantity denoted by p_m. Its magnitude is p_m = 2q_m l and the direction is from south pole to north pole. Its SI unit is A m².

4. State Coulomb's inverse law.

Answer: The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them. Mathematically, F = (μ₀/4π)(q_mA × q_mB / r²).

5. What is magnetic susceptibility?

Answer: Magnetic susceptibility measures how easily and how strongly a material can be magnetised. It is defined as the ratio of the intensity of magnetisation (M) induced in the material to the magnetising field (H). Mathematically, χ_m = M/H. It is a dimensionless quantity.

6. State Biot-Savart's law.

Answer: The magnitude of magnetic field dB at a point P at distance r from a small current element varies directly as the strength of the current I, the magnitude of length element dl, the sine of the angle θ between dl and r̂, and inversely as the square of the distance r. Mathematically, dB = (μ₀/4π)(I dl sinθ / r²).

7. What is magnetic permeability?

Answer: Magnetic permeability is the measure of the ability of the material to allow the passage of magnetic field lines through it, or the measure of the capacity of the substance to take magnetisation. The relative permeability μ_r is the ratio of absolute permeability of the medium to the permeability of free space: μ_r = μ/μ₀.

8. State Ampere's circuital law.

Answer: Ampère's law states that the line integral of magnetic field over a closed loop is μ₀ times the net current enclosed by the loop. Mathematically, ∮B·dl = μ₀ I_enclosed, where I_enclosed is the net current linked by the closed loop C.

9. Compare dia, para and ferro-magnetism.

Answer: Diamagnetic materials have negative susceptibility, relative permeability slightly less than unity, and are repelled by magnetic fields. Paramagnetic materials have small positive susceptibility, relative permeability slightly greater than unity, and are weakly attracted. Ferromagnetic materials have large positive susceptibility, very large relative permeability, and are strongly attracted into magnetic fields. Susceptibility of paramagnetic materials follows Curie's law (χ ∝ 1/T), while ferromagnetic materials follow Curie-Weiss law above Curie temperature.

10. What is meant by hysteresis?

Answer: Hysteresis is the phenomenon of lagging of magnetic induction (B) behind the magnetising field (H) during the cyclic variation of the magnetising field. The word hysteresis means 'lagging behind'. The energy lost per unit volume per cycle is equal to the area of the hysteresis loop.

11. Define magnetic declination and inclination.

Answer: Magnetic declination (D) is the angle between the magnetic meridian and the geographical meridian at a point. Magnetic inclination or dip (I) is the angle subtended by the Earth's total magnetic field B_E with the horizontal direction in the magnetic meridian at that point.

12. What is resonance condition in cyclotron?

Answer: The resonance condition in cyclotron is that the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_oscillator. This ensures the ion is always accelerated at the right moment when it crosses the gap between the Dees.

13. Define ampere.

Answer: One ampere is defined as that constant current which when passed through each of the two infinitely long parallel straight conductors kept side by side parallelly at a distance of one metre apart in air or vacuum causes each conductor to experience a force of 2 × 10⁻⁷ newton per metre length of conductor.

14. State Fleming's left hand rule.

Answer: Stretch out forefinger, the middle finger and the thumb of the left hand such that they are in three mutually perpendicular directions. If the forefinger points in the direction of magnetic field and the middle finger in the direction of electric current, then the thumb will point in the direction of the force experienced by the conductor.

15. Is an ammeter connected in series or parallel in a circuit? Why?

Answer: An ammeter is connected in series in a circuit. Because an ammeter must measure the current flowing through the circuit, it must be placed in series so that the same current passes through it. Also, an ammeter has very low resistance so that it does not appreciably change the current in the circuit.

16. Explain the concept of velocity selector.

Answer: When a charged particle moves through crossed electric field E and magnetic field B (perpendicular to each other), a particle with a specific speed v = E/B experiences equal and opposite electric and magnetic forces, making the net force zero, and passes through undeflected. This speed is independent of mass and charge. By proper choice of E and B, particles of a particular speed can be selected; such an arrangement is called a velocity selector.

17. Why is the path of a charged particle not a circle when its velocity is not perpendicular to the magnetic field?

Answer: When the velocity is not perpendicular to the magnetic field, it is split into two components: one parallel to the field and one perpendicular. The component parallel to the field remains unchanged (no force), while the perpendicular component causes circular motion due to Lorentz force. The combination of these two motions results in a helical path around the field lines, not a circular path.

18. Give the properties of dia/para/ferromagnetic materials.

Answer: Diamagnetic: (1) Magnetic susceptibility is negative. (2) Relative permeability slightly less than unity. (3) Field lines are repelled. (4) Susceptibility is nearly temperature independent. Paramagnetic: (1) Susceptibility is positive and small. (2) Relative permeability slightly greater than unity. (3) Field lines attracted into material. (4) Susceptibility inversely proportional to temperature (Curie's law). Ferromagnetic: (1) Susceptibility is positive and large. (2) Relative permeability is large. (3) Field lines strongly attracted. (4) Susceptibility inversely proportional to temperature above Curie temperature (Curie-Weiss law).

19. What happens to the domains in a ferromagnetic material in the presence of external magnetic field?

Answer: In the presence of an external magnetic field, two processes take place in ferromagnetic domains: (1) The domains having magnetic moments parallel to the field grow bigger in size. (2) The other domains (not parallel to the field) are rotated so that they align with the field. As a result of these mechanisms, there is a strong net magnetisation in the direction of the applied field.

20. How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?

Answer: (i) Ammeter: A galvanometer is converted into an ammeter by connecting a low resistance (shunt resistance S) in parallel with the galvanometer. The shunt value is S = I_g R_g / (I − I_g). The ammeter is always connected in series in the circuit. (ii) Voltmeter: A galvanometer is converted into a voltmeter by connecting a high resistance R_h in series with the galvanometer, where R_h = (V/I_g) − R_g. The voltmeter is always connected in parallel across the circuit element.

III. Long Answer Questions

1. Discuss Earth's magnetic field in detail.

Answer: Earth itself behaves like a gigantic bar magnet. The north pole of a magnetic compass needle is attracted towards the magnetic south pole of Earth, which is near the geographic north pole. The branch of physics dealing with Earth's magnetic field is called Geomagnetism or Terrestrial magnetism.

Three elements are required to specify Earth's magnetic field: (a) Magnetic declination (D): The angle between magnetic meridian and geographical meridian at a point. At higher latitudes it is greater; near equator it is smaller. For Chennai it is −1°16'. (b) Magnetic dip or inclination (I): The angle subtended by Earth's total magnetic field B_E with the horizontal direction in the magnetic meridian. For Chennai it is 14°28'. (c) Horizontal component B_H: The component of Earth's field along the horizontal direction in the magnetic meridian.

The relations are: B_H = B_E cos I, B_V = B_E sin I, and tan I = B_V/B_H. At the magnetic equator, I = 0°, so B_H = B_E and B_V = 0. At the magnetic poles, I = 90°, so B_V = B_E and B_H = 0.

2. Deduce the relation for the magnetic field at a point due to an infinitely long straight conductor carrying current.

Answer: Consider an infinitely long straight conductor YY' carrying steady current I. To find the magnetic field at point P at perpendicular distance a from the wire, consider a small line element dl at a point A on the conductor.

Using Biot-Savart's law, the magnetic field due to element dl is dB = (μ₀/4π)(I dl sinθ / r²) n̂, where n̂ is the unit vector perpendicular to the plane. By applying trigonometric relations and integrating over the entire conductor from −90° to +90°, the total magnetic field is:

B = μ₀I / 2πa

In vector form, B⃗ = (μ₀I/2πa) n̂, where n̂ is the tangential unit vector. The magnetic field forms concentric circles around the conductor. This result is also obtained directly using Ampère's circuital law by constructing a circular Ampèrian loop of radius r: ∮B·dl = μ₀I, giving B(2πr) = μ₀I, so B = μ₀I/2πr.

3. Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current.

Answer: Consider a current-carrying circular loop of radius R with current I. To find the magnetic field at point P on the axis at distance z from the centre O, consider two diametrically opposite current elements of length dl at C and D.

By Biot-Savart's law, since the current element is perpendicular to the position vector (θ = 90°), each element contributes dB = μ₀I dl / 4π r², where r = √(R² + z²). Due to symmetry, the horizontal components cancel and only the vertical components (dB sinφ) contribute, where sinφ = R/√(R² + z²).

Integrating over the entire loop (circumference = 2πR):

B = μ₀IR² / 2(R² + z²)^(3/2)

For a coil of N turns: B = μ₀NIR² / 2(R² + z²)^(3/2). At the centre (z = 0): B = μ₀NI / 2R.

4. Compute the torque experienced by a magnetic needle in a uniform magnetic field.

Answer: Consider a bar magnet of length 2l and pole strength q_m placed in uniform magnetic field B at angle θ. Each pole experiences a force of magnitude q_mB in opposite directions, so the net translatory force is zero.

These two equal and opposite forces form a couple. The moment of force (torque) about the midpoint O: τ = (force) × (perpendicular distance) = q_mB × 2l sinθ = q_m(2l)B sinθ.

Since p_m = q_m × 2l (magnetic dipole moment):

τ = p_m B sinθ

In vector notation: τ⃗ = p⃗_m × B⃗. The torque is maximum (τ = p_mB) when θ = 90° (magnet perpendicular to field) and zero when θ = 0° or 180° (magnet along field).

5. Calculate the magnetic field at a point on the axial line of a bar magnet.

Answer: Consider a bar magnet NS with pole strength q_m and half-length l. The magnetic field at point C on the axial line at distance r from centre O is found by placing a unit north pole at C.

Field due to north pole: B_N = (μ₀/4π) q_m/(r−l)² (away from N, towards C)

Field due to south pole: B_S = (μ₀/4π) q_m/(r+l)² (towards S, away from C)

Net field: B_axial = B_N − B_S = (μ₀/4π) q_m [1/(r−l)² − 1/(r+l)²]

= (μ₀/4π) × (2p_m r) / (r²−l²)²

For a short magnet (r >> l): B_axial = μ₀/4π × 2p_m/r³, directed from S to N (same as magnetic moment direction).

6. Obtain the magnetic field at a point on the equatorial line of a bar magnet.

Answer: Consider a bar magnet NS with pole strength q_m and half-length l. The magnetic field at point C on the equatorial line at distance r from centre O is computed by placing a unit north pole at C.

Both poles are equidistant from C: distance = √(r² + l²). The magnitudes B_N = B_S = (μ₀/4π) q_m/(r²+l²). The vertical components cancel; horizontal components add up:

B_equatorial = 2 × (μ₀/4π) q_m/(r²+l²) × cosφ, where cosφ = l/√(r²+l²)

= (μ₀/4π) × p_m / (r²+l²)^(3/2)

For a short magnet (r >> l): B_equatorial = μ₀/4π × p_m/r³, directed from N to S (opposite to magnetic moment). Note: B_axial = 2 × B_equatorial.

7. Find the magnetic field due to a long straight conductor using Ampere's circuital law.

Answer: Consider a straight conductor of infinite length carrying current I. Since the wire is cylindrically symmetric, construct a circular Ampèrian loop of radius r centred on the conductor.

From Ampère's circuital law: ∮B⃗·dl⃗ = μ₀I_enclosed

Since B is uniform in magnitude along the Ampèrian loop and parallel to dl at every point (angle = 0°): B∮dl = μ₀I

The circumference of the circular loop is 2πr, so: B(2πr) = μ₀I

B = μ₀I / 2πr

In vector form, B⃗ = (μ₀I/2πr) n̂, where n̂ is the unit vector along the tangent to the Ampèrian loop. The magnetic field lines are concentric circles around the conductor.

8. Discuss the working of cyclotron in detail.

Answer: Principle: When a charged particle moves perpendicular to the magnetic field, it experiences magnetic Lorentz force which acts as centripetal force, making the particle move in a circle.

Construction: Two semicircular metal containers called Dees (hollow D-shaped objects) are enclosed in an evacuated chamber kept in a uniform magnetic field perpendicular to the plane of the Dees. The source S is placed in the gap between the Dees. The Dees are connected to a high-frequency alternating potential difference.

Working: A positive ion from source S is accelerated towards the Dee with negative potential. Inside the Dee, the magnetic force causes circular motion. After a semicircle, the ion reaches the gap; the polarity of Dees is reversed and the ion is accelerated again with greater velocity. The radius of circular path increases with velocity: r = mv/qB. The particle moves in a spiral path of increasing radius, and on reaching the edge is deflected by a deflector plate towards the target.

Resonance condition: f_particle = f_oscillator = qB/2πm. The kinetic energy gained is K.E. = q²B²r²/2m. Limitations: speed is limited; electrons and uncharged particles cannot be accelerated.

9. What is tangent law? Discuss in detail.

Answer: Tangent Law: When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields. If B is one field and B_H is the other (Earth's horizontal component), and the needle deflects by angle θ, then: B = B_H tan θ.

This law is used in the Tangent Galvanometer (TG). When current I passes through the coil of N turns and radius R, the field at centre is B = μ₀NI/2R. This field acts perpendicular to B_H. By tangent law: μ₀NI/2R = B_H tan θ. Therefore, I = (2RB_H/μ₀N) tan θ = K tan θ, where K = 2RB_H/μ₀N is the reduction factor.

The TG is a moving magnet galvanometer used to measure small currents. The horizontal component of Earth's field is B_H = μ₀NI / (2R tan θ).

10. Derive the expression for the torque on a current-carrying coil in a magnetic field.

Answer: Consider a rectangular coil PQRS of length a and breadth b, carrying current I, placed in uniform magnetic field B. The unit normal n̂ to the coil makes angle θ with B.

Force on arm PQ: F_PQ = IaB sin(90°) = IaB (upward). Force on arm RS: F_RS = IaB (downward). These are equal and opposite but not collinear — they form a couple.

Force on arms QR and SP: F_QR = F_SP = IbB sinθ, but they are equal, opposite and collinear — they cancel.

Torque due to couple (F_PQ and F_RS): τ = IaB × b sinθ = I(ab)B sinθ = IAB sinθ, where A = ab is the area of coil.

In vector form: τ⃗ = p⃗_m × B⃗ = IAB sinθ (direction along AB). For N turns: τ = NIAB sinθ. Maximum torque when θ = 90° (coil parallel to B); zero torque when θ = 0° (coil perpendicular to B).

11. Discuss the conversion of galvanometer into an ammeter and also a voltmeter.

Answer: Galvanometer to Ammeter: An ammeter measures current and must have low resistance. A low resistance (shunt S) is connected in parallel with the galvanometer of resistance R_g. The shunt value: S = I_g R_g / (I − I_g), where I_g is the full-scale deflection current and I is the total current. The effective ammeter resistance R_a = SR_g/(S+R_g), which is very small. The deflection indicates the total current I. An ideal ammeter has zero resistance.

Galvanometer to Voltmeter: A voltmeter measures potential difference and must have high resistance. A high resistance R_h is connected in series with the galvanometer. The value: R_h = V/I_g − R_g, where V is the range of voltmeter. The voltmeter resistance R_v = R_g + R_h, which is very large. The galvanometer is connected in parallel across the element. An ideal voltmeter has infinite resistance. In order to increase the range of voltmeter n times: R_h = (n−1)R_g.

12. Calculate the magnetic field inside and outside of the long solenoid using Ampere's circuital law.

Answer: Consider a solenoid of length L with N turns carrying current I. The number of turns per unit length n = N/L.

Inside the solenoid: Construct a rectangular Ampèrian loop abcd. The line integral ∮B·dl has contributions only from side ab (length h), since the field outside is zero and the sides bc and da are perpendicular to B. Thus B·h = μ₀(nh)I (current enclosed = n×h×I turns). Therefore: B = μ₀nI (uniform and parallel to axis inside).

Outside the solenoid: The magnetic field outside a long solenoid is negligibly small (approximately zero), because the field lines are confined within the solenoid. This is verified using Ampèrian loops outside — the net enclosed current is zero for loops entirely outside, giving B = 0.

Thus, the solenoid produces a strong uniform field μ₀nI inside and zero field outside.

13. Derive the expression for the force between two parallel, current-carrying conductors.

Answer: Consider two long parallel conductors A and B separated by distance r, carrying currents I₁ and I₂ respectively in the same direction.

The magnetic field at distance r due to conductor A: B₁ = μ₀I₁/2πr (directed into the page at B, by right-hand rule).

The Lorentz force on a length dl of conductor B in field B₁: dF = I₂ dl × B₁. Force per unit length on conductor B: F/l = μ₀I₁I₂/2πr (directed towards A — attractive).

Similarly, the magnetic field due to conductor B at conductor A: B₂ = μ₀I₂/2πr. Force per unit length on A: F/l = μ₀I₁I₂/2πr (directed towards B — attractive).

Conclusion: The force between two parallel current-carrying conductors is attractive if currents are in the same direction and repulsive if currents are in opposite directions. The force per unit length is: F/l = μ₀I₁I₂/2πr.

14. Give an account of magnetic Lorentz force.

Answer: When an electric charge q moves with velocity v⃗ in a magnetic field B⃗, it experiences a magnetic force called Lorentz force: F⃗_m = q(v⃗ × B⃗). The general Lorentz force including electric field is F⃗ = q(E⃗ + v⃗ × B⃗).

Properties: (1) F_m is proportional to B, v, sinθ, and q. (2) Direction of F_m is perpendicular to both v⃗ and B⃗. (3) Direction on negative charge is opposite to that on positive charge. (4) If v⃗ is parallel to B⃗ (θ = 0°), F_m = 0. (5) Lorentz force does no work on the particle (F⃗·v⃗ = 0), so it does not change the kinetic energy — it only changes the direction of velocity. (6) Tesla is defined: B = 1 T when unit charge moving with unit velocity experiences unit force.

15. Compare the properties of soft and hard ferromagnetic materials.

Answer:

Property	Soft Ferromagnetic	Hard Ferromagnetic
Magnetisation after field removal	Disappears	Persists
Hysteresis loop area	Small	Large
Retentivity	Low	High
Coercivity	Low	High
Susceptibility & permeability	High	Low
Hysteresis loss	Less	More
Uses	Solenoid core, transformer core, electromagnets	Permanent magnets
Examples	Soft iron, Mumetal, Stalloy	Carbon steel, Alnico, Lodestone

16. Derive the expression for the force on a current-carrying conductor in a magnetic field.

Answer: Consider a small segment of conductor of length dl, cross-sectional area A, carrying current I in magnetic field B⃗. The free electrons drift opposite to the current direction with drift velocity v⃗_d.

The relation between current and drift velocity: I = neAv_d (where n = free electron density, e = electron charge).

Average force on each electron: f = −e(v⃗_d × B⃗). Number of electrons in volume Adl: N = nAdl.

Total Lorentz force on the segment: dF⃗ = N × f = nAdl × (−e)(v⃗_d × B⃗).

Since current element I dl⃗ = −enAv⃗_d dl: dF⃗ = I dl⃗ × B⃗

For a straight conductor of length l: F⃗ = I l⃗ × B⃗, with magnitude F = BIl sinθ. Maximum force (F = BIl) when conductor is perpendicular to B (θ = 90°); zero force when parallel to B (θ = 0°).

17. Explain the principle and working of a moving coil galvanometer.

Answer: Principle: When a current-carrying loop is placed in a uniform magnetic field, it experiences a torque.

Construction: A rectangular coil PQRS of insulated copper wire with many turns is wound on a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil. The coil is suspended freely between the pole pieces of a horse-shoe magnet by a phosphor-bronze strip (upper end) and a hair spring (lower end). A small plane mirror is attached to the suspension strip to measure deflection using lamp and scale. Electric current is passed through suspension strip and spring.

Working: The horse-shoe magnet has hemispherical poles producing a radial field. Sides QR and SP are always parallel to B (no force); sides PQ and RS are always perpendicular to B. The deflecting torque for N turns: τ = NABI. The restoring torque due to the spring: τ = Kθ. At equilibrium: NABI = Kθ, so I = (K/NAB)θ = Gθ, where G = K/NAB is the galvanometer constant. Current is directly proportional to deflection. Current sensitivity = θ/I = NAB/K = 1/G.

IV. Numerical Problems

1. A bar magnet having a magnetic moment p⃗_m is cut into four pieces i.e., first cut into two pieces along the axis of the magnet and each piece is further cut along the axis into two pieces. Compute the magnetic moment of each piece.

Answer: When cut along the axis, pole strength is halved but magnetic length is unchanged. First cut (along axis) gives two pieces each with magnetic moment = p_m/2. Second cut (again along axis) halves the pole strength again. Each final piece has magnetic moment = p_m_new = (1/4) p⃗_m.

2. A conductor of linear mass density 0.2 g m⁻¹ suspended by two flexible wires as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s⁻².

Answer: For tension to be zero, the magnetic force must balance gravity. F = BIl = mg (per unit length: BI = (m/l)g). Linear mass density = 0.2 g m⁻¹ = 0.2 × 10⁻³ kg m⁻¹. So I = (m/l)g / B = (0.2 × 10⁻³ × 10) / 1 = 2 × 10⁻³ A = 2 mA. Direction of current: using Fleming's left hand rule with B into the page and force upward, the current flows horizontally (determined by the specific setup shown in the figure).

3. A circular coil with cross-sectional area 0.1 cm² is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3 A and plane of the loop is perpendicular to the direction of magnetic field. Calculate (a) total torque on the coil (b) total force on the coil (c) average force on each electron in the coil due to the magnetic field. (The free electron density for the material of the wire is 10²⁸ m⁻³).

Answer: (a) When the plane of the loop is perpendicular to B, the normal to the loop is parallel to B, so θ = 0°. Torque τ = NIAB sinθ = NIAB sin0° = zero. (b) Total force on a current loop in a uniform magnetic field = zero (net force is always zero in uniform field). (c) The average force on each electron: F = evB (where e = 1.6 × 10⁻¹⁹ C). Using I = neAv_d, v_d = I/(neA) = 3/(10²⁸ × 1.6 × 10⁻¹⁹ × 0.1 × 10⁻⁴) = 1.875 × 10⁻⁴ m/s. Force = ev_dB = 1.6 × 10⁻¹⁹ × 1.875 × 10⁻⁴ × 0.2 ≈ 0.6 × 10⁻²³ N.

4. A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. If the bar magnet is oriented at an angle 30° with the external field experiences a torque of 0.2 N m. Calculate: (1) the magnetic moment of the magnet (2) the work done by the magnetic field in moving it from most stable configuration to the most unstable configuration and also compute the work done by the applied magnetic field in this case.

Answer: (1) τ = p_mB sinθ → 0.2 = p_m × 0.8 × sin30° = p_m × 0.8 × 0.5. So p_m = 0.2/0.4 = 0.5 A m². (2) Most stable configuration: θ = 0° (U_min = −p_mB = −0.5 × 0.8 = −0.4 J). Most unstable configuration: θ = 180° (U_max = +p_mB = +0.4 J). Work done = U_max − U_min = 0.4 − (−0.4) = W = 0.8 J. Work done by applied magnetic field = −W_magnetic = W_mag = −0.8 J.

5. A non-conducting sphere has a mass of 100 g and radius 20 cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turn concentric with the sphere. This sphere is placed on an inclined plane such that plane of coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in vertically upward direction. Compute the current I required to rest the sphere in equilibrium.

Answer: I = 2/π A. For equilibrium on the inclined plane, the torque due to magnetic force on the current loop must balance the torque due to gravity. Using τ_magnetic = NIAB = Nmg(R sinα) and solving for I with A = πR² gives I = 2/π A.

6. Calculate the magnetic field at the centre of a square loop which carries a current of 1.5 A, length of each side being 50 cm.

Answer: For a square loop, the magnetic field at the centre is B = 4 × (μ₀I/4π) × (2/a) × √2 = (4μ₀I√2)/(2πa). With I = 1.5 A and a = 0.5 m: B = (4 × 4π × 10⁻⁷ × 1.5 × √2)/(2π × 0.5) ≈ 3.4 × 10⁻⁶ T.

📝 Summary

1. Magnetism has a rich history rooted in the study of magnetite (Fe₃O₄), and magnets are widely used in modern devices such as motors, loudspeakers, mobile phones, hard discs, and generators, making them essential to everyday life.
2. In 1820, H.C. Oersted discovered that a current-carrying wire deflects a magnetic compass needle, unifying electricity and magnetism into the single branch of physics known as electromagnetism.
3. Earth itself behaves like a gigantic bar magnet, and its magnetic field plays a key role in navigation — both for humans using compasses and for birds and animals (such as the Zebra finch) that use magnetic sensing through proteins like Cry4 in their retinas.
4. Magnetism exists at all scales, from tiny particles like electrons to the entire universe, and the study of magnetic effects of electric current includes important laws such as Biot-Savart's law, Ampère's circuital law, and the Tangent law used in Tangent Galvanometers.
5. Key concepts covered in this unit include magnetic dipoles, torque on bar magnets, classification of magnetic materials (diamagnetic, paramagnetic, and ferromagnetic), hysteresis, the Lorentz force, cyclotron motion, and the working of a Moving Coil Galvanometer.