5.0 Preview

This chapter deals with magnets and its properties. A magnet is a substance that exhibits the phenomenon of attraction of small bits of iron towards itself.

We shall begin our discussion with properties of a Bar magnet and magnetic field lines around it.

The resemblance of magnetic field lines for a bar magnet and a solenoid suggests that a bar magnet may be thought of as a large number of circulating currents. Thus, it is analogous to a solenoid. We will be obtaining magnetic field at a point from one such current carrying solenoid.

Next, we shall study about the dipole in a uniform magnetic field— a dipole is a pair of equal and oppositely charged or magnetized poles separated by a distance. Then, we will look at Gauss law which states that, the net magnetic flux through any closed surface is always zero. You would be interested in knowing about Earth’s magnetism; the branch of physics which deals with the study of magnetism of earth is called terrestrial magnetism or geomagnetism.

Next, we shall study about an important phenomenon called, dynamo effect. Magnetic field around the earth is due to the electric currents produced due to convective motions of metallic fluids in the outer core of the earth. This is known as dynamo effect.

Further, we will study about two important angles, magnetic declination and angle of dip. We shall then define some important terms regarding earth’s magnetic field; like: Magnetization, Magnetic permeability, Relative permeability, Magnetic induction or flux density, Magnetic intensity and Magnetic susceptibility. Based on these magnetic properties and based on the behaviour of magnet when placed in a magnetic field, we shall study the three classifications made by Faraday: Diamagnetic substances, Paramagnetic substances and Ferromagnetic substances.

We will see that when a sample of ferromagnetic material is placed in an external magnetic field, the lines are highly concentrated inside the material. In other words, it is strongly magnetized. When the magnetization persists in the ferromagnetic materials despite the external magnetic field being removed, the materials are called, hard magnetic materials; whereas, if the magnetization disappears when the external magnetic field is removed, the materials are called soft magnetic materials.

Then, we shall study an interesting phenomenon known as Hysteresis as the lagging behind of intensity of magnetization with respect to the magnetizing field strength when a specimen of a ferromagnetic material is subjected to a cycle of magnetization.

Finally, we will study about Permanent magnets and electromagnets in some detail; Permanent magnets are Substances which retrain their ferromagnetic property for a long period of time at room temperatures; an electromagnet is a type of magnet in which the magnetic field is induced by passing electric current through a metal element.

5.1 Introduction

The word magnet is derived from the name of an island in Greece called Magnesia where magnetic ore deposits were found, as early as 600 BCE. The ore so found was named magnetite. The phenomenon of attraction of small bits of iron towards the ore was referred to as magnetism. The ore that shows this property was called magnet.
Properties of magnet

1. When freely suspended, the magnet always settles in north-south direction.

2. Natural magnet has attractive and directive properties.

Note Box 1. The magnetic needles were used in navigation of ships. 2. The magnetite iron is also called lode stone means leading stone. 3. Magnets and magnetic materials are used to store information as in Debit cards and Credit cards.

Known facts of magnetism

1. Earth behaves as a magnet with field pointing approximately from geographic south to north.
2. The freely suspended magnet rests in north-south direction. The end which points to the geographic south is called South Pole.
3. The repulsive force comes into picture when like poles are brought together. Unlike poles attract each-other.
4. Magnet attracts small pieces of iron, cobalt and nickel. Attraction is maximum near the poles.
5. Magnetic poles always exist in pairs.
6. It is possible to make magnets out of iron and its alloys.
7. The sure test of magnetism is repulsion.
8. The force of attraction and repulsion between two magnetic poles is directly proportional to the product of the pole strength and inversely proportional to the square of distance between them.

Questions for section 5.1

1. State some properties of magnet.
2. State some of the facts of magnet.

5.2 The bar magnet

Let us begin our study by examining iron filings sprinkled on a sheet of glass placed over a short bar magnet. The arrangement of iron filings is in an arranged circular shape. The pattern of iron filings suggests that the magnet has two poles similar to the positive and negative charge of an electric dipole.

As we know one pole is designated the North pole and the other, the South pole. When suspended freely, these poles point approximately towards the geographic north and south poles, respectively. A similar pattern of iron filings is observed around a current carrying solenoid.

The magnetic moment of a bar is defined as the product of the pole strength $q_m$ and magnetic length 2l $ m = q_m x 2l $

5.2.1 Magnetic field lines

Some of the properties of magnetic field lines are:

1. They form continuous closed loops.
2. The tangent drawn to the field line at a given point represents the direction of net magnetic field at that point.
3. Larger the number of field lines, stronger the field.
4. The magnetic field lines do not intersect because of the unique direction of the magnetic field at each point.

5.2.2 Bar magnet as an equivalent solenoid

The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid.

Let us assume, as shown in the figure that a be the radius and 2l be the length of solenoid with centre O and n be the number of turns per unit length. The magnetic field at a point P on the axis of the solenoid can be calculated at a distance r from the centre O.

Let us consider a circular element of thickness dx of the solenoid at a distance x from O.

The number of turns in this element = n dx

When the current I flows through the solenoid, the magnitude of the magnetic field at P due to this circular element is $ dB = \frac { μo}{4 π} \frac { 2 π(n \space dx ) I a^2 }{ [ (r-x)^2 +a^2 ]^{3/2}} $

When the point P is at very large distance from O ie r>>a and r>>x Then $ [(r-x)^2+a^2]^{3/2} = r^3 $ As the range of x varies from -l to +l, the total magnetic field at P due to the current carrying solenoid is $$ B = \int{-l}^{+l} \frac { μo}{4 π} \frac {2 π nIa^2}{r^3} dx = \frac { μ_o}{4 π} \frac {2 π nIa^2}{r^3} \int{-l}^{+l} dx $$
$$ = \frac { μo}{4 π} \frac {2 π nIa^2}{r^3} (x){-l}^{+l} $$
$$ = \frac { μ_o}{4 π} \frac {2 π nIa^2}{r^3} (l-(-l)) = \frac { μ_o}{4 π} \frac {4 π nIa^2l}{r^3} $$
$$ m =NI π a^2 = ( n×2l )I π a^2$$
$$ B = \frac { μ_o}{4 π} \frac {2m}{r^3 } $$

The above equation is similar to the magnetic field on the axial line of a bar magnet. So, for all practical purposes, a finite solenoid carrying current is equivalent to a bar magnet.

5.2.3 The dipole in a uniform magnetic field

What is a dipole?

A dipole is a pair of equal and oppositely charged or magnetized poles separated by a distance.
The pattern of iron filings, i.e., the magnetic field lines gives us an approximate idea of the magnetic field B. We may at times be required to determine the magnitude of B accurately. This is done by placing a small compass needle of known magnetic moment m and moment of inertia I and allowing it to oscillate in the magnetic field.

Concept box Moment of Inertia – It is the tendency of a body to resist angular acceleration.

The torque on the needle is τ = mB

In magnitude τ = mB sinθ

Here τ is restoring torque and θ is the angle between m and B.

Therefore, in equilibrium $ I \frac {d^2 θ }{dt^2} = -mB \space sin θ $

Negative sign with mB sinθ implies that restoring torque is in opposition to deflecting torque. For small values of θ in radians, we approximate sin θ ≈ θ and get
$$ I \frac {d^2 θ }{dt^2} ≈ mB \space θ $$
$$ or \space \frac {d^2 θ }{dt^2} = – \frac {mB}{I} θ $$

This represents a simple harmonic motion. The square of the angular frequency is $ ω^2 = mB/I $ and the time period is,
$$ T = 2 π \sqrt { \frac { I}{mB}} $$
$$ or \space B = \frac {4 π^2I}{mT^2} $$

An expression for magnetic potential energy can also be obtained on lines similar to electrostatic potential energy. The magnetic potential energy $U_m$ is given by
$$ U_m = \int τ θdθ $$
$$ = \int mB \space sinθ = – mB \space cosθ $$

$$= -m.B$$

We have emphasised in Chapter 2 that the zero of potential energy can be fixed at one’s convenience. Taking the constant of integration to be zero means fixing the zero of potential energy at θ = 90º, i.e., when the needle is perpendicular to the field. Equation (5.6) shows that potential
energy is minimum (= –mB) at θ = 0º (most stable position) and maximum (= +mB) at θ = 180º (most unstable position).

Example 5.1 In Fig. 5.2, the magnetic needle has magnetic moment $ 6.7 × 10^{–2} Am^2 $and moment of inertia $ I = 7.5 × 10^{–6} kg m^2$. It performs 10 complete oscillations in 6.70s. What is the magnitude of the magnetic field?

Solution The time period of oscillation is, $ T = \frac {6.70}{10} = 0.67 s $

From Eq. (5.5)
$$ B = \frac { 4 π^2 I}{mT^2} $$
$$ = \frac {4×(3.14)^2×7.5×10^{-6}}{6.7×10^{-2}×(0.67)^2} $$
$$=0.0 1 T$$

Example 5.2 A short bar magnet placed with its axis at 30º with an external field of 800 G experiences a torque of 0.016 Nm. (a) What is the magnetic moment of the magnet? (b) What is the work done in moving it from its most stable to most unstable position? (c) The bar magnet is replaced by a solenoid of cross-sectional area $ 2 × 10^{–4} $ m2 and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid.

Solution

(a) From Eq. (5.3),
$$ τ = m \space B \space sin θ , θ =30^o , hence sin θ= \frac{1}{2} $$

Thus ,
$$ 0.016 = m ×2/800 = 0.40 \space A m^2 $$

(b) From Eq. (5.6), the most stable position is θ = 0º and the most unstable position is θ = 180º. Work done is given by
$$ W = U_m(θ=180^o)-U_m(θ=0^o) $$
$$ = 2 \space m \space B = 2×0.40×800×10^{-4} = 0.064 J $$

(c) From Eq. (4.30), $ m_s$ = NIA. From part (a), $ m_s = 0.40 A m^2 $
$$ 0.40 = 1000 × I × 2 × 10^{–4} $$
$$ I = 0.40 × 10^4/(1000 × 2) = 2A $$

Example 5.3
(a) What happens if a bar magnet is cut into two pieces: (i) transverse
to its length, (ii) along its length?
(b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why?
(c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid?
(d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B.]

Solution

(a) In either case, one gets two magnets, each with a north and south pole.

(b) No force if the field is uniform. The iron nail experiences a nonuniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.

(c) Not necessarily. True only if the source of the field has a net nonzero magnetic moment. This is not so for a toroid or even for a straight infinite conductor.

(d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see whichone of the two bars is a magnet, pick up one, (say, A) and lower one of its ends; first on one of the ends of the other (say, B), and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the
end to the middle of B, then A is magnetised.

5.2.4 The electrostatic analogue

Comparison of previous equations with the corresponding equations for electric dipole, suggests that magnetic field at large distances due to a bar magnet of magnetic moment m can be obtained from the equation for electric field due to an electric dipole of dipole moment p, by making the following replacements:
$$ E \ce { -> } B ,p \ce { ->} m , \frac {1}{4 π ε_o} \ce { ->} \frac { μ_o}{4 π} $$

In particular, we can write down the equatorial field (BE) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet:
$$ B_E = \frac { μ_o m}{4 πr^3 } $$

Likewise , the axial field $(B_A)$ of a bar magnet for r >> l is :
$$ B_A = \frac { μ_o}{4 π} \frac {2m}{r^3} $$

Example 5.2 A short bar magnet placed with its axis at 30º with an external field of 800 G experiences a torque of 0.016 Nm. (a) What is the magnetic moment of the magnet? (b) What is the work done in moving it from its most stable to most unstable position? (c) The bar magnet is replaced by a solenoid of cross-sectional area $ 2 × 10^{–4} $ m2 and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid.

Solution

(a) From Eq. (5.3),
$$ τ = m \space B \space sin θ , θ =30^o , hence sin θ= ½ $$
Thus ,
$$ 0.016 = m ×2/800 = 0.40 \space A m^2 $$

(b) From Eq. (5.6), the most stable position is θ = 0º and the most unstable position is θ = 180º. Work done is given by
$$ W = U_m(θ=180^o)-U_m(θ=0^o) $$
$$ = 2 \space m \space B = 2×0.40×800×10^{-4} = 0.064 J $$

(c) From Eq. (4.30), $ m_s$ = NIA. From part (a), $ m_s = 0.40 A m^2 $
$$ 0.40 = 1000 × I × 2 × 10^{–4} $$
$$ I = 0.40 × 10^4/(1000 × 2) = 2A $$

Example 5.4 What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5.0 cm at a distance of 50 cm from its mid-point? The magnetic moment of the bar magnet is 0.40 A $m^2$, the same as in Example 5.2.

Solution From Eq. (5.7)
$$ B_E = \frac {μ_om}{4 π r^3} = \frac {10^{-7}×0.4}{\big( 0.5 \big)^3 } = \frac {10^{-7}×0.4}{0.125} =3.2×10^{-7} T $$

From Eq. (5.8), $$ B_A = \frac {μ_o2m}{4 π r^3} = 6.4× 106{-7} T $$

Example 5.5 Figure 5.5 shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.

(a) In which configuration the system is not in equilibrium?
(b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium?
(c) Which configuration corresponds to the lowest potential energy among all the configurations shown?

Solution

Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression [Eqs. (5.7) and (5.8)]:

$$ B_P = – \frac { μ_o}{4 π } \frac {m_p}{r^3} $$ ( on the normal bisector )
$$ B_P = \frac { μ_o2}{4 π } \frac {m_p}{r^3} $$ ( on the axis )

Where $ m_P$ is the magnetic moment of the dipole P. Equilibrium is stable when $m_Q$ is parallel to $B_P$, and unstable when it is anti-parallel to $B_P$.

For instance for the configuration $Q_3$ for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence $Q_3$ is stable.

Thus,

(a) $ PQ_1$ and $PQ_2$

(b) (i) $PQ_3$, $PQ_6$ (stable); (ii) $PQ_5$,$ PQ_4$ (unstable)

(c)$ PQ_6$

Questions for section 5.2

1. Explain the properties of magnetic fields of lines.
2. Explain how bar magnet and solenoid are equivalent to each other.
3. Explain the concept of potential energy of a magnetic dipole in a magnetic field.
4. What is electrostatic analogue of magnetism?

5.3 Magnetism and gauss’s law

According to Gauss law,

The net magnetic flux through any closed surface is always zero.

That means the number of magnetic lines leaving the closed surface is equal to the number of magnetic lines entering it.

Let us consider a uniform magnetic field b and let ΔS be a small area element of this surface with n along its normal as shown in the figure. The magnetic flux through this area element is

$$ ΔΦB = \vec B \vec {ΔS} $$ The total flux through the closed surface is the sum of the magnetic flux through all the elements on the closed surface. $$ i.e., \Sigma{all} \space ΔΦB = \Sigma{all} \space \vec B. \vec {ds} $$

According to Gauss’s law in magnetism

$$ ΦB = \Sigma{all} \space \vec B. \vec{ds} = 0 $$

According to Gauss’s law in electronics the total electric flux across a closed surface in an electric field is given by
$$ ΦE = \Sigma{all} \space \vec E . \vec {ds} = \frac {q}{ ε_o} $$

We have $ Φ_E $ only if the charge enclosed by the closed surface is zero ( ie q = 0) or the closed surface encloses electric dipoles . According to Gauss’s law in magnetism B =0 .This implies that the simplest magnetic element is a dipole or a current loop. Isolated magnetic poles or mono poles do not exist.

Example 5.6 Many of the diagrams given in Fig. 5.7 show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.

Solution

(a) Wrong. Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor, as described in Chapter 4.

(b) Wrong. Magnetic field lines (like electric field lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges.

(c) Right. Magnetic lines are completely confined within a toroid. Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field.

(d) Wrong. Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops.

(e) Right. These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the net flux of the field is zero.

(f ) Wrong. These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. The difference between Fig. [5.7(e) and (f)] should be carefully grasped.

(g) Wrong. Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric field lines.

Example 5.7

(a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point?
(b) Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?
(c) If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified?
(d) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire?
(e) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero?

Solution
(a) No. The magnetic force is always normal to B (remember magnetic force = qv × B). It is misleading to call magnetic field lines as lines of force.
(b) (b) If field lines were entirely confined between two ends of a straight solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field B through any closed surface must always be zero. For a toroid, this difficulty is absent because it has no ‘ends’.
(c) (c) Gauss’s law of magnetism states that the flux of B through any
closed surface is always zero $ \oint_s = B.ds = 0 $
If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, $ \int_s= B.ds = μ_oq_m $ , where qm is the (monopole) magnetic charge enclosed by S.]
(d) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straightwire, this force is zero.)
(e) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.

Questions for section 5.3

1. State the Gauss law.
2. Derive an equation for total electric flux across a closed surface in an electric field.

5.4 The earth’s magnetism

• The branch of physics which deals with the study of magnetism of earth is called terrestrial magnetism or geomagnetism.
• The strength of earth’s magnetic field keeps varying from place to place in the order of 10-5 telsa. It extends to several kilometres into the space around the earth.
• The belief is that the magnetic field around the earth is due to the electric currents produced due to convective motions of metallic fluids in the outer core of the earth. This is known as dynamo effect.
• The earth’s magnetic field is like a huge magnetic dipole inclined at 11.30 with respect to axis of rotation of earth.
• The magnetic pole near the geographic North Pole NG is called the north magnetic pole and similarly south magnetic pole.
• In reality, the magnetic pole of earth behaves like south-pole of the bar magnet inside the earth and vice versa.

5.4.1 Magnetic declination and dip

Let us consider a point on the earth the longitude circle determines the geographic north-south direction, the line of longitude towards the North Pole being the direction of true north. The vertical plane containing the longitude circle and the axis of rotation of the earth is called the geographic meridian.

One can define magnetic meridian of a place as the vertical plane which passes through the imaginary line joining the magnetic north and the south poles. This plane would intersect the surface of the earth in a longitude like circle. A magnetic needle, which is free to swing horizontally, would then lie in the magnetic meridian and the north pole of the needle would point towards the magnetic north pole.

Since the line joining the magnetic poles is titled with respect to the geographic axis of the earth, the magnetic meridian at a point makes angle with the geographic meridian. This, then, is the angle between the true geographic north and the north shown by a compass needle. This angle is called the magnetic declination or simply declination.

The declination is greater at higher latitudes and smaller near the equator.

There is one more quantity of interest. If a magnetic needle is perfectly balanced about a horizontal axis so that it can swing in a plane of the magnetic meridian, the needle would make an angle with the horizontal. This is known as the angle of dip (also known as inclination). Thus, dip is the angle that the total magnetic field $B_E$ of the earth makes with the surface of the earth.

Figure shows the magnetic meridian plane at a point P on the surface of the earth. The plane is a section through the earth. The total magnetic field at P can be resolved into a horizontal component HE and a vertical component $Z_E$. The angle that $B_E$ makes with $H_E$ is the angle of dip, I.

In most of the northern hemisphere, the north pole of the dip needle tilts downwards. Likewise in most of the southern hemisphere, the south pole of the dip needle tilts downwards. To describe the magnetic field of the earth at a point on its surface, we need to specify three quantities, viz., the declination D, the angle of dip or the inclination I and the horizontal component of the earth HE. These are known as the element of the earth. Representing the vertical component by $Z_E$, we have

$$Z_E = B_E \space sin I $$

$$ H_E = B_E \space cos I $$

This gives,

$$ tan \space I = \frac {Z_E}{H_E} $$

Example 5.8 The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment.

Solution From Eq. (5.7), the equatorial magnetic field is,

$$ B_E = \frac {μ_om}{4 π r^3} $$
We are given that $B_E ~ 0.4 G = 4 × 10^{–5} T $. For r, we take the radius of the earth $6.4 × 10^6$ m. Hence,
$$ m = \frac {410^{-5}×(6.4×10^6)^3}{ μ_o/4 π} = 4×10^2×(6.4×10^6)^3 $$ ($μ_o/4 π =10^{-7}$)
$$ = 1.05 × 10^{23} A m^2 $$
This is close to the value 8 × 1022 A m2 quoted in geomagnetic texts.

Example 5.9 In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is $60^o$. What is the magnetic field of the earth at this location?

Solution
It is given that $H_E$ = 0.26 G. From Fig. 5.11, we have
$$ cos \space 60^o = \frac {H_E}{B_E} $$
$$ B_E = \frac { H_E}{ cos \space 60^o} $$
$$ = \frac {0.26}{\big( 1/2 \big) } $$

Questions for section 5.4

1. What is dynamo effect?
2. What is magnetic declination?
3. What is dip?
4. What is isocyclic lines and aclinic lines?

5.5 Magnetization and magnetic intensity

The earth abounds with a bewildering variety of elements and compounds. In order to understand these, we have to clearly understand the following terms.

1) Magnetization (M)

Magnetization of a substance is defined as the net magnetic moment per unit volume.

The dipole moment of all the electrons revolving around the nucleus of an atom adds up vertically to give a magnetic moment $m_{net}$ for the material.

$$ M = \frac { m_{net}}{V} = \frac {qm×2l}{A×2l} = \frac {qm}{A} $$

Unit is $Am^{-1}$ and is also called pole strength per unit area.

2) Magnetic permeability

Magnetic Permeability is defined as the ability of the medium to allow the magnetic field lines to pass through it.

3) Relative permeability

The Relative Permeability of a material is defined as the ratio of magnetic flux density in a medium to the magnetic flux density that would be present if the medium was replaced by vacuum.
$$ μ_r = \frac {B}{B_o} $$

4) Magnetic induction or flux density

Magnetic flux density at any point is defined as the density at any point in a medium is the magnetic flux per unit area around the point along the normal.

5) Magnetic intensity (H)

Magnetic intensity is defined as the degree to which a magnetic field can magnetise a substance.

It is also known as Magnetizing force. It mainly depends upon the source of magnetic field.

Let us consider a long solenoid of n-turns per unit length. Let I be the current through it.

The magnetic field inside the solenoid is given by

B = µnI

B = µH

Where H is called magnetic intensity.

Magnitude of magnetic intensity is equal to the number of the ampere turns per unit length of solenoid to produce magnetic field.

If the solenoid is filled with vacuum then,
$$ B = µ_0 \space n \space I $$

6) Magnetic susceptibility (χ)

It represents how easily a substance is magnetized. Experimentally it is observed that

M is directly proportional to H.

$$ χ = \frac {M }{ H} $$

Magnetic susceptibility is defined as the ratio of magnetic induction to the magnetic intensity.

Relation between Magnetic permeability and Magnetic susceptibility

Consider a long solenoid or toroid of n turns per unity per unity length carrying current I .The magnetic field in the interior of the solenoid is given by

$$ B_o = µ_0 \space n I = B = µ_0 H $$

If the solenoid is filled with a material which gets Magnetized , the magnetic field of solenoid is greater than $B_o$ . The net magnetic field B in the solenoid is
$$ B = B_o +B_m , \space where $$
$B_m$ – magnetic field due to magnetization
$$ B_m = µ_0 M $$
$$ B = µ_0 (H+M) = µ_0 ( I+ \frac {M}{H}) = µ_0 \space H (l + χ ) $$
If µ is the permeability of the material inside the solenoid then
$$ B = µ H $$
$$ µ H = µ_0 H ( 1+ χ) \space µ = µ_o (1+ χ) $$

Example 5.10 A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current $I_m$.

Solution
(a) The field H is dependent of the material of the core, and is
H = nI = 1000 × 2.0 =$ 2 ×10^3$ A/m.
(b) The magnetic field B is given by
$$ B = μ_r μ_0 H $$
$$ = 400 × 4π ×10^{–7} (N/A^2) × 2 × 10^3 $$ (A/m)
= 1.0 T
( c ) Magnetisation is given by
$$ M = (B– μ_0 H)/ μ_0 $$
$$= (μ_r μ_0 H–μ_0 H)/μ_0 = (μ_r – 1)H = 399 × H$$
$$ ≅ 8 × 10^5 A/m $$
The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus $ B = μ_r n_0 (I + I_M)$. Using I = 2A, B = 1 T, we get $I_M$ = 794 A.

Questions for section 5.5

1. Define magnetization.
2. Define magnetic permeability.
3. Define relative permeability.
4. Define flux density.
5. What is magnetic intensity? Explain with an equation.
6. Define magnetic susceptibility.
7. Derive a relation between magnetic permeability and susceptibility.

5.6 Magnetic properties of materials

Based on the magnetic properties, faraday classified the substances into the following 3 categories.
1) Diamagnetic substances
2) Paramagnetic substances
3) Ferromagnetic substances

5.6.1 Diamagnetic substances

Diamagnetic substances are those substances which have a tendency to move from stronger to weaker part of the external magnetic field. That is, they are repelled by magnet.
Some of the examples are copper, gold, diamond, hydrogen etc.
Supposing a small piece of diamagnetic material is placed in a non-uniform magnetic field, it has a tendency to move from stronger to the weaker regions of the field. The dipole moment of such an atom is zero.

Fact box When a magnetic field is applied, a small value of magnetic dipole is induced in a direction opposite to the magnetic field. This is the reason for the material being repelled by the magnet. ·       Because the susceptibility of a diamagnetic substance has a small magnetic value, the relative permeability $μ_r$ is less than 1. When the metals are cooled at very low temperatures, they become super-conductors. These super-conductors are perfect examples for perfect diamagnetism. When super-conductors are placed in the magnetic field, they are completely expelled. This phenomenon of perfect magnetism in the super-conductor is called Meissner effect. The susceptibility does not change with temperature.

5.6.2 Paramagnetic substances

Paramagnetic substances are those which get weakly magnetised when placed in an external magnetic field. They have a tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get weakly attracted to a magnet.

Examples are Aluminium, Oxygen, Lithium, Platinum etc.

• When a paramagnetic substance is placed in an external magnetic field, the lines get concentrated as shown in the figure and the field inside the material is slightly increased.
• The atoms of a paramagnetic substance possess a small value of permanent dipole moment.
• When such a material is placed in a strong magnetic field, the atomic dipole moment aligns parallel to the direction of external magnetic field. So, it is weakly magnetised in the direction of magnetic field.
• In non-uniform magnetic field, it tends to move from weaker to stronger regions of the field.
• The susceptibility has a small value, so the relative permeability is slightly greater than 1.
• The susceptibility of substances is inversely proportional to the absolute temperature. This is called Curie’s law.

5.6.3 Ferromagnetic substances

Ferromagnetic substances are those which gets strongly magnetised when placed in an external magnetic field. They have strong tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get strongly attracted to a magnet.

• The individual atoms which possess permanent magnetic dipole moment interact with each-other in a common direction called domain.
• Each domain contains about $ 10^{11}$ atoms and has a net magnetization.
• The net magnetic moment of the assembly is zero.
• When the external magnetic field is applied, the domains orient themselves along the direction of the magnetic field and grow in size at the expense of the domain. The number of domains pointing in the direction of the magnetic field increases and the ferromagnetic substance gets fully magnetized.

Fact box When a sample of ferromagnetic material is placed in an external magnetic field, the lines are highly concentrated inside the material. Hence, it is strongly magnetized.

Hard magnetic materials

When the magnetization persists in the ferromagnetic materials when the external magnetic field is removed, such materials are called Hard magnetic materials.

Example: Alnico

Soft magnetic materials

When the magnetization disappears if the external magnetic field is removed, such materials are called soft magnetic materials.
Example: cobalt, iron

• The susceptibility of the ferromagnetic material has a large positive value and the relative permeability is very high. This is the reason for them being strongly magnetized.
• The properties depend on temperature.
  Rise in temperature leads to decrease in susceptibility.
  At very high temperature, it becomes a paramagnet.
• The temperature at which a ferromagnetic substance becomes a paramagnet is called Curie temperature.

Definition box The phenomenon of lagging behind of intensity of magnetization with respect to the magnetizing field strength when a specimen of a ferromagnetic material is subjected to a cycle of magnetization is called Hysteresis.

The relation between B and H in ferromagnetic materials is complex. It is often not linear and it depends on the magnetic history of the sample. Figure depicts the behaviour of the material as we take it through one cycle of magnetisation.

Let the material be un-magnetised initially. We place it in a solenoid and increase the current through the solenoid. The magnetic field B in the material rises and saturates as depicted in the curve Oa. This behaviour represents the alignment and merger of domains until no further enhancement is possible. It is pointless to increase the current beyond this.

Next, we decrease H and reduce it to zero. At H = 0, B „ 0. This is represented by the curve ab. The value of B at H = 0 is called retentivity or remanence. In figure, BR ~ 1.2 T, where the subscript R denotes retentivity. The domains are not completely randomised even though the external driving field has been removed.

Next, the current in the solenoid is reversed and slowly increased. Certain domains are flipped until the net field inside stands nullified. This is represented by the curve bc. The value of H at c is called coercivity. In figure Hc ~ . As the reversed current is increased in magnitude, we once again obtain saturation. The curve cd depicts this. The saturated magnetic field Bs ~ 1.5 T.

Next, the current is reduced (curve de) and reversed (curve ea). The cycle repeats itself. Note that the curve Oa does not retrace itself as H is reduced. For a given value of H, B is not unique but depends on previous history of the sample. This phenomenon is called hysterisis. The word hysterisis means lagging behind.

Example 5.11 A domain in ferromagnetic iron is in the form of a cube of side length 1μm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is $7.9 g/cm^3$. Assume that each iron atom has a dipole moment of $ 9.27×106{–24} A $m^2$.

Solution The volume of the cubic domain is $V = (10^{–6} m)^3$ = $10^{–18} m^3$ = $ 10^{–12} cm^3$ .Its mass is volume × density = $7.9 g cm^{–3} × 10^{–12} cm^3 $=$ 7.9 × 10^{–12} $ g .It is given that Avagadro number (6.023 × $10^{23}$) of iron atoms have a mass of 55 g. Hence, the number of atoms in the domain is
$$ N = \frac { 7.9×10^{-12}×6.023 × 10^{23}}{55} $$
$$ = 8.65 × 10^{10} $$ atoms

The maximum possible dipole moment $m_{max}$ is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned.
Thus,
$$ m {max} =( 8.65×10^{10}×(9.27×10^{-24}) $$ $$ = 8.0 × 10^{–13} A m^2 $$ The consequent magnetisation is $$ M{max }= m_{max}/Domain volume $$
$$ = 8.0 × 10^{–13} Am^2/10^{–18} m^3 $4
$$ = 8.0 × 10^5 Am^{–1} $$

Questions for section 5.6

1. With a neat diagram, explain the properties and nature of diamagnetic substances.
2. With a neat diagram, explain the properties and nature of paramagnetic substances.
3. With a neat diagram, explain the properties and nature of ferromagnetic substances.
4. What are hard and soft magnetic materials?
5. Explain hysteresis with a diagram.

5.7 Permanent magnets and electromagnets

Definition box Substances which retrain their ferromagnetic property for a long period of time at room temperatures are called Permanent Magnets.

The permanent magnets have the following properties.

1. High retentivity so that the magnet is strong.
2. High coercivity so that it is not easily demagnetized.
3. High permeability so that it can be magnetized easily.

Example: steel alnico, ticonal

Note box: An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass current through the solenoid. The magnetic field of the solenoid magnetizes the rod. Soft iron is most suitable to make magnets. As placed in the set up as shown in the figure, when the current is switched off, the magnetism is also switched off and soft iron has low retentivity.

The electromagnets must possess the following properties
1) Low retentivity so that the magnet is strong.
2) Low coercivity so that it is not easily demagnetized.
3) High permeability so that it can be magnetized easily.

Uses

• Electric bells
• Loud speakers
• Telephone diaphragms
• Giant electromagnets in cranes to lift iron and steel.

Questions for section 5.7

1. What are permanent magnets?
2. Explain with a setup, how soft iron can be made into a magnet.
3. What are the required properties of material of a permanent magnet?
4. What are the required properties of electromagnets?
5. State the uses of permanent and electromagnets.