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Electromagnetism
You need to cut up some
chunks of steel
...
That’s not
even the crazy part though
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Three cars
on a single fingernail
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One fingernail
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That
description alone makes this
the coolest cutting method
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1 Introduction
When current flows in a conductor, the atoms in the conductor all line up in a definite
direction, producing a magnetic field
...
To reverse all the atoms requires that power be expended, and this power is
lost
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Hysteresis
loss is common to all ac equipment; however, it causes few problems except in motors,
generators, and transformers
...
When current flows through a conductor, there is
emergence of magnetic environment around the conductor due to arrangement of dipoles
in that conductor
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Following are some magnetic field patterns and right hand grip rule to find the direction
of magnetic field lines (the polarity of induced magnetic field)
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2 Basic Terminologies
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1 MAGNETIC FIELD LINE
A line of force where if a permanent magnet is placed will experience a force of
attraction of repulsion
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2
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A magnetic field consists of several field lines, where number of lines define the strength
of magnetic field
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2
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In case of solenoid, thumb gives direction of magnetic field and curled fingers give
direction of current
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3 Magnetic Field Patterns
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1
STRAIGHT WIRE:
Consider following case when wire is passing through a card board
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Apply grip rule to
verify the direction of magnetic fields
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CIRCULAR COIL

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2

Current out of the
board

Current into the
board
B

A
B

A

If we see the coil from top, at terminal A current seems to be coming out of card board
and at B current seems to be coming into the board
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Dot with a circle (out of page)(i
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e
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The central line is straight as strength on both sides is equal
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3
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In solenoid several individual magnetic
fields of loops are combined to form a net magnetic field as shown in fig
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Look at the field carefully; the lines are parallel and straight within solenoid and
diverging at the ends
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4 FORCE ON A CURRENT CARRYING CONDUCTOR:

U-shaped
Permanent
magnet

Linear magnetic field

Following figure shows a U-shaped permanent magnet through which a current carrying
wire is passing
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A current
carrying wire placed between the poles where only shaded region is a magnet
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This field produces a force called magnetic field force and pushing the wire out of
the magnet as shown
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Let’s first have a close look at the intersection of both magnetic fields i
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linear and
circular
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Current seems
to be coming out of board if seen from one side
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Linear
Field
Magnetic field force
Circular
Field

Look at the diagram carefully
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Now nature’s basic tendency activates
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So the field lines always avoid interaction or intersection so all field lines are collected on
right hand side
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In this arrangement a force is introduced on the wire pushing the wire towards right and
hence wire is kicked out of linear magnetic field
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Check it by reversing the direction of current
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1 FACTORS AFFECTING MAGNETIC FIELD FORCE
Following factors affect the magnitude of magnetic field force acting on the wire
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LENGTH OF WIRE (EXPOSED)
Magnetic field force is proportional to the length of wire because with this factor number
of concentric circles of circular magnetic field increase so more force is exerted on the
wire
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MAGNITUDE OF ELECTRIC CURRENT:
With increase in current the strength of circular magnetic field increases hence more
interaction causes more magnetic field force
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Units of B is Tesla
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MAGNITUDE OF MAGNETIC FIELD STRENGTH
With the increase in magnetic flux density of field strength the force on the current
carrying wire also increases
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5 MAGNETIC FLUX DENSITY / MAGNETIC FIELD
STRENGTH

Magnetic field force

Every field has certain strength e
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, gravitational and electrostatic which you already
studied
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From the above relationship
F
B M
IL
So by definition“ B is force acting per unit length per unit current”
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5
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Now move to the next step
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2 FLEMING’S LEFT HAND RULE
If indication finger, middle finger and thumb of left hand are placed perpendicular to
each other, then indication finger gives the direction of magnetic field lines, middle
finger gives the direction of current and thumb gives the direction of magnetic field force
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The
discussed case involves the direction of current perpendicular to magnetic field lines
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𝐵
𝜃

Now the force is exerted due to B sin  as B cos is parallel to current
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e
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6 AMPERE’S SWIMMING RULE
This rule is helpful in determining the nature of circular magnetic fields
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One magnetic compass is placed
below the wire and the other is above the wire
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For the case when compass is above the wire, rule is still valid but now the person is
considered to swim on his back
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Look at the
following figure if current in the wire is out of the page then

If a magnetic compass is placed on magnetic field line, it will continuously change its
direction along the circumference of the circle which shows that in circular magnetic
field instantaneous polarity of magnetic fields continuously changes along the
circumference of the circle, so that the magnetic flux density B which is a vector quantity
is always along the tangential line of the circle as shown in below
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A
vector can’t be consider as clockwise or anticlockwise but always as right, left, up, down,
out or into the page
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7 FORCE ON A MOVING CHARGE IN MAGNETIC FIELD
Below is an arrangement when a wire is placed in a magnetic field
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From Fleming’s left hand
rule, the direction of force is upwards
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F  BIL
From
L
v   L  vt
We know that
t
So
F  BI v  t 
Q
I
And
t
Q
F  B v  t 
So
t 
Therefore 𝐹 = 𝐵𝑄𝑣
Where
B  Magnetic flux density
Q  Charge (C)
v  Speed of charge ( ms 1 )
Consider a situation as given below, the velocity of the charge is exhibiting certain angle
 with magnetic flux density
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e
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If we consider electrons then whatever the direction is given by the thumb from
Fleming’s left hand rule; must be reverted
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7
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Following factors affect the magnitude of magnetic field force acting on the wire
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SPEED OF THE CHARGED PARTICLE
Magnetic field force is proportional to the speed of charged particle in the wire
Mathematically FM  v  1
B
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Mathematically FM  q  2
Combining (1) and (2)
𝐹∝ 𝑞× 𝑣
Therefore
𝐹 = 𝐵𝑞𝑣
Where B is magnetic flux density or magnetic field strength
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This is another way to derive the relationship which have already derived
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MAGNITUDE OF MAGNETIC FIELD STRENGTH

With the increase in magnetic flux density of field strength the force on the moving
charge also increases
...

We discussed a magnetic field force on a charged particle moving in a magnetic
field
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7
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8
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Following diagram gives a complete picture of
the setup
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The right
hand side BC of the frame is place in a solenoid and left hand side AD carries a pointer
with scale and a rider of known mass
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When
current enters in the solenoid as shown, the solenoid becomes electromagnet
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Current divides in two parts
one toward B and the other towards A
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Now find the direction of magnetic field in solenoid using grip rule (as shown the
diagram)
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This is done by adjusting known masses or changing the position of wedges
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By altering magnitude of current using rheostat and known
masses, a point comes when again the frame is balanced
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8
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Before going into the details of Hall probe let’s discuss Hall Effect
first that is the working principle of a hall probe
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If the electrons are moving inside a conductor, the field pushes them to one edge of the
conductor
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This is in fact exactly what happens, but it is most noticeable in thin, wide conductors and
semi-conductors such as silicon
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Consider the passage of free electrons through a conductor as in following figure
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The upper edge is positively charged
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The charge builds to a level where the magnetic field force on each
electron is balanced by the electric field force due to the electric field between positive
and negative across the conductor
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In other words the potential difference between the edges continues to
build until it has created an electric field strong enough to prevent an electron from
magnetic effect
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Following diagram renders the
situation after few seconds
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A
 V  A v t  6
Put in equation (5)
Total number of charges N  n A v t   n A v t
If the charge of 1 carrier is q then

QT  N  q
 n Av t  q
As

QT n A v t q

t
t
 n Av q
So
I
v
n Aq
I

Put in equation (4)
 I 
Vh  B d 
 n Aq 



So
BId
Vh 
nq
This is the way to calculate hall potential difference mathematically
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A hall probe is a slice of
doped semiconductor; such material can be doped to have a much lower concentration of
charge carriers than in metals so they produce much bigger hall voltage than metals
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Another two wires
connected across the edge of the slice allow the hall potential difference to be measured
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With a constant current hall potential difference is proportional to the magnetic field
strength
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3 CALIBRATION OF HALL PROBE UNIT
Suppose we first place hall probe slice perpendicular to a known magnetic flux density B0
and note the value of hall potential difference and then in unknown B and again note the
value of V h
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Then place hall probe
perpendicular to B0 , then𝑉ℎ 𝑜 is noted so if suppose B0  10 T and𝑉ℎ 𝑜 = 1𝑉then
B  10Vh
Now we just need to place hall probe in unknown magnetic flux density B
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Problem is solved…
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IMPORTANT
While solving designing questions in paper 5, use the name “calibration hall probe unit”
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You must remember the name and its diagram
so that you can use while giving designing experiment
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7
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10
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Figure 1-5(A) is actually
a partial cutaway view showing the construction of a simple coil
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Notice that the two ends of the coil are identified as
X and Y
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The combined influence of all the turns
produces a two-pole field similar to that of a simple bar magnet
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In the above figure field lines are parallel to
each other within a solenoid
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B  I  1
N
 2 
l
Combining (1) and (2),
IN
B
L
So
 IN
B 0
L
For 1 m wire
B

B  0 I N

Where

 0  Permeabili ty of free space  4  10 7 H / m
H  Henry Unit of inductance of coil 

This has the same concept of permittivity of free space (𝜀 𝑜 ) used in electrostatics
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VARIATION OF B INSIDE THE SOLENOID
Following graph explains the variation of magnetic flux density B
...
Remember that at center field lines are parallel so concentrated
2
whereas at the end they are variable because diagonally diverted

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2 CIRCULAR COIL

In this case

B  I  1
B  N  2
Combining (1) and (2)

BIN
 IN
B 0
2r
Where
0  Permeability of free space
I  Current
N  No
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3

STRAIGHT WIRE

In this case

B  I  1

 2
2 r
Combining (1) and (2) we get
B

1

B

I

2 r
I
B 0
2 r
Where
2 r  Circumference of the circle
Remember that greater the circumference further will be the point from wire and hence
less magnetic flux density is observed
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11 FORCE BETWEEN TWO CURRENT CARRYING WIRES:
Following figure gives picture of two parallel wires having current in the same direction
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11
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Now consider their magnetic field pattern
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Both fields will combine to form a resultant magnetic field of shape of digit 8 as shown
below
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The problem can be understood in a more general way as given
below
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With reference to top view situation is as below
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Now apply Fleming’s left hand rule on I 1 and I 2 one by one, you will find magnetic field
force due to I 1 on I 2 and due to I 2 on I 1 as given below
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F12  Force due to I 1 on I 2
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IMPORTANT:
This can be understood from Newton’s 3rd law
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Now let’s starts mathematical working as we know that
F  BI L
So

F12  B1  I 2  L

 1

And already we know that for straight wire
 I
B 0
2 r
hence
 I
B1  0 1
2 r
And
 I
B2  0 2
2 r
So from equation (1)

 I 
F12   0 1  I 2  L
 2 r 


 II
F
 12  0 1 2  2 
L
2 r
Now for the force F21
F21  B2  I 1  L

0 I 2
 I1  L
2 r
 II L
F21  0 1 2
2 r
F21  0 I 1 I 2

 3
L
2 r
F21 

If you compare (2) and (3) then they are same, so generally force per unit length is
F  0 I1 I 2

L
2 r
MPORTANT:

1
...

L
So
F
 I1 I 2
L
2
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If two wires exert force on each other they should attract but we don’t see this in
high tension wires because the value of  0 is too low, currents are normally in the
ranges of 70  100 A and the distance between them is 50  70 cm
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7
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2

CASE # 2

Following diagram shows two wires having current in opposite direction

𝑟

From top view
Both magnetic fields combine to produce a resultant magnetic field as given below
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Or

If two parallel current carrying wires separated at a distance of 1 m have current of 1 A then they
will exert a force of 2  10 7 Nm 1 on each other

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