It is an arrangement of eletric field (E) and magnetic field (B) perpendicular to each other. When charged particles enter that region, particles with a certain velocity can pass through that region.
v = E/B
The speed is independent of charge and mass.

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

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.

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.

Magnetic field is the region or space around every magnet within which its influence will be felt by keeping another magnet in that region.
\(\vec { B } =\frac{1}{q_m}\vec{E}\)
Its unit is N A^-1m^-1

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
\(=\frac{F}{q}=\frac{0.6}{10^{23}} \)
\(F =0.6 \times 10^{-23} \mathrm{~N} \)

Ampere's circuital law states that the line integral of magnetic field over a closed loop is μ₀, times net current enclosed by the loop.
\(\oint _{ c }^{ }{ \vec { B } \vec { dl } } \) = μ₀I _enclosed.

Biot-Savart's law states that, the magnitude of magnetic field \(d\vec { B } \) at a point P at a distance of r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length of element \(\vec { dl } \)
(iii) directly as the sine of the angle θ between \(\vec { dl } \) and \(\hat { r } \).
(iv) inversely as the square of the distance r between the point P and length of element \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)

(i) When a substance is placed in a uniform magnetising field, the substance gets magnetised.
(ii) The total magnetic field inside the specimen is equal to the sum of the magnetic field produced in vacuum due to the magnetising field and the magnetic field due to the induced magnetism of the substance.

The magnitude of the force between two poles is given by
\( F =k\frac { { q }_{ m_A }{ q }_{ { m }_{ B } } }{ { r }^{ 2 } } \)
(Given : F = 9 × 10^–3 N, r = 10 cm = 10 × 10^–2 m
Since q_mA = q_mB = q_m, we have
9 x 10^-3 = 10^-7 x \(\frac { { q }_{ m }^{ 2 } }{ { \left( 10\times { 10 }^{ -2 } \right) }^{ 2 } } \Rightarrow { q }_{ m }\) = 30NT^-1 

B_H = 0.15 G and B_V = 0.26 G
tan I = \(\frac { 0.26 }{ 0.15 } \Rightarrow I=ta{ n }^{ -1 }(1.732)=60°\)
The resultant magnetic field of the Earth is
\(B=\sqrt { { B }_{ H }^{ 2 }+{ B }_{ V }^{ 2 } } =0.3G\)

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

Magnetic field lines can be entirely confined within the core of a toroid since the toroid has no ends. θ solenoid is open ended and the field lines inside it which are parallel to the length of the solenoid cannot form closed curves inside the solenoid.

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) Magnetic field lines are continuous closed curves. The direction of magnetic field lines is from the North pole to the South pole outside the magnet and South pole to the North pole inside the magnet.
(ii) The direction of a magnetic field at any point on the curve is known by drawing a tangent to the magnetic line of force at that point.
(iii) Magnetic field lines never intersect each other.
(iv) 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 thin out.

(i) Obey inverse square law, so they are long-range fields.
(ii) obey the principle of superposition and are linear with respect to source in magnitude,
E ∝ q
B ∝ Idl

Magnetic field \(\overset { \rightarrow }{ B } ={ 0.500\hat { i } T } \)
Velocity of the particle
\(\hat { v } \) = (1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) ms^-1
Charge of the proton q = 1.67 x 10^-19 C
Mass of the proton m = 1.67 x 10^-27kg
(a) The force experienced by the proton is \(\overset { \rightarrow }{ F } \) = q(\(\overset { \rightarrow }{ v } \) x \(\overset { \rightarrow }{ B } \) )
= 160 x 10^-19 x  ((1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) x (0.500 \(\hat { i } \)))
\(\overset { \rightarrow }{ F } \)= 1.60 x 10^-14 \(\hat { j } \)  N
Therefore, from Newton’s second law,
\(\overset { \rightarrow }{ a } =\frac { 1 }{ m } \overset { \rightarrow }{ F } =\frac { 1 }{ 1.67\times { 10 }^{ -27 } } (1.60\times { 10 }^{ -14 })\hat j\)
\(=9.58\times { 10 }^{ 12 }\hat jm{ s }^{ -2 }\)
(b) Trajectory is helical Radius of helical path is
\(R=\frac { { mv }_{ z } }{ \left| q \right| B } =\frac { 1.67\times { 10 }^{ -27 }\times 2.00\times { 10 }^{ 5 } }{ 1.60\times { 10 }^{ -19 }\times 0.500 } \)
= 4.175 x 10^-3m = 4.18mm
Pitch of the helix is the distance travelled along x-axis in a time T, which is P = v_x T
But time, \(T=\frac { 2\pi }{ \omega } =\frac { 2\pi m }{ \left| q \right| B } =\frac { 2\times 3.14\times 1.67\times { 10 }^{ -27 } }{ 1.60\times 1{ 0 }^{ -19 }\times 0.500 } =13.1\times { 10 }^{ -8 }s\)
Hence, pitch of the helix is
\(p={ v }_{ x }T=(1.95\times { 10 }^{ 5 })(13.1\times { 10 }^{ -8 })=25.5\times { 10 }^{ -3 }m=25.5mm\)
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.

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 = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

\(tan \ I=\frac{B_V}{B_H}=1\)
∴ I = 45^o

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

Electric field between the plates \(= \frac{σ}{ε_0}\)
Electric force on an electron \(= e\frac{σ}{ε_0}\)
Magnetic force on an electron, F = BIl
But, \(I= \frac{e}{t}\)
∵Electron moves in a straight line. So,
E_F = M_F
\(e\frac{σ}{ε_0}=B(\frac{e}{t})l\)
\(\therefore t = ε_0\frac{lB}{σ}\)

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

Number of turns, N = 20
Radius of circular coil, r = 10 cm = 0.1 m
Magnetic field, B = 0.1T
Angle between area vector and magnetic field, θ = 0°
Current, I = 5.0A       (A = πr² = Area)
(i) Torque on the coil, t = NABI sinθ)
= 20 x 5 x  π(0.1)² x (0.1)² sin0°
= 0
(ii) The forces on the planar loop are in pairs
i.e., F₁ = -F₂ and F₃ = -F₄
∵ they are equal and opposite to each other, the total force on the coil is zero.
(iii) Number density of electrons, n = 10²⁹/m³
Area, A,= 10^-5 m²
Magnitude of force, F = e (v_d x B)
F = BeV_d sinθ ⇒ I = nAe v_d
v_d =\(\frac { I }{ nAe } \)
F = e\(\left( \frac { I }{ nAe } \right) \)Bsin90⁰
= \(\frac { IBsin{ 90 }^{ 0 } }{ nA } \)
F = \(\frac { 0.1\times 5\times 1 }{ 10^{ -5 }\times { 10 }^{ 29 } } \)  = 5 x 10^-25N

l = 0.2m
N = 120, I = 25A
Magnetic field in the interior of the solenoid,
B = μ₀nI = μ₀\(\left( \frac { N }{ l } \right) \)I
= 4π x 10^-7 x \(\frac { 120\times 2.5 }{ 0.2 } \)
= 1.85 x 10^-3T

Two coils are lying in perpendicular planes and having a common centre.
The current carrying by P and Q is I and.\(\sqrt{2}\) I
The magnetic field at the centre of P, due to its current

\(\vec { { B }_{ Q } } =\frac { { \mu }_{ 0 }I }{ 2R } \)
The magnetic field at the centre of Q, due to its current \(\sqrt{2}\) I.
\(\vec { { B }_{ Q } } =\frac { { \mu }_{ 0 }\sqrt { 2 } I }{ 2R } \)
∴ \(\vec { { B }_{ net } } =\vec { B_{ p } } +\vec { B_{ Q } } \)
=\(\frac { { \mu }_{ 0 }I }{ 2R } +\frac { { \mu }_{ 0 }\sqrt { 2 } I }{ 2R } \)
\(|\vec { { B }_{ net } } |=\sqrt { \left( \frac { { \mu }_{ 0 }I }{ 2R } \right) ^{ 2 }+\left( \frac { { \mu }_{ 0 }\sqrt { 2 } I }{ 2R } \right) ^{ 2 } } \)
=\(\frac { { \mu }_{ 0 }I }{ 2R } \times 2=\frac { { \mu }_{ 0 }I }{ R } \)
∴ tanθ =\(\frac { |\vec { { B }_{ p } } | }{ |\vec { { B }_{ Q } } | } =\frac { \frac { { \mu }_{ 0 }I }{ R } }{ \frac { { \mu }_{ 0 }\sqrt { 2 } I }{ 2R } } \)
tanθ = \(\frac { 1 }{ \sqrt { 2 } } \)
θ₂ tan^-1\(\left( \frac { 1 }{ \sqrt { 2 } } \right) \)
θ = 31⁰.

(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.

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.

(i) 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.
(ii) 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.
(iii) Under the action of two magnetic fields, the needle comes to rest making angle with B_H , such that
B = B_H  tan \(\theta\)    .........(1)
Where B ⇒ magnetic field produced by current
B_H ⇒ horizontal component of earth's magnetic field
Construction:
(i) Copper coil of wire wound on a non-magnetic circular frame such as brass or wood. Compass box is kept at centre.
(ii) This compass box consists of pivoted magnet and aluminum pointer.
(iii) This compass box is having circular scale graduated with four quadrants.
Working:
(i) Two magnetic fields are perpendicular to each other.
(ii) Magnetic induction due to the current in the coil acting to normal to the plane of the coil.
(iii) Magnetic induction at the centre of the coil,
\(B=μ_o\frac{NI}{2R}\)   .....(2)
Sub. eqn.(1) in eqn. (2)
\(B_H tan \theta =μ_o\frac{NI}{2R}\)
\(B_H =μ_o\frac{NI}{2R}\frac{1}{tan\theta}\)

Let YY' be an infinitely long straight conductor carry current I. 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).
According to Biot Savart law, the magnetic field at a point P due to current element Idl is,
\({ d \vec B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { Idl sin \theta} }{ { r }^{ 2 } }\hat n \).
To apply trigonometry, draw a perpendicular AC to the line BP as shown in Figure.
In triangle ΔABC, \(\sin \theta=\frac{\mathrm{AC}}{\mathrm{AB}}\)
∴ AC = AB sinθ
\(\text { But, } A B  =d l \Rightarrow A C=d l \sin \theta\)
Let dΦ be the angle subtended between AP and BP
ie., \(\angle \mathrm{APB}=\angle \mathrm{APC}=d \phi\)
In a triangle \(\triangle \mathrm{APC}, \sin (d \phi) \simeq A C / A P\)
Since, dΦ is very small, \(\sin (d \phi) \simeq d \phi\)
But,  \(\mathrm{AP}  =r \Rightarrow A C=r d \phi \)
\(\therefore \mathrm{AC} =d l \sin \theta=r d \phi \)
\(\therefore d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{r^2}(r d \phi) \hat{n}=\frac{\mu_0}{4 \pi} \frac{I d \phi}{r} \hat{n}\)
Let Φ be the angle between AP and OP
\(\text {In a } \triangle \mathrm{OPA}, \cos \phi =\frac{\mathrm{OP}}{\mathrm{AP}}=\frac{\mathrm{a}}{\mathrm{r}}  \)
\(r  =\frac{a}{\cos \phi}  \)
\(\text {Now, } d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{a / \cos \phi} d \phi . \hat{n}  \)
\(d \vec{B}  =\frac{\mu_0 I}{4 \pi a} \cos \phi d \phi \hat{n}\)
The total magnetic field at P due to the conductor YY' is
\(\vec { B } = \int _{- \Phi _{ 1 } }^{ { \Phi }_{ 2} }d\vec B =\int _{ -\Phi _{ 1 } }^{ { \Phi }_{ 2 } }\frac { { \mu }_{ 0 }I }{ 4\pi a }{ cos\phi d\phi } \hat { n }\)
\(=\frac { { \mu }_{ 0 }I }{ 4\pi a }[{ sin\phi ]^{\phi_2} _{\phi_-1}} \hat { n }\)
\( \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } (sin{ \Phi }_{ 1 }+sin{ \Phi }_{ 2 })\hat { n } \)
For infinitely long conductor, Φ₁ =  Φ₂ = 90^o
\(\therefore \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } \times 2\hat{n}\Rightarrow\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi a } \hat { n } \)