Physics
Revision on chapter4
Part2
Third; Important Give Reasons :
1) On switching on a circuit containing an induction coil, the current
intensity does not reach its maximum value quickly
Due to formation an induced backward e.m.f by self induction retard the growth of the
original current
2) The ohmmic resistance is made of double wounded wires .
In order to, avoid the self induction inside the resistance.
3) A bar of soft Iron will not be magnetized if a wire carrying a current is
doubly wounded a round it.
To make the direction of current in a branch of the coil is opposite to the direction of
it in the second branch so the two magnetic field, produced are equal and opposite
so , cancel each other .
4) The transformer does not operate with direct current (can not step up or
step down D.C. voltage.
Because the D.C. voltage generate a magnetic field of constant value and
direction i.e
= 0 so that there is no an induced e.m.f is generated
5) The core of the transformer is made of thin insulated sheets of siliconic soft
iron.
To minimize the harm effect of the eddy currents since its resistivity is very
high
6) There is no an ideal transformer (100 %)
Because loss in energy is carried out, due to eddy currents , mechanical energy inside
the core, and thermal energy in the wires .therefore it is impossible to get rid of all of
these factors completely
7) The transformer does not operate in spite that its primary coil is
connected to the a.c. source, otherwise after the secondary coil is loaded.
Due to formation backward induced e.m.f by self induction in the primary coil which equal
in magnitude and opposite in direction to the original e.m.f
8) The coil of the electric motor continues in rotating when it is
perpendicular to the lines of the magnetic flux, in spite that the torque
of electromagnetic couple in this position equals zero
Due to inertia which acts to continue the rotating of the coil in the same circular
direction this permit the two halves of the cylinder interchange positions relative to the
two brushes and thus the current in the coil is reversed once more.
1st.s 02
Fourth;important Comparisions
a) The states of generation backward e.m.f and forward e.m.f by mutual
induction.
a))Forward E.M.F.
Backward E.M.F.
a) When the circuit of the primary coil is a) When the circuit of the primary coil is
switched off.
switched on.
b) When the primary coil is for apart from b) When the primary coil is plunged into the
the secondary coil.
secondary coil.
c) When the current intensity that flow c) When the current intensity that flow
across the primary coil being decreases.
through the primary coil being increases.
b)
Fleming's left hand rule and Fleming's right hand rule with respect to the
application of each.
1- Fleming” s left hand
2- Fleming” s right hand
It is used to define the direction of
It is used to define the direction of induced
motion (force) of a wire carrying current current created in a wire moves in magnetic field
placed in magnetic field.
Extend the forefinger, the thumb & the rest
fingers of the left hand mutually
perpendicular to each other if the forefinger
points inthe field direction and the rest
fingers points in the current then the thumb
will points in the motion direction.
Extend the forefinger, the thumb & the rest fingers of
the right hand mutually perpendicular to each other if
the forefinger points in the field direction and the
thumb points in the motion direction then the rest
fingers will points in the current direction.
c) Lenz,s rule and amper,s right hand rule from the point of view of the
function of each of them.
3- Amper”s right hand rule
6-Lenz”s rule
It is used to define the direction of
magnetic field due to current carrying
wire. And the
polarity of the coil.
It is used to define the direction
of induced current created in a
coil.
Imagine you grasp the wire with your right
hand such that the thumb points in the
direction of current. The rst of fingers around
the wire gives the direction of the magnetic
field.
Whenever an e.m.f is induced in a coil
the direction of the induced current
must be in a direction such that to
oppose the change in magnetic flusx
producing it.
d) A.C generator and D.C motor from the point of view of its function,the
scientific base of action and the structure of each of them.
j)
Function
Scientif
ic base
of
action
The
strcture
A.C generator (Dynamo)
Conversion of mechanical energy into electrical
energy in the form of alternating current &
electrometive force
Based electro magnetic induction as the motion of
a coil to cut the magnetic field lines with
available rate producing induced E.M.F and
induced current
1-Field magnet.
2-Armator: is arectangular coil suspended
between the poles of the magnet.
3-Two slip rings x, y:
The two rings touch two
graphite brushers F1,
F2 to conduct the
produced current to the
external circuit.
Electric Motor
Conversion of electrical energy into mechanical
energy in the form of rotational motion
Based on magnetic torque acting on a coil
carrying current placed in magnetic field
1- Armator coil made from insulated copper
wrapped around
insulated soft iron core
connected with two
halves of splited
cylinder x, y touches by
two graphits brushes
F1, F2, the two brushes
connect the electric
current to the coil.
2- Magnet.with concaved shaped poles
Fifth; Important proves
1) Deduce a relation for calculating the induced e.m.f
generating in a straight wire of length (L) moving with
constant velocity (V) perpendicular to a uniform
magnetic field of flux density (B).
When a straight wire of length (l) is placed normal to a uniform magnetic
filed of magnetic flux density B as
shown in the fig. The wire is moved in
a direction perpendicular to the field at
velocity v, so that it is displaced a
distance x in time t. the change in
area is A =l. x
∴ The change in flux is: m = BA =
Bl x
The e.m.f. is determined from the relation
i. e .m .f .  
m
l
Bl x
e .m .f .   t   Blv
The minus sign is required according to Lenz's rule. The e.m.f. is
e .m .f .  Blv
-If the angle between the direction of velocity and the direction of the magnetic flux = , then
e .m .f .  Blv sin
2) Prove that, the Instantaneous value of e.m.f in
dynamo is given by
(e.m.f)= NAB ω sin t.
When the armator rotate within the field, an induced E.M.F will generate across each
langitudinal side equals.
e .m .f  BLV sin 
So that total E.M.F is given by:
e .m .f  2BLV sin 
since the coil produce circular or periodical motion, therefore the value of the linear
V  r
velocity (V) must be converted into angular velocity( ω) i.e
  2BL r sin
A  (2r )(L )
  B A sin 
Letting N: is the number of turns of the coil
\
e.m .f = NA B w sin q "
Note: "  t""  2 "
The angle  confined between the direction of filed and the perpendicular direction on
the plane of generator coil.
Sixth ; Miscellaneous questions;
A-mention one application only for each of the following
1) Eddy current.
2) Induction furnaces
3) Fleming's left hand rule .…………………………………….
4) Lenz's rule
5) Fleming's right hand rule ……………………………………
6) The metallic cylinder splitted into two halves insulated from each Other
and connected to the coil of dynamo orThe commutator in
dynamo…………
7) The two carbon brushes in the dynamo,
8) The step up transformer ………………………………………
9) Transformer
10) Electric motor ……………………………………………….
11) The reverse current induced in the motor's coil during its motion
…………
12) The iron core formed of thin discs isolated from each other in the
motor.
1st.s 05
1st.s 98
2nd.s 02
1st.s 97
2nd.s 98
1st.s 99
B-Mention the parameters (factors) acting on the following
physical quantities then write the relation between these
factors?
1) The induced e.m.f. generated in the coil by electromagnetic
induction. Mention the relation between the e.m.f. and
such factors.
2) The induced e.m.f. generated in the coil of the dynamo.
Mention the relation between the e.m.f. and such
factors. ………………………….
3) Induced e.m.f in straight wire Mention the relation
between the e.m.f. and such factors.
4) The
self
induction
coefficient.
……………………………………….
5) Mutual induction coefficient..
6) Eddy currents.
1st.s 00
2nd.s 99
C- Mention the scientific base of action in which each of the
following depend
a)
b)
c)
d)
e)
f)
g)
Flourscent flash lamp.
A.C generator (Dynamo………………………..
Ohmmeter.
The electric transformer…………………………..
Ruhmkorff,s coil
Electric motor
Induction furnace. …………………………………
2nd.s 01
1st.s 01
1st.s 05
1st.s 04
D- What will happen with interpretation when……..;
1) In the opposite figure a coil of
insulated copper wire is
connected to a centre-zero
galvanometer. What happens to
the galvanometer needle in each
of the following cases…
a- bar magnet is pushed into the coil?
b- The magnet remains stationary
inside the coil?
c- The magnet is pulled out again?
d- The magnet is pushed in faster
than before
e- The number of turns of the coil is
increased and the magnet moved
as in a?
2)In the circuit shown in the
opposite figure,
coil (1) is connected in series
with a cell, a switch (k) and an
ammeter.
Coil
(2)
is
connected to a Centre Zero
sensitive galvanometer. State
and explain what will be
observed on the reading of
the
ammeter
and
the
galvanometer in the following
two cases:
1st.s 07
a- The instant of closing the switch
(k).
b- A rod of soft-iron is inserted
into, both coils then the switch
(k) is closed.
c- iii)Mention the name of an
apparatus whose operation
principle is based on the above
experiment
1st.s 01
2) In the opposite figure a coil of
insulated
copper
wire
is
connected to a battery, rheostat
And switch S. The coil above a
small permanent magnet placed
on one pan balance State and
explain what will be observed on
the reading of the balance when,
a. the switch is closed?
b. the switch remains closed but
resitance R decreases gradually?
c. When the current direction is
reversed through the coil circuit.
4)The replacement of the two
metallic rings of the alternating
current generator by a metallic
cylinder splitted into two halves
insulated from each
other…………………………
1st.s 06
…………………
5)Closing the primary coil circuit
in the shown transformer while
opening the secondary coil circuit.
Answer the
questions
following
1. Draw the figure in front of you in your
paper and answer the following :
August (2001)
i- What is the kind of magnetic pole
produced in the terminal (B) of the coil?
ii-What is the effect of putting the soft
iron cylinder inside the coil on the value
of thee instantaneous deflection of the
galvanometer? Explain that
2. In The opposite figure the N pole of
the permanent magnet was thrust
into the centre of the aluminum ring.
The ring moved to the right,
indicating that it behaves like a
magnet.
a) State the polarity of the face of the ring
nearest the permanent magnet.
b) Draw a diagram to show the direction
in which the induced current flows in the
ring. Explain why this current is
produced.
1st.s 04
3. The figure shows a horizontal wire
AB placed between the poles of a
horseshoe magnet and connected
by flexible leads to a sensitive
galvanometer.
a)Explain why, if the wire AB is moved
slowly to and fro in a horizontal plane,
there is no deflection of the galvanometer
pointer, but if a similar movement takes
place in a vertical plane the pointer
moves to and fro.
b)In which direction would you expect
the current to flow when AB is moving
vertically upwards? (Give a reason for
your answer.)
.
Ninth; The problems
1. A solenoidal coil of 200 turns, the crosssectional area of each turn equals 2 cm2, it is
placed normally to a magnetic field of flux
density 0.6 w/m2. calculate the induced e.m.f.
for the cases when:
a)
Flux
density
increases to 0.8 tesla in 210-3 sec.
b)
Flux
density
decreases to 0.4 tesla in 0.210-3
sec.
c)
The field vanishes in
0.1 sec.
d)
The coil is turned
back in 0.1 sec
a)  = A (B2 – B1 )= 210-4 (0.8 – 0.6) = 410-5 Wb.

4  105  200
e .m .f . 
N 
 4 Volts
t
2  103
 4  105  200 

e .m .f .  
N  
  40 Volts
4
t
2

10


b)
c) e .m .f .   N
 2  104  0.6 

   200  
  0.24 Volts
t
0.1


…………………………………………
……………………………
a-
B
 NI
L
4  22  107  2  700

 16  104 Tesla
7  0.1
b-
e .m .f .   N
 BAN

 0.11 Volt
t
t
c-
0.112  L
I
2
L
t
0.01
 L = 5610 henry
-5
2. An alternating current dynamo, its armature consists of 100
turns and its rotates in a magnetic field so that it makes 3600
cycles/minutes. The magnetic flux
density is 0.3 tesla, if the
cross-section of its coil is 0.1m2, calculate:
a) The maximum induced e.m.f.
b) The instontenceus e.m.f after 1/270 sec. From the beginning of the
motion in the perpendicular direction on the field.
c) The mean e.m.f during a half cycle.
a)
(e.m.f)o = NAB = 100 x 0.1 x 0.3 x 2x
(e.m.f)1 =
b)
22 3600

7
60
132
 60  1131.4 volt
7
(e.m.f)1 = 0 sin (2 x 180 x 60 x
1
) = 0 sin 80
270
 (e.m.f) = 565.7 volt
c)
1
1
1
Time of half cycle  =



0.12 0.360 120
(e.m.f) = -N
t
 100
 1 


 120 
Calculate the charge quantity which passes through a
galvanometer of resistance 200 and connected to a
circular coil of resistance 400 and contains 1000
turns, and the coil is wounded around a wooden
cylinder if radius 1cm, the coil is placed in a magnetic
flux of density 0.0113 wb/m2 and parallel to the
cylinder's axis, then the flux is reduced to zero
suddenly.
Solution
  NA 
e .m .f .   N

t
t
3.
e.m.f. = IR
ANB
 IR 
t
IRt = ANB
Q = It
-6
 Q = 5.910 coulomb.
 QR = ANB
4. An electric current of intensity 2 amp. In a coil of 400 turn,
the produce flux is 10-4 weber. Calculate the average
produced e.m.f. in the coil when the current vanishes in
0.08 sec. also calculate the self induction of the coil.
Solution
104  400

 0.5 Volt
e .m .f .  
N e .m .f .  
t
8  102
I
1
2
e .m .f .   L
 L 
2
t
8  102
 L = 0.02 Henry
5. An A.C dynamo, the length of its coil is 40cm, its width is
30cm and it has 300 turns produces a current with frequency
50/11Hz and an effective potential 200 2 volt. Calculate:
a) The maximum value of the potential between the poles of the
generator.
b) The magnetic flux density acting on the coil.
c) The maximum value of the p.d when the coil rotates around an
axis parallel to its length with a velocity of 24m/s.
a)Veff = Vmax x 0.707
200 2 = Vmax x 0.707
Vmax = 400V
b)Vmax = BNAW
400 = B x 300 X 30 X 40 x 10-4 x 2 x
22
7
x
50
11
400  7 1110 4
B
 0.388 tesla
300  30  40  2  22  50
V
24
ω

r
15 10 2
Vmax = BNA = 300 x 0.388 x 30 x 40 x 10-4 x
24
15 10  2
6. A step down transformer decreases the potential from 2400 to 120
volts, the number of turns of its primary coil was 4000 turns. Clculate
the number of turns of its secondary coil if the produced power from
the transformer is 13500 watt and its effieciency is 90% then find the
current passing in the primary and secondary coils.
η
Vs Is Vs N p
90 120 4000
  


Vp I p Vp N s 100 2400 Ns
Ns = 222 turns
 Power = Vs Is
 13500 = Is x 120  Is = 112.5A
η
I x Vs
90
13500



Ip Vp
100 2400  Ip
Ip = 6.25A