- When a current flows through a conductor, it produces a magnetic field around it. The magnitude of the magnetic field depends on the current and distance from the conductor. - According to Faraday's law of electromagnetic induction, a changing magnetic flux induces an electromotive force (emf) in any closed circuit. The induced emf opposes the change in flux according to Lenz's law. - Transformers work on the principle of mutual induction to change the voltage of an alternating current (AC) while keeping the frequency the same. They have a primary coil and secondary coil with different numbers of turns.

1. REVIEW LECTURE-8
UNIT-8
(Electromagnetism and
Electromagnetic induction)

2. Magnetic Effect of Current:
 When an electric current is passed through a conductor
then a magnetic field is produced around the conductor.
Magnetic field (B):
 “The region or space around a magnet in which its
magnetic effects are experienced, is defined as magnetic
field.”
 It is a vector quantity.
 Unit intensity of magnetic field (B) is weber m–2
or
Newton/ampere x meter or tesla
MAGNETIC FIELD

3. Magnitude of Magnetic Field:
𝐵 ∝ 𝐼
𝐵 ∝
1
𝑟
Direction of magnetic field:
Right Hand Rule:
 “If the wire is grasped in fist of right hand with the
pointing in the direction of the conventional current, the
fingers of the hand will circle the wire in the direction of
magnetic field”

MAGNETIC FIELD DUE TO CURRENT IN A LONG STRAIGHT WIRE

4. Magnetic flux
 The scalar product of magnetic field strength (B) and the
vector area (A) is
𝜙 = 𝐵. 𝐴= BA cos
 If the angle between B and A is 0o
 max = BA
 If the angle between B and A is 90o min = 0
 At angle of 30o magnetic flux is 86.6% of its maximum
value.
 At angle of 45o magnetic flux is 70.7% of its maximum
value.
o
MAGNETIC FLUX AND MAGNETIC FLUX DENSITY

5.  The unit of magnetic flux  is weber.

Magnetic flux density:
 “The magnetic induction B is the flux per unit area of a
surface perpendicular to B is also called as flux density.”
𝐵 =
𝜙
𝐴
 weber metre–2
 1 tesla = 104 gauss

6.  Current carrying conductor experiences a force when
placed in a uniform magnetic field that is given by;
𝐹𝑚=I 𝐿 × 𝐵
 𝐵 =
𝐹𝑚
ILsin𝜃
 So, magnetic field is defined as magnetic force acting on
a conductor of 1m carrying 1A current and placed  to
magnetic field.
 If Fm = 1N, I = 1A, L=1m & θ = 90°, then B= 1 tesla
 1T=
𝑁
𝐴𝑚
1T = 10+4 G
FORCE ON A CURRENT CARRYING CONDUCTOR
IN A UNIFORM MAGNETIC FIELD

7.  When a charged particle of charge q is moving with
velocity v in a magnetic field B at an angle , then force
acting on the particle:
F = qvB sin
 In vector form 𝐹𝑚 = 𝑞 𝑣 × 𝐵
 Direction of force can be found by using right hand palm
rule:
Fingers of right hand is in direction of magnetic field and
thumb is in direction of velocity of charged particle then
palm will demonstrate the direction of force. ( Rule is
applicable only for positive charges)
FORCE ACTING ON A CHARGED PARTICLE IN A
UNIFORM MAGNETIC FIELD:

8.  If v = 0, q = 0,  = 0o, 180o then F = 0
 If  = 90o F = qvB
 If  = 30o F = qvB/2
PATH FOLLOWED BY CHARGE PARTICLE IN MAGNETIC FIELD:
 If  lies between 0o and 90o, then the path of the particle is helical or helix.
 If  = 90o, then F = qvB sin 90o or Fmax = qvB
So charged particle will move in circular path.

9. MOTION OF A MOVING CHARGED PARTICLE IN A
MAGNETIC FIELD:
 𝑟 =
𝑚𝑣
𝑞𝐵
=
𝑝
𝑞𝐵
=
2𝑚𝑘𝐸
𝑞𝐵
r is called radius of circular path or cyclotron radius.
 𝑓 =
𝜔
2𝜋
=
𝑞𝐵
2𝜋𝑚
(Where f is called cyclotron frequency)
 𝑇 =
2𝜋𝑚
𝑞𝐵
 𝐾. 𝐸 =
𝑟2𝑞2𝐵2
2𝑚

10. e/m RATIO OF AN ELECTRON (1.7588 x 1011 C/kg):
e/m of a charged particle can be given as;
𝑒
𝑚
=
𝑣
Br
 Where r is determined by Thomson’s apparatus while
velocity is determined by following two methods:
Potential difference method:
𝑒
𝑚
=
2V𝑜
𝐵2𝑟2
Velocity selector method: v=
𝐸
𝐵
𝑒
𝑚
=
𝐸
𝐵2𝑟

11. 3 A of current is flowing in a linear conductor having a
length of 40 cm. The conductor is placed in a magnetic
field of strength 500 gauss and makes an angle of
30∘with the direction of the field. It experiences a force
of magnitude
(a) 3×10−4newton
(b) 3×10−2newton
(c) 3×104newton
(d) 3×102newton
QUESTION-1

12. A uniform electric field and a uniform magnetic field are
produced, pointed in the same direction. An electron is
projected with its velocity pointing in the same direction
(a) the electron velocity will increase in magnitude
(b) the electron velocity will decrease in magnitude
(c) the electron will turn to its left
(d) the electron will turn to its right
QUESTION-2

13. A beam of ions with velocity 2×105m/s enters normally
into a uniform magnetic field of 4×10−2tesla. If the
specific charge of the ion is 5×107C/kg, then the radius
of the circular path described will be
(a) 0.10 m
(b) 0.16 m
(c) 0.16 m
(d) 0.25 m
QUESTION-3

14.  “When the magnetic flux linking a conductor changes, an e.m.f is induced in the
conductor, this phenomenon is known as electromagnetic induction”. Induction is the
change in flux linking the conductor (or coil)
 The product of number of turns (N) of the coil and the magnetic flux (Φ) linking the coil is
called flux linkages i.e.
Flux linkages = NΦ
ELECTROMAGNETIC INDUCTION

15.  The induced emf is equal to the negative rate of change
of magnetic flux. Thus, if  be the change in flux in time
interval t, then the induced emf in the circuit is
𝜀 = −
𝛥𝜙
𝛥𝑡
, N number of turns, 𝜀 = −𝑁
𝛥𝜙
𝛥𝑡
 The negative sign shows the induced emf opposes the
change in magnetic flux.
 If rate of change of magnetic flux be in weber/sec, the
induced emf will be in volt.
 If the coil contains N turns 𝜀 = −𝑁
𝛥𝜙
𝛥𝑡
=
𝛥 𝑁𝜙
𝛥𝑡
FARADAY’S LAW

16.  According to Lenz’s rule the induced current produced in
a closed circuit always flows in such a direction that it
opposes the cause due to which it has been produced itself.
 When the north pole of a magnet moves
towards a stationary loop,
 When the N-pole of the magnet is moved away
from the loop,
Because in each case induced current opposes the motion
of magnet hence some mechanical work has to be done on
the system to move the magnet against this opposing
force. According to law of conservation of energy, this work
is obtained in the coil in the form of heat energy.
LENZ’S LAW AND CONSERVATION OF ENERGY

17.  Current generator is a device, which converts mechanical
energy into electrical energy in the presence of magnetic
field.
 Principle: Faraday’s Law of electromagnetic induction
 Main parts of A.C generator
Pole pieces (U-shape magnet) with concave poles.
Armature (assembly of coil on iron cylinder)
Slip rings (as connector)
Carbon brush (external supply)
ALTERNATING CURRENT GENERATOR

18.   = N ω AB Sin ω t
when θ = 90o max = N ω AB
then  = max Sin ωt If ω = 2πf, then
 = max Sin2πft
In terms of potential difference, V = Vo Sin 2πft
In terms of current, I = Io Sin2πft.

19. It is device which increases/decreases input electric
potential
Principle: Mutual induction
It only operate on A.C
Transformer does not change frequency
Construction:
Primary coil, Secondary coil, Iron core
Turns Ratio:
Voltage turns ratio
𝑉𝑠
𝑉𝑝
=
𝑁𝑠
𝑁𝑝
Current turns ratio
𝐼𝑠
𝐼𝑝
=
𝑁𝑝
𝑁𝑠
,
𝑁𝑠
𝑁𝑝
=
𝑉𝑠
𝑉𝑝
=
𝐼𝑝
𝐼𝑠
TRANSFORMER

20. Types of Transformer:
Step up Ns > Np, Vs > Vp, Is < Ip
𝑁𝑠
𝑁𝑝
> 1
Step down Ns < Np, Vs < Vp, Is > Ip
𝑁𝑠
𝑁𝑝
< 1
Losses in Actual Transformer
Copper Losses in Winding
Flux Losses
Iron Losses in Core
Iron losses are of two types: Eddy current loss and
hysteresis loss.

21. The turns ratio of a transformer is
𝟏
𝟐
. If a dry cell of emf
1.5 volt is connected with primary coil then emf establish
in secondary coil is?
(a) 0.75 V
(b) 3 V
(c) 1.5 V
(d) Zero
QUESTION-1

22. A coil having 500 square loops each of side 10 cm is
placed normal to a magnetic flux which increases at the
rate of 1.0 tesla/second. The induced e.m.f. in volts is
(a) 5
(b) 1
(c) 0.1
(d) 0.5
QUESTION-2

23. The north pole of a long horizontal bar magnet is being
brought closer to a vertical conducting plane along the
perpendicular direction. The direction of the induced
current in the conducting plane will be
(a) horizontal
(b) vertical
(c) clockwise
(d) anticlockwise
QUESTION-3

24. The alternating current has frequency of 106 Hz, in such
a way that time period for completion of cycle is
(a) 1μs
(b) 1.5μs
(c) 106sec
(d) 1sec
QUESTION-4