(i) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
(ii) A galvanometer is converted into a volmeter by connecting high resistance R_s in series with the galvanometer.

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 helix around the field lines.

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

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

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's force. This force is known as magnetic Lorentz force
\(\vec { F } =q(\vec { v } \times \vec { B } )\)
In general, if the charge is moving in both the electric and magnetic fields, the total force experienced by the charge is given by \(\vec { F } =q(\vec { v } \times \vec { B } )q\vec { E } \) . It is known as Lorentz force.

If 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 wire points in the direction of the magnetic field lines produced.

The magnetic flux density can also be defined as the number of magnetic field lines crossing unit area kept normal to the direction of line of force. 
Its unit is Wb m^-2 or telsa.

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.

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 field B = 0.2 T
Current flowing through the coil I = 3A
Cross sectional area of a circular coil A =  0.1 cm²
∴ A = 0.1 x 10^-4 m²
(a) Total torque on the coil:
\(\tau=\text { IBA } \cos \theta\)
Here, the plane of the loop is perpendicular to the direction of magnetic field
\(\therefore \theta=90^{\circ} \)
\(\therefore \tau=\text { IBA } \cos 90^{\circ} \)    (∵ cos90^o = 0)
= 3 x 0.2 x 0.1 x 10^-4 x 0
= zero
∴ Total Torque on the coil = zero
(b) Total force on the coil \(F=B q v \sin \theta \)
\(Here\ \theta=0 \)
\(\therefore F=0.2 \times \mathrm{qv} \times \sin \theta=\text { zero } \)
Total force on the coil = zero
c) Average force: F = Bqv
Charge density \(=\sigma=\frac{q}{A} \)
∴ Charge  \(q =\sigma A \)
\(=10^{28} \times 0.1 \times 10^{-4} \)
\(q=10^{23} \mathrm{C} \)
Force = Bll and I = 1 m
F = 0.2 x 3 x 1 = 0.6 N
∴ Average force on each electron
\(=\frac{F}{q}=\frac{0.6}{10^{23}} \)
\(F =0.6 \times 10^{-23} \mathrm{~N} \)

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

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

Velocity of the particle is \(\vec { v } =v\hat { j } \)
(a) Magnetic field is along positive y-direction, this implies \(\vec B=B\hat { j } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { j } )=\vec 0\)
So, no force acts on the particle when it moves along the direction of magnetic field.
(b) Since the magnetic field points in positive z - direction, this implies, \(\vec { B } =B\hat { k } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { k } )=qvB\vec i \)
Therefore, the magnitude of the Lorentz force is qvB and direction is along positive x - direction.
(c) Magnetic field is in zy - plane and making an angle θ with the velocity of the particle, which implies \( {\vec B } =Bcos\theta \hat { j } +Bsin\theta \hat { k } \)

From Lorentz force,
\({ \vec F }_{ m }=q(v\hat { j } )\times (Bcos\theta \hat { j } +Bsin\theta \hat k)\) 
\(=qvBsin\theta \hat { i } \)

The total flux emanating from the closed surface S enclosing the dipole is zero. So,
\({ \Phi }_{ B }=\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero,
\(\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
This is similar to Gauss’s law in electrostatics.

 (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, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\) 

The magnetic field at the center of a circular coil
B = \(\frac { { \mu }_{ 0 }NI }{ 2R } \)
When current I is doubled and radius R is halved,
B' = \(\frac { { \mu }_{ 0 }N\times 2I }{ 2\left( \frac { R }{ 2 } \right) } \)  = 4B
∴ magnetic field becomes four times the original field.

w = \(\frac { qB }{ m } \) independence of angle of the entrance with the magnetic field.
W₁ : w₂ = 1 : 1

If the magnetic force \(\vec { F } =q(\vec { v } \times \vec { B } )\)
(i) The force on a charged particles is perpendicular to both \(\vec { v } \) and magnetic field \(\vec { B } \).
(ii) No, This is since the charged particle moves on a circular path.
\(\vec { F } =q(\vec { v } \times \vec { B } )\)
The particle does not gain energy.

(i) Horizontal component:
The Earth's magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of the magnetic compass rests horizontally at an angle of dip, I = 0⁰ as shown in the figure.
B_H = B_E
B_v = 0
This implies that the horizontal component is maximum at the equator and the vertical component is zero at the equator.

(ii) Vertical component: The Earth's magnetic field is perpendicular to the surface of the Earth (i.e., vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, I = 90° as shown in Figure
Hence,
B_H = 0
B_v = B_E
This implies that the vertical component is maximum at poles and the horizontal component is zero at poles.

M = 100 g = 100 x 10^-3 kg
M = 0.1 kg
g = 10 m/s²
N = 5
B = 0.5 T
R = 20 cm = 20 x 10^-2 m
When, the sphere is in translational equilibrium
\(\mathrm{f}_{\mathrm{s}}-\mathrm{Mg} \sin \theta =0 \ldots(1) \)
\(\mathrm{f}_{\mathrm{s}} =\mathrm{Mg} \sin \theta \)
When, the sphere is in rotational equilibrium,
Torque ፒ = μB sin θ
For Equilibrium, f_sR - ፒ = 0
\(f_{s} R-\mu B \sin \theta=0 \ldots(2)\)
eqn (1) subs. eqn (2)

\(M g \sin \theta R-\mu B \sin \theta =0 \)
\(\sin \theta(M g R-\mu B) =0 \)
\(M g R =\mu B \)
\(\text { Now } \mu =N I \pi R^{2} \)
\(\therefore \operatorname{MgR} =N I \pi R^{2} B \)
\(\therefore I =\frac{M g}{N \pi R B} \)
\(I =\frac{0.1 \times 10}{5 \times \pi \times 20 \times 10^{-2}\times 0.5 }\)
\(=\frac{1}{\pi \times 50 \times 10^{-2}}=\frac{1 \times 10^{2}}{\pi \times 50}=\frac{100}{\pi \times 50}\)
\(\mathbf{I} =2 / \pi \mathrm{A} \)

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T
Deflection is 
\(\theta =60°\Rightarrow tan60°=\sqrt { 3 } =1.732\)
\(I=\frac { 2R{ B }_{ H } }{ { \mu }_{ ° }N } tan\theta \)
\(=\frac { 2\times 0.12\times 25\times 1{ 0 }^{ -6 } }{ 4\times 1{ 0 }^{ -7 }\times 3.14\times 100 } \times 1.732=0.82\times 1{ 0 }^{ -1 }A\)
I = 0.082 A

Let  \(\vec{p}_m\)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, the magnetic dipole orient parallel (θ = 0^o) to the magnetic field with energy,
U_parallel = U_minimum = -p_mBcos 0
U_parallel = -p_mB
since cos 0^o = 1
Otherwise, the magnetic dipole orients anti-parallel (θ = 180^o) to the magnetic field with energy,
U _{anti-parallel} = U _maximum = -p_mBcos 180
\(\Rightarrow \) U_{anti-parallel} = P_mB
since cos 180^o = -1

Magnetic flux density = \(\frac{\mu_{o}I}{2a}\)
\(=\frac{4 \pi \times 10^{-7} \times 5}{2 \times 20 \times 10^{-2}}\)
= 1.57 x 10^-5 T

\(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T\)
Using an equation,
Lorentz force, \(\vec{F}=Il\hat{j}\times\vec B\)
We can get,
Lorentz force \(F=\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)

Lorentz force F = Bqv
Energy w = qV
Energy is equal to kinetic energy,
\(qV=\frac{1}{2}mv^2\)
\(v=\sqrt { \frac {2qV }{ m } } \)
\(\therefore Lorentz \ force \ F= Bq\times \sqrt \frac{2qV}{m}=\sqrt { \frac { 2{ B }^{ 2 }{ q }^{ 3 }V }{ m } } \)

B = 60G,   A = 2 x 10^-4m²
N = 40,     I = 4mA = 4 x 10^-3A
θ = 16⁰
I =\(\frac { K\theta }{ BAN } \)
K = \(\frac { BANI }{ \theta } \)
= \(\frac { 40\times 60\times 2\times 10^{ -4 }\times 4\times 10^{ -3 } }{ 16 } \)
= 1.2 x 10^-4 Nm/degree

(i) Consider a charged particle of charge 'q' having mass m enters into a region of a uniform magnetic field \(\vec { B } \) with velocity \(\vec { v } \).
(ii) Such that velocity is perpendicular to the magnetic field and velocity \(\vec{v}\).
(iii) The charged particle moves in a circular orbit.
(iv) Lorentz force.

\(\vec { F } =q(\vec { v } \times \vec { B } )\)
In magnitude F = qVB
(v) This Lorentz force acts as centripetal force for the particle to execute circular motion. Therefore,
qvB = m\(\frac { { v }^{ 2 } }{ r } \)
The radius of the circular path is
r = \(\frac { mv }{ qB } =\frac { p }{ qB } \)               ..........(1)
(vi) where p = mv is the magnitude of the linear momentum of the particle. Let T be the time taken by the particle to finish one complete circular motion, then
T = \(\frac { 2\pi r }{ v } \)                  .............(2)
Hence substituting (1) in (2), we get,
T = \(\frac { 2\pi m }{ qB } \)        .............(3)
(vii) Equation (3) is called the cyclotron period. The reciprocal of time period is the frequency f, which is
f = \(\frac { 1 }{ T } \)
f = \(\frac { qB }{ 2\pi m } \)                  ...........(4)
In terms of angular frequency ω,
ω = 2πf = \(\frac { q }{ m } \)B          ...........(5)
(viii) Equations (4) and equation (5) are called cyclotron frequency or gyrofrequency.
(ix) Time period and frequency depend only on charge-to-mass ratio (specific charge) and independent of velocity or radius.

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.

Consider a rectangular loop abcd. Then from Ampere's circuital law,
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } \) = μ₀ I_enclosed = μ₀ x (total current enclosed by Amperian loop)
The left hand side of the equation is
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } =\int _{ a }^{ b }{ \vec { B } .\vec { dl } } +\int _{ b }^{ c }{ \vec { B } .\vec { dl } } +\int _{ c }^{ d }{ \vec { B } .\vec { dl } } +\int _{ d }^{ a }{ \vec { B } .\vec { dl } } \)
Elemental lengths bc and da are perpendicular to magnetic field.
\(\therefore \int _{ b }^{ c }{ \vec { B } .\vec { dl } } =\int _{ b }^{ c }{ |\vec { B } ||\vec { dl } | } cos 90^o = 0\)
similarly,
\(\int _{ d }^{ a }{ \vec { B } .\vec { dl } } = 0\) 
Since the magnetic field outside the solenoid is zero,
\(\int _{ c }^{ d }{ \vec { B } .\vec { dl } } =0\) and
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } } =BL \quad \quad (\because \theta =0^o)\)
Let I be the current passing through the solenoid of N turns, then
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } }  = BL = μ_o NI ⇒ B = μ_o\frac {NI}{L}\)
The number of turns per unit length is given by \(\frac { N }{ L } \) = n, Then
B = \(\mu_o \frac { nLI }{ L } \) = μ₀nI
Since n is a constant for a given solenoid and μ₀ is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

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.