This document discusses electricity and defines key concepts related to electric current. It defines current as the rate of flow of electric charge and gives its SI unit as the ampere. It describes conventional current as the flow of positive charges and electric current as the flow of negative charges. It also discusses different types of current sources and the effects of electric current, including heating, chemical, and magnetic effects.

1. GROUP - 6

2.  A form of energy resulting from the existence of
charged particles (such as electrons or protons), either
statically as an accumulation of charge or dynamically
as a current.
OR
 The set of physical phenomena associated with the
presence and flow of electric charge.

3.  Time rate of flow of charge through any cross section
of a conductor is called current
 Mathematically, it can be expressed as
I=Δq
Δt
 Where, Δq is the charge passing through any cross
section of a conductor
 And Δt is the time taken by the charge
 The SI unit of current is ampere

4.  Conventional Current
The current which passes from a point of high
potential to a point of low potential as if it is
represented by the movement of positive charges.
 Electric Current
The current which passes from a point of low potential
to a point of high potential as if it is represented by a
movement of negative charges

5.  When two conductors at different potential are
connected through a metallic wire, then the current
will start to flow.
TYPES OF CURRENT SOURCES
1. Cells convert the chemical energy into electrical
energy
2. Electric Generators convert mechanical energy into
electrical energy
3. Thermo-couples convert heat energy into electrical
energy
4. Solar Cells convert sunlight into electrical energy

6. 1. Heating Effect
• Current flows through the conductor due to motion of
the free electrons
• The free electrons collide with the atoms of metal
• The electrons lose their kinetic energy and transfer it to
the atoms of metal
• The vibrational K.E of the atoms is increased, which
produces heat in the wire.
• Mathematically, the heat can be measured by
H=I²Rt
• Applications: Heater, toaster,
electric iron, etc.

7. 2. Chemical Effect
 There are certain liquids through which electricity can
pass through due to some chemical reaction
 The study of this process is called electrolysis
 The chemical changes produced during the electrolysis
are due to the chemical effects of the current
 Depends upon the nature of the liquid and the quantity
of electricity passed through liquids
 Applications: Electroplating
and purification of metals

8. 3. Magnetic Effect
 When a current passes through the wire, a magnetic
field is produced around it
 The strength of the magnetic field depends upon the
magnitude of current and the distance from current
carrying conductor.
 Applications: Used to detect and measure current,
used in the machines involving electric motors

9. The branch of physics which inter-relates
electricity and magnetism

10.  In 1820, Prof. Hans Oersted described the magnetic field due
to current in a wire
EXPERIMENT
• Take a copper wire that passes vertically
through a horizontal piece of cardboard
• Place small magnetic compass needles
on the card board along a circle with centre at the wire
• All the compass needles point in the direction of the north-
south. When a heavy current passes through wire , the
compass needles set themselves along the tangent to the circle.
• Reverse the direction of the current , the direction of needles is
also reversed
• When the current through the wire is stopped, all the needles
again point in the north-south direction

11. CONCLUSIONS
1. A magnetic field is set up around current carrying
conductor
2. The lines of forces are circular and their direction depends
upon the direction of current
3. The magnetic field lasts only as long as the current is
flowing through it
DIRECTION OF MAGNETIC FIELD
 RIGHT HAND RULE
If the wire is grasped in the fist of the right
hand with the thumb pointing in the direction
of current , then the curled fingers indicate
the direction of magnetic field

12. MAGNETIC FLUX
 The number of magnetic lines of forcing passing
through certain element of area is called magnetic flux
 Mathematically,
Φᵦ= B.A
Φᵦ =BACosθ
The SI unit of magnetic flux is Weber or Nm/A

13. MAGNETIC FLUX DENSITY (Magnetic
Induction)
 The magnetic flux per unit area of a surface is perpendicular
to the magnetic field is called magnetic flux density
 Mathematically,
B= Φᵦ
A
 SI unit of magnetic flux density is Wbmˉ² or NAˉ¹mˉ¹ or tesla

14.  A force in experienced by a current carrying conductor
placed in a uniform magnetic field. In actual, the
magnetic field exerts the force on the charged particles
moving in the conductor.
 Mathematically,
sin
)(
qvBF
BvqF




15.  Consider a segment of wire of length L
and area of cross section A placed in a
magnetic field B
 Let I be the current flowing through
the wire, then magnetic fore acting
on the conductor is,
Fв=I( L × B ) (1)
 Let,
• Area of cross section of wire = A
• Length of conductor = L
• Volume of the wire segment = AL
• Number of charge carrier per unit volume = n
• Total number of charge carriers in the wire segment = nAL

16. • Velocity of each charge = v
• Charge on a charge carrier = q
• Total charge on nAL charge carriers = ΔQ = nALq
As current is defined as
I = ΔQ
Δt
OR I = nALq [As Δt = L/v]
L/v
OR I = nAqv (2)
Using this value of current in Eq. (1), we get
F = nAqv (L × B) (3)

17.  As the direction of segment L̂ is similar to v, it can be
written as
L̂ = v̂
So, L = LL̂ = Lv̂
 Thus Eq. 3 becomes,
OR
OR
]vvˆv[)Bv(nAqLF
)Bvˆv(nAqLF
)BvˆL(nAqvF
L
L
L









18.  This is the force exerted on nAL charges
 Now the force experienced by a single charge is,
(4)
 This is the general equation which holds for any charge
carrier moving in a magnetic field
)Bv(qF
nAL
)Bv(nAqL
F
nAL
F
F L









19.  For an Electron
F = -e ( v × B )
 For a Proton
F = +e ( v × B )
 Direction of force can be determined by right hand
rule. The direction of force is given by the direction of
vector ( v × B )
 Special Cases
• When θ=90˚, F=qvB
• When θ=0˚, F=0
• When θ=180˚, F=0
• When charge particle v=0 , F=0

20.  When the charge particle q is moving with velocity v
in a region where there is an electric and magnetic
field, then the total force F is the vector sum of electric
and magnetic force
 Mathematically,
 Electrical force does work only while the magnetic
force deflects the charged particle. It performs no work
]EqF[)Bv(qEqF
FFF
e
Be







21.  Consider a current carrying rectangular
coil which is capable of rotating
about its axis
 Suppose it is placed in uniform magnetic
field B with its plane along the field.
 The force acting on a conductor of length L
in uniform magnetic field is,
F = I (L×B)
F = ILBsinθ
 Where θ is the angle b/w L and B
 Divide the length of rectangular coil into four segments
AB, BC, CD and DA

22.  Forces acting on the sides of the coil can be calculated
as:
• Force on segment AB
As AB is anti-parallel to magnetic field, θ=180˚
• Force on segment BC
As BC is perpendicular to magnetic field, θ=90˚
• Force on segment CD
As CD is parallel to magnetic field, θ=0˚
• Force on segment DA
As DA is perpendicular to magnetic field, θ=90˚
 From right hand rule, its direction is out of the paper
0)0(180sin  ILBILBF 
ILBILBILBF  )1(90sin2

0)0(0sin  ILBILBFCD

ILBILBILBF  )1(90sin1


23.  As F₁ and F₂ are equal and opposite, they form a couple
which tends to rotate the coil about its axis
 Now,
Torque due to couple=(magnitude of either force) ×
(couple arm)
τ = ILB × a
 Where a is the perpendicular distance b/w two forces called
couple arm and is equal to AB or CD and La is the area of
the coil. (i.e. La=A)
 So, τ = IBA
 If the field makes an angle α with the plane of the coil, then
the couple arm becomes acosα
τ = ILB × acosα
τ = IB(La) cosα
τ = IBA cosα
 If there are N number of turns of a coil, then
τ = NIBA cosα

24.  Special case
• The torque acting on current carrying coil will be
maximum when the plane of coil is parallel to the
magnetic field i.e. when α=0˚, so
τ max= N IB A cos 0˚= N IB A
 NUMERICAL PROBLEM
• A coil of 0.1 m × 0.1 m and of 200 turns carrying a
current of 1.0 mA is placed in a uniform magnetic field
of 0.1 T . Calculate the maximum torque acting on the
coil
• Area of coil = A = 0.1 m × 0.1 m = 0.01 m²
• Number of turns = N = 200 turns
• Current = I = 1.0 × 10¯³ A
• Magnetic field = B = 0.1 T
• Maximum torque = τ max = ?
• Formula : τ = NIBA cosα
• Answer : 0.2 × 10¯³ OR 2.0 × 10¯⁴ Nm

25.  If the magnetic field through a circuit changes, an emf
and a current are induced in the circuit
 Applications:
 Used in power generating system
 Transformers use this principle to produce emfs
 ATM card also uses this principle
 Any appliance we plug into a wall socket uses induced
emf
 INDUCED EMF AND INDUCED CURRENT
 If a conductor moves through a magnetic field then due
to change in magnetic flux, an emf is induced called
induced emf. If the circuit is closed, it will cause an
electric current called induced current

26.  EXPERIMENTS TO PRODUCE INDUCED EMF AND
CURRENT
 Case 1
When the coil is stationary and
the bar magnet is moved
 Case 2
When area of the coil is changed
in a uniform magnetic field
 Case 3
By rotating the coil of constant
area in uniform magnetic field

27.  The phenomenon in which changing of current in one coil
induces an emf in another coil
 Explanation
 The coil connected with a battery through a switch and a
rheostat is called primary coil
 The one connected to the galvanometer is called secondary
coil
 When variation is produced in the rheostat, current in the
primary coil changes thus magnetic flux changes.
 This produces an induced emf in the secondary coil
 According to Faraday’s Law, induced emf in secondary coil
is
εs= -Ns Δφs (1)
Δt

28.  As M is constant, so
εs= -MΔIp (5)
Δt
 Above Eq. shows that induced emf in the secondary coil
is directly proportional to rate of change of current in the
primary coil
 From Eq. 5,
 Hence, mutual inductance can be defined as
The emf induced in the secondary coil when the rate of
change of current in the primary coil is one ampere/sec
 Its SI unit is VsA¯¹ known as henry (H)
t
M s






29.  If φs is flux passing through one turn of coil then flux through
whole coil is Nsφs
 This magnetic flux is directly proportional to current Ip
flowing through the primary, so
(2)
(3)
 Where M is the constant of proportionality known as mutual
inductance of two coils
 From Eq. 1,
εs= - Δ(Ns φs) (4)
Δt
 Using Eq 2 in 4 ,
εs= - Δ(MIp)
Δt
P
SS
PSS
PSS
I
N
M
MIN
IN







30.  The phenomenon in which a changing current in a coil
induces an emf in itself
 Explanation
 Consider a coil connected in series
with a battery & a rheostat
 Due to varying the resistance of the
rheostat ,the current in the coil is changed
 Changing flux produces an induced emf in the coil
 Let φ be the flux passing through one loop of the coil then
the total flux for N no. of turns will be Nφ
 Since flux (φ=B.A) is proportional to magnetic field, and
magnetic field (B = µoI) is proportional to current. So,

31. Nφ I
Nφ = LI
L = Nφ (1)
I
 Where L is the proportionality constant of self
inductance. According to faraday’s Law,
εL= -N Δφ (2)
Δt
Or εL = -Δ (Nφ/Δt)
Or εL = -Δ (LI) [ As Nφ=LI]
Δt
 But L is constant, so
εL = - L ΔI (3)
Δt


32.  This equation shows that emf is directly proportional to
the rate of change of current in the coil.
 Thus from Eq. 3,
 Hence self inductance can be defined as,
 The ratio of induced emf produced in a coil to rate of change
of current in the coil , in the same coil
 Its SI unit is VsA¯¹ known as henry (H)
t
L L





33.  Transformer is an electrical device which changes a
given AC voltage into a larger or smaller AC voltage
 PRINCIPLE
• Mutual induction b/w two coils
wound on same iron core
 CONSTRUCTION
• It consists of two copper coils which are magnetically
linked to each other
• Primary coil : The coil to which AC power is supplied
• Secondary coil : The coil which delivers power to the output
circuit
 WORKING
• Let no. of turns of the primary coil = Np
• No. of turns of the secondary coil = Ns
• Alternating emf applied to the primary coil = Vp

34.  Rate of change of flux in the primary coil = Δφ/Δt
 Due to this change of flux, a back emf is also induced in
the primary which opposes the applied voltage. Thus,
Self-induced emf in primary coil is
Self induced emf = - Np Δφ
Δt
 If the resistance of the coil is negligible, then the back
emf is equal and opposite to applied voltage Vp so,
Vp = - back emf [Vp =ε + IR, If R=0; Vp = ε]
Vp = Np Δφ (1)
Δt
 Emf induced across the secondary coil is
Vs = Ns (Δφ/Δt) (2)

35.  Transformation Ratio
Ns Δφ
 Vs = Δt [Dividing Eq.1 and 2]
Vp Np Δφ
Δt
 OR Vs = Ns
Vp Np
 Efficiency of a Transformer can be expressed by,
output power
input power
 Efficiency of a transformer can be improved by
i. The core should be made of iron
ii. The core should be assembled by laminated sheets
iii. Resistance of primary and secondary coil should be
minimum
 100

36.  In a ideal transformer, the output power is nearly equal to
input power
 Principle for Transmission
 Power loss in transformers can be due to one of the
following reasons:
• Eddy Currents: These currents cause
energy loss in the core due to heat produced
in it. It can be reduced by using laminated
core with insulation b/w the lamination
of the sheet
• Hysteresis Loss: Due to repeated magnetization
and demagnetization of core due to flow of AC.
It can be reduced by using soft iron core
p
s
s
p
sspp
I
I
V
V
IVIV
outputPowerinputPower




37.  TYPES OF TRANSFORMERS
 STEP UP TRANSFORMER
 A transformer in which voltage across the
secondary is greater than primary voltage
 Condition: No. of turns in the secondary
is greater than no. of turns in primary coil
i.e. Ns > Np
 STEP DOWN TRANSFORMER
 A transformer in which voltage across the
secondary is less than primary voltage
 Condition: No. of turns in the secondary
coil is less than no. of turns in primary coil
i.e. Ns < Np

38.  NUMERICAL PROBLEM
 An ideal step down transformer is connected to main supply of
240 V. It is desired to operate a 12 V, 30 W lamp. Find the
current in the primary coil and the transformation ratio
 Primary voltage =Vp = 240 V
 Secondary voltage = Vs = 12 V
 Output Power = Po = 30 watt
 Current in primary = Ip = ?
 Transformation Ratio = ?
 Formulae: Ip = Po/Vp & (Ns/Np )= (Vs/Vp)
 Answers : Ip= 0.125 A
Ns = 1
Np 20