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. Therefore, the study of hysteresis loop will help us in selecting proper and suitable material for a given purpose.

The branch of physics which deals with the Earth's magnetic field is called Geomagnetism or Terrestrial 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.

Magnetic susceptibility is defined as the ratio of the intensity of magnetisation (\(\vec { M } \)) induced in the material due to the magnetising field |\(\vec H\)|.
\( \chi _{ m }=\frac { |\vec { M } | }{ |\vec { H } | } \).

Magnetic dipote moment \(\vec { p }_{ m } \) is defined as the product of its pole strength and magnetic length. It is a vector quantity denoted by \(\vec { p }_{ m } \)
\(\vec { p }_{ m } ={ q }_{ m }\vec { d } \)

Electric field, E = 6.0 x 10⁶ N C^-1 and magnetic field, B = 0.83 T.
Then.
\(v=\frac { E }{ B } =\frac { { 6.0\times 10 }^{ 6 } }{ 0.83 } =7.23\times { 10 }^{ 6 }{ ms }^{ -1 }\)
When an electron goes with this velocity, it shows null deflection. Since the accelerating potential is 200 V, the electron acquires kinetic energy because of this accelerating potential. Hence,
\(\frac { 1 }{ 2 } mv^{ 2 }=eV \)
\(v=\sqrt { \frac { 2eV }{ m } }\)
Since the mass of the electron, m = 9.1 x 10⁻³¹kg and charge of an electron, \(\left| q \right| =e=1.6\times { 10 }^{ -19 }C.\) The velocity acquired by the electron due to accelerating potential 200 V is
\({ v }_{ 200 }=\sqrt { \frac { 2\left( 1.6\times { 10 }^{ -19 } \right) \left( 200 \right) }{ \left( 9.1\times { 10 }^{ -31 } \right) } } =8.39\times { 10 }^{ 6 }m{ s }^{ -1 }\)
Since the speed v₂₀₀ > v, the electron is deflected towards direction of Lorentz force. So, in order to have null deflection, the potential, we have to supply is
\(v=\frac { { 1mv }^{ 2 } }{ 2\quad e } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 7.23\times { 10 }^{ 6 } \right) ^{ 2 } }{ 2\times \left( 1.6\times { 10 }^{ -19 } \right) } \)
V = 148.65 V

The susceptibility of material X is
X_m,x = \(\frac { \left| \overset { \rightarrow }{ M } \right| }{ \left| \overset { \rightarrow }{ H } \right| } =\frac { 500 }{ 1000 } =0.5\)
The susceptibility of material Y is
X_{m,y =} \(\frac { \left| \overset { \rightarrow }{ M } \right| }{ \left| \overset { \rightarrow }{ H } \right| } =\frac { 2000 }{ 1000 } =2\)
Since, susceptibility of material Y is greater than that of material X, material Y can be easily magnetized than X.

 (a) 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 \({ q }_{ m }^{ ' }=\frac { { q }_{ m } }{ 2 } \) but magnetic length does not change. So, the magnetic moment is
\({ p }_{ m }^{ ' }={ q' }_{ m }2l\)
\({ p }_{ m }^{ ' }=\frac { { q }_{ m } }{ 2 } 2l=\frac { 1 }{ 2 } ({ q }_{ m }2l)=\frac { 1 }{ 2 }p_ m\)
In vector notation, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\)
(b) a 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
\({ p }_{ m }^{ ' }={ q }_{ m }\times \frac { 1 }{ 2 } (2l)=\frac { 1 }{ 2 } ({ q }_{ m }.2l)=\frac { 1 }{ 2 } { p }_{ m }\) 
In vector notation, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\) 

Magnetic moment μ = 0.6 J/T
Angle of inclination with magnetic field θ = 30°
Torque acts on bar magnet ፒ = 0.051 J
Magnitude of magnetic field B = ?
ፒ = \(\vec { \mu } \times \vec { B } \) = μB sinθ
B = \(\frac { \tau }{ \mu sin\theta } =\frac { 0.051 }{ 0.6\times sin{ 30 }^{ 0 } } \)
= \(\frac { 0.051 }{ 0.6\times \frac { 1 }{ 2 } } =\frac { 2\times 0.051 }{ 0.6 } =\frac { 0.102 }{ 0.6 } \)
B = 0.17T
(ii) The position of maximum energy corresponds to a position of stable equilibrium.
The energy (v) = -mB cosθ
when θ = 0°, v = -μB = minimum energy Hence when the bar magnet is placed parallel to the magnetic field, it is the state of stable equilibrium.

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. 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 be 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. 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 parallel to the magnetic field and experience force in opposite directions. Due to this, torque is produced.

For single turn, the deflection torque is,
て = bF = bBIl = (lb)BI
て = ABI
since, area of the coil A = lb
For coil with N turns, we get
て = NABI               ........(1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the moment of restoring couple is proportional to the amount of twist θ. Thus
て = Kθ      ............(2)
where K is the restoring couple per unit twist or torsional constant of the spring.
At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equations (1) and (2), we get,
NABI = Kθ
⇒ I =\(\frac { K }{ NAB } \) θ      ...........(3)
(or) I = Gθ
where G = \(\frac { K }{ NAB } \) is called galvanometer constant or current reduction factor of the galvanometer.
Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galvanometer we use are pointer type moving coil galvanometer.

In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

Dipole moment, \(\vec{p}_m=n\times I\times\vec{A}\)
\(\vec{p}_m\) = 50 x 3 x 3.14 x 25 x 10^-4 ≈ 1.2 A m²

Magnetic filed at the centre of a circular
loop, B = \(\frac{μ_oI}{2\pi R}\)
From the figure, R =\(\frac{2r}{\pi}\)
\(\therefore B'=\frac{μ_oI}{2\pi \times\frac{2r}{\pi}}=\frac{μ_oI}{4r}\)
\(B'=\frac { { \mu }_{ ° }I }{ 4r } \bigotimes \)

Right-hand thumb rule is used :
This rule is used to determine the direction of the magnetic field. If we rotate aright-handed screw using a screwdriver, then the direction of current is the same as the direction in which the screw advances and the direction of rotation of the screw gives the direction of the magnetic field.

When an electric charge q is moving with velocity \(\vec { v } \) in the magnetic field \(\vec { B } \), it experiences a force, called magnetic force \(\vec { { F }_{ m } } \). After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field \(\vec { { F }_{ m } } \).
\(\vec { { F }_{ m } } =q(\vec { v } \times \vec { B } )\)             ...........(1)
In magnitude, F_m = qvB sinθ         .......(2)
The equations (1) and equation (2) imply
(i) \(\vec { { F }_{ m } } \) is directly proportional to the magnetic field \(\vec { B } \).
(ii) \(\vec { { F }_{ m } } \) is directly proportional to the velocity \(\vec { v } \).
(iii) \(\vec { { F }_{ m } } \) is directly proportional to sine of the angle between the velocity and magnetic field.
(iv) \(\vec { { F }_{ m } } \) is directly proportional to the magnitude of the charge q.
(v) The-direction of \(\vec { { F }_{ m } } \) is always perpendicular to \(\vec { v } \) and B as \(\vec { { F }_{ m } } \) in the cross product of \(\vec { v } \) and \(\vec { B } \).

(vi) The direction of \(\vec { { F }_{ m } } \) on a negative charge is opposite to the direction of \(\vec { { F }_{ m } } \) on positive charge provided other factors are identified as shown in Figure.
(vii) If the velocity \(\vec { v } \) of the charge, q is along the magnetic field \(\vec { B } \) then, \(\vec { { F }_{ m } } \) is zero.