TOPIC 1: ELECTROMAGNETISM
This is the production of a magnetic field by current flowing in a conductor.
The magnetic effect of current was discovered by Ousted in 1820. The verified magnetic effect of current by the following simple experiment.
Figure below shows a conducting wire AB Above a magnetic needle parallel to it.
So long as there is no current in the wire, the magnetic needle remains parallel to the wire i.e. there is no deflection in the magnetic needle.
As soon as the current flows through the wire AB, the needle is deflected.
Magnetic needle
When the current in wire AB is Reversed the needle is deflected in the opposite direction
This Deflection is a convincing proof of the existence of a magnetic field around a current carrying conductor.
On increasing the current in the wire AB the deflection of the needle is increased and vice versa.
This shows that magnetic field strength increases with the increase in current and vice versa
It is clear from Worsted’s experiment that current carrying conductor produces a magnetic field around it.
The larger the value of current in the conductor the stronger is the magnetic field and vice versa.
Magnetic field
Is the region around a magnet where magnet effect can be experienced.
OR
Is the space around a current carrying conducting (magnet) where magnetic effects can be experienced.
The Direction of a field at a point is taken to be the direction
in which a north magnetic pole would move more under the
influence of field if it were placed at that point.
The magnetic field is represented by magnetic lines of force which form closed loops.
The magnetic field disappears as soon as the current is switched off or charges stop morning.
Magnetic flux Φ
is a measure of the number of magnetic field lines passing through the region.
The unit of magnetic flux is the Weber (Wb)
The flux through an area A on figure below the normal to
which lies at angle ðœƒ to a field of flux density B
Is a quantity which measures the strength of the magnetic field
It is sometimes called magnetic
It is a vector quantity
The SI unit of Magnetic flux density is Tesla (T) or Wb/m^{2}
Magnetic flux density is simply called magnetic field B
B = θ/A
FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD
Consider a positive charge +Q moving in a uniform magnetic field with a velocity
Let the Angle between and be θ as shown
It has been found experimentally the magnetic field exerts a force F on the charge.
The magnitude F of this force depends on the following factors
(i) F α θ
(ii)F α B
(iii)
Combining the factors we get
Where K is a constant of proportionality
The unit of B is so defined that K = 1
Equation (a) can be written in a vector form as:
F = the force of the particle (N)
B = the magnitude of the magnetic flue density of the field T
Q = the charge on the particle
V= the magnitude of the velocity of the particle
Definition of
From
F = BQVsinâ¡θ
If V = 1, Q = 1, θ= 90 then
F = Sin90
F=B
Magnetic field ( ) at a point in space is equal to the force
experienced by a unit charge moving with a unit velocity perpendicular
to the direction of magnetic field at that point
Right Hand Grip Rule
Grip the wire using the right hand with the thumb pointing in the direction of the current the other fingers unit point in the direction of the field.
– For an electron (negatively charged) entering the magnetic field as shown below
The Direction of positive charge will be exactly opposite. Applying Right hand Grip Rule it is clear that Direction of force on the electron will be vertically upward
For a positively charged particle, it will be vertically downward
Direction of magnetic field means from N –pole to Spole.
SOME CASES OF MAGNETIC FORCE F
Consider an electric charge Q moving with a velocity V through a magnetic field B. then the magnetic force F on the charge is given by
F = BQV
(i) When = 0^{o }or 180^{0 }
F = BQV
F= BQVSin0^{0 }or F = BQV
F= 0
Hence a charged particle moving parallel(or Anti parallel) to the direction of magnetic field experiences no force
(ii) When =90^{0}
F = BQV
F = BQV
=1
F = BQV
Hence a force experienced by charged particle is maximum when it is moving perpendicular to the direction of magnetic field.
(iii) When V=O, the charge particle is at rest.
F = BQV
F=BQ(0)
∴F=O
If a charged particle is at rest in a magnetic field it experiences no force.
(iv) When Q = O
F = BQV
F = 0
Hence electrically neutral particle (eg neutron) moving in a magnetic field experiences no force.
The magnetic force F acts perpendicular to velocity V (as well as B)
This means that a uniform magnetic field can neither speed up nor slow down a moving charged particle; it can charge only the Direction of V and not magnitude of V
Since the magnitude of V does not charge the magnetic force does not change the kinetic energy of the charged particle.
UNITS AND DIMENSIONS OF
The SI unit of magnetic field B is Tesla
Now
F = BQV
If Q = 1C, V =1m/s, Q= 90^{0} F= 1N
B = 1T
Hence the strength of magnetic field at a point is 1T if a charge of 1C when moving with a velocity of 1m/s at right angles to the magnetic field, experiences a force of 1N at that points.
Magnetic field of earth at surface is about 10 ^{– 4}T. On the other hand, strong electromagnets can produce magnetic fields of the order of 2T.
Dimensions of
Worked Examples
1. A proton is moving northwards with a velocity of m/s
in a magnetic field of 0.1Tdirected eastwards. Find the force on the
proton. Charge on proton = 1.6 x 10 ^{19}C.
Solution
F = BQV
B= 0.1T
V= m/s
F=0.1 X 1.6 X 10^{19 }X 5 X 10^{6}X Sin 90
Q = 1.6 X 10^{19}C
= 90^{0}
 An electron experiences the greatest force as it travel at 3.9 x10^{5 }m/s in a magnetic field when it is moving westward. The force is upward and is of magnitude N what is the magnitude and direction of the magnetic field. Solution
The conditions of the problem suggest that the electron is moving at right angles
To the direction of the magnetic field
F = BQV , F = 8.7 x 10 ^{13}N
Q= 1.6 X10 ^{19}C
V=3.9X10^{5}m/s
B = 13.14T
By right hand rule per cross product, the direction of the magnetic field is towards northward.
 An α – particle of mass 6.65 x 10^{27 }kg is travelling at right angles to a magnetic field with a speed of 6×10^{5}m/s. The strength of the magnetic field is 0.2T.calculate the force on the – particle and its acceleration. Solution
Force on α – particle F = BQV
M = 6.65 X10^{27}Kg
V = 6 x 10^{5}m/s
B = 0.2T
= 90^{0 }
^{ } F = BQV
= (0.2 x 2x 1.6×10^{19}) x x Sin90Ëš
Acceleration of α – particle
F= mÉ‘
É‘= =
 A copper wire has 1.0 x 10^{29} free electrons per cubic meter, a cross sectional area of 2mm^{2} and carries a current of 5A . The wire is placed at right angle to a uniform magnetic field of strength 0.15T. Calculate the force the acting on each electron.
Solution
I = neA
Drift velocity =
n= 1×10^{29}m^{3} e = 1.6×10^{19}c A= 2mm^{2} = 2×10^{6}m^{2}
I = 5A
Force on each electron F= BQ Sin
Q= 1.6 x 10^{19}c
B= 0.15T
Q=90^{0}
BIOT –SAVART LAW
The Biot – Savart law states that the magnitude of magnetic flux density dB at a point P which is at a distance r from a very short length dl of a conductor carrying a current I is given by.
where is the Angle between the short length dl and the line joining it to point P
K is a constant of proportionality its value depends on the medium in which the conductor is situated and the system of units adopted.
For free space vacuum or air
This equation is known as Biot –Savart Law and gives the magnitude of the magnetic field at a point due to small current element
Current element
Is the product of current (I) and length of very small segment ( ) of the current carrying conductor.
Current element =
Current element produces magnetic field just as a stationary charge produces an electric field the current element is a vector.
Its Direction is Tangent to the element and acts in the direction of current flow in the conductor
Biot Savart law holds strictly per steady currents
Direction of B
The direction of is perpendicular to the plane containing and by right hand rule for the cross product the field is directed inward.
Special cases
(i) When = 0^{0} or 180^{0}
i.e Point P lies on the axis of the conductor
Hence there is no magnetic field at any point on the thin current carrying conductor minimum value.
(ii) When = 90^{0}
When point P lies at a perpendicular position w .r. t current element
Hence magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis.
Important point about Biot – Savant law
(i) Biot – Savant law is valid per symmetrical current distributions.
(ii) Biot – Savant law cannot be proved experimentally because it is not possible to have a current carrying conductor of length dl
(iii) Like coulomb’s law in electrostatics, Biot Savant law also obeys inverse square law
(iv) The Direction of dB is perpendicular to the plane containing and
(v) This law is also called Laplace’s law and inverse square law’
BIOT – SAVART LAW VERSUS COULOMB’S LAW IN ELECTROSTATICS
According to coulomb’s law in electrostatics, the eclectic field due to a charge element dQ at a distance r is given by
According to Biot – Savart law the magnetic field due to a current element at a distance r is given by
From the above two equations we note the following points of Similarities and Dissimilarities.
Similarities
(i) Both laws obey inverse square law
(ii) Both the fields(magnetic field and Electro static field) obey superposition principles
(iii)Both the fields are long range fields.
Dissimilarities
(i) The Electric field is produced by a scalar source i.e. Electric charge . However the magnetic field is product by a vector source i.e. current
(ii) The Direction the Electric field is along the displacement vector i.e. The line joining the source and field point. However the direction of magnetic field is perpendicular to the plane containing current element and displacement vector
(iii) In Biot –Savant law the magnitude of magnetic field dB α Sin Where is the Angle between current element and displacement vector However there is no angle dependence in coulomb’s law for electrostatics
MAGNETIC FIELD AT THE CENTER OF CURRENT CARRYING CIRCULAR COIL
Consider a circular coil of radius r and carrying current I in the Direction shown in figure
Suppose the loop lies in the plane of paper it is desired to find the magnetic field at the centre O of the coil
Suppose the entire circular coil is divided into a large number of current elements each of length
According to Biot – Savant law, the magnetic field at the centre O of the coil due to current element is given by
……………
The direction of dB is perpendicular to the plane of the coil and is Directed inwards
Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the centre O can be found by integrating equation…………(i)
L Total length of the coil = 2 r
If the coil has N turns each carrying current in the same direction then contribution of all turn are added up.
B=
MAGNETIC FIELD DUE TO INFINITELY LONG CONDUCTOR
The flux density dB at P due to the start length dl given by equation as
^{ }
From the figure (A)
,
r =
= a cot
= a
Substituting for and gives
The total flux density B at P is the sum of the flux densities of all the short lengths and can be found by letting d →O and integrating over the whole length of the conductor.
The limits of the integration are and 0 because these are values of ðœƒ at the ends of the conductor
FLUX DENSITY AT ANY POINT ON THE AXIS OF A PLANE CIRCULAR
Circular coil with its plane perpendicular to that of the paper
The flux density dB at p due to the short length dl of the coil at X, where X is in the plane of the paper, is given by equation as
By symmetry, when all the short lengths are taken into account the components of magnitude sum to zero.
Each short length produces a component of magnitude Sin α parallel to the axis and all those components are in the direction shown
The total flux density is therefore in the direction of Sin α and its magnitude B is given by
The radius vector XP of each small length is perpendicular to it, so that =90^{0} and there pore Sin = 1
Since,
= 2 (the circumference of the coil)
, But =
For a coil of N Turns
When S= r
Also from the figure
AMPERE’S CIRCUITAL LAW
States that the line integral of magnetic field around any closed path in vacuum/air is equal to times the total current (I) enclosed by that path
I = current enclosed by that path.
Ampere’s law is an alternative to Biot – Savart law but it is useful for calculating magnetic field only in situations with considerable symmetry.
This law is true for steady currents only.
In order to use law it is necessary to choose a path for which it is possible to determine the value of the line integral
It is because there are many situations where there is no such path that the law is of limited use.
Hence the application of ampere law
(i) Magnetic field due to constraining conductor carrying current
(ii)Magnetic field due to solenoid carrying current
(iii)Magnetic field due toroid
MAGNETIC FIELD DUE TO STRAIGHT CONDUCTOR CARRYING CURRENT
Consider a long straight conductor carrying current I in the direction as shown in the figure below
It is desired to find the magnetic field at a point p at a perpendicular distance r for the conductors
Applying Ampere’s circuital law to this closed path
SOLENOID
Is a long coil of wire consisting of closely packed loops
Or
Is a cylindrical coil having many numbers of turns
By long solenoid we mean that the length of the Solenoid is very large as compared to its Diameter.
Figure below shows the magnetic field lines due to an air cored solenoid carrying current
Inside the solenoid the magnetic field is uniform and parallel to the solenoid axis.
Outside solenoid the magnetic field is very small as compared to the field inside and may be assumed zero.
It is because the same no of field line that are concentrated inside the solenoid spread out into very faster space outside
Magnetic flux density due to an Axis of an in finely long Solenoid
Consider the magnetic flux density at P due to a section of the solenoid of length
n = number of turns per unit length.
N= number of turns the section can be treated as a plane circular coil of N turns in which case dB is given by
Since dx is small, the section can be treated as a plane circular coil or N turns in which case dB is given by
From the figure
Also
Substituting for and dx gives,
The flux densities at P due to every section of the Solenoid are all in the same direction and therefore the total flux density B can be found by letting dB→o and integrate over the whole length of the solenoid.
The limits of integration are and 0 because these values of β at the end of the solenoid.
If the Solenoid is Ironcored of relatively permeability magnitude of magnetic field inside the Solenoid is
From
At points near the ends of an air cored Solenoid, the magnitude of magnetic field is
The magnetic field outside a solenoid is zero
Also in a current carrying long solenoid the magnetic field produced does not depend upon radius of the Solenoid.
TOROID
Toroid is a solenoid that bent into the form of the closed ring.
The magnitude field B has a constant magnitude every where inside the
toroid while it is zero in the open space interior and exterior to the
toroid.
If any closed path is inside the inner edge of the toroid then ther is
no current enclosed. Therefore, by Ampere’s circular law B=0.
Magnetic field due to toroid
Consider the diagram below
Let r = mean radius of toroid
I = Current through toroid
n = number of turns permit length
B = magnitude of magnetic field inside the toroid
Then
FORCE ON A CURRENT CARRYING CONDUCTOR PLACED IN A MAGNETIC FIELD
We know that a moving charge in a magnetic field experiences a force
Now electric current in a conductor is due to the drifting of the force electrons in a definite direction in the conductor
When such a current carrying conductor is placed in a uniform magnetic field, each free electron experiences a force.
Since the free electrons are constrained in the conductor, the conductor itself experiences a force.
Hence a current carrying conductor placed in magnetic field experiences a force F.
Consider a conductor of the length L and area of cross section a placed at an angle ðœƒ to the direction of uniform of magnetic field B.
– is the angle between the plane of the conductor. The magnetic force experienced by the moving charge in a conductor is F = BQV Sin
For steady current I =
Q =I t
F=
The velocity for direct current is constant
V=
F = B I t
F= Force on the conductor (N)
B= Magnitude of the magnetic flux density of the field (T)
I = Current in the Conductor (A)
L= length of the conductor (M)
The current in the conductor I
Special cases
Thus if current carrying conductor is placed parallel to the direction of the magnetic field of the conductor will experience no force.
ii.)
F = BIL
Hence current carrying conductor will experience maximum force when it placed at right angles to the direction of the field.
One Tesla
Is the magnetic flux density of a field in which a force of IN acts on a 1M length of a conductor which is carrying a current of IA and is perpendicular to the field.
B = Tesla
The Direction of the force
Experiment shows that the force is always perpendicular to the plane which contains both the current and the external field at the site of the conductor
The direction of the force can be found by using Fleming’s left hand rule
Fleming’s left hand rule
States that if the first and the second fingers and the thumb of the left hand are placed comfortably at right angles to each other, with the first finger pointing in the direction of the current then thumb points in the direction of the force i.e. Direction in which Motion takes place If the conductor is free to move.
Maxwell’s Corkscrew rule
States that if a right handed corkscrew is turned so that its point travels along the direction, the direction of rotation of corkscrew gives the direction of the magnetic field.
FORCE BETWEEN TWO PARALLEL CONDUCTORS CARRYING CURRENTS
When two parallel current carrying conductors are close together, they exert force on each other.
It is because one current carrying conductor is placed in the magnetic field of the other
If currents are in the same direction the conductor attract each other and If currents are in the opposite directions conductors repel each other
Thus like currents attract, unlike currents repel.
Consider two infinitely long straight parallel conductors X and Y carrying currents I_{1} and I_{2 }respectively in the same direction.
Suppose the conductors are separated by a distance rin the plane of the paper.
As each conductor is in the magnetic field produced by the other, therefore each conductor experiences a force
The current carrying conductor Y is placed in the magnetic field produced by conductor X
Therefore force act on the conductor Y. The magnitude of the magnetic field at any point P on the conductor Y due to current I, in the conductor X is
By right hand grip rule ; the direction of B is perpendicular to the place of the paper and is directed inwards.
Now conductor Y carrying current is placed in the magnetic field produced by conductor X
Therefore force per unit length of conductor Y will experience a force given by
=
According to FLHR, force on conductor Y acts in the place of the paper perpendicular to Y and is directed towards to the conductor X.
Similarly, the Force on conductor X per unit length is = B_{y}I_{1}L
But
Hence when two long parallel conductors carry currents in the same direction they attract each other. The force of attraction per unit length is
This shows that the attraction between two parallel straight conductors carrying currents in the same direction in terms of magnetic field lines of conductors
It is clear that in the space between X and Y the two fields are in opposition and hence they tend to cancel each other
However in the space outside X and Y the two fields assist each other. Hence resultant field distribution will be
If two straight current carrying conductors of unequal length are held parallel to each other then force on the long conductor is due to the magnetic field of the short conductor
I_{1} = Current through short conductor
l = Length of short conductor
I_{2} = Current through long conductor
L = Length of long conductor
If r is the separation distance between these parallel conductors
Force on Long conductor = force on short conductor
Force on each conductor is the same in magnitude but opposite in direction (Newton’s third law)
DEFINITION OF AMPERE
Force between two current currying conductors per unit length
If And r =1m then
Ampere
Is that steady current which when it is flowing in each of two infinitely long, straight parallel conductors which have negligible areas of cross – section and are 1m apart in a vacuum, causes each conducts to exert a force of N on each mete of the other.
WORKED EXAMPLES
 The plane of a circular coil is horizontal it has 20 turns each of 8cm radius A current of 1A flows through it which appears to be clockwise from a point vertically above it. Find the Magnitude of the magnetic field at the centre of the coil.
Solution
The magnitude of the magnetic field at the centre of the coil carrying current is given by,
As the currents appears to be clockwise from appoint vertically above the coil the direction of the field will be vertically downward (By R.H.G.R)

A wire placed along the SouthNorth direction carries currents of 5A
from South to North. Find the magnetic field due to a 1cm piece of wire
at a point 200cm NorthEast from the place.
Solution
By RHGR, The field is vertically vertical downwards
 A coil of radius 10cm and having 20 turns carries a current of 12A in a clockwise direction when seen from east. The coil is in North – South plane. Find the magnetic field at the centre of the coil.
Solution
The magnitude of the magnetic field at the centre of the coil
The electron of hydrogen atom moves along a circular path of radius 0.5 x 10^{10 }with the uniform speed of 4 x 10^{6 }m/s. Calculate the magnetic field produced by electron at the centre ( e= 1.6 x 10^{9}c)
Number the revolution made by the electron in 1 second is
Current =
I = 1.27 X 10^{16} X 1.6 X10^{19}
1_{S}
I = 2.04 X 10^{3}A
Magnetic field produced by the electron at the centre is
 A circular coil of 100 turns has a radius of 10cm and carries a current of 5A Determine the magnetic field
(i) At the centre of the coil
(ii) At a point on the axis of the coil at a distance of 5cm from the centre of the coil.
Solution
(i) Magnetic field at the centre of the coil is
= 4 x 10^{7} TA ^{1}
N = 100 turns
I = 5A
r = 10×10^{2}m
B = 4 x 10^{7}x 100 x S
2 X 0.1
B= 3.14 X10^{3} T
The magnetic field of the centre of the coil = 3.14 X10^{3} T
(ii) Magnetic field on the axis of the coil at a distance X from the centre is
= 4 x 10^{7} TA ^{1}
N = 100 turns
I = 5A
r = 10 x 10^{2}
x = 0.05m
 An electric current I is flowing in a circular wire of radius at what dose from the centre on the axis of circular wire will the magnetic field be 1/8^{th} of its value at the centre?
Solution
Magnetic field B at the centre of the circular coil is
Suppose at a distance X from the centre on the axis of the circular coil the magnetic field is
 In Bohr’s model of hydrogen atom the electron circulates around nucleus on a path of radius 0.51Å at a frequency of 6.8x is rev/second calculate the magnetic field induction at the centre of the orbit.Solution
The circulating electron is equivalent to circular current loop carrying current I given by
I = 1.6
I = 1.1 A
Magnetic field at the centre due to this current is
= 14T
 A long straight wire carries a current of 50A. An electron moving at 10^{7}ms is 5cm from the wire
Find the Magnetic field acting on the electron velocity is directed
(i) Towards the wire
(ii) Parallel to the wire
(iii) Perpendicular to the directions defined by I and ii
Solution
The magnetic field produced by current carrying long wire at a distance r
The field is directed downward perpendicular to the plane of the paper
( i) The velocity V_{1} _{ }is towards the wire. The angle between V_{I }and B is 90^{0} force on electron
F= BQV
F = 2x 10^{4} x 1.6×10^{19}x10^{7}x Sin 90^{0}
F = 3.2 x 10^{16} N
(ii) When the electron is moving is moving parallel to the
wire ,angle between V2 and B is again 90Ëš Therefore, force is
again
3.2×10^{16}N
(iii) When the electron is moving perpendicular to the directions defined by (i) and (ii) the angle between V and B is O
F = O
 A solenoid has a length of 1 .23 m and inner diameter 4cm it has five layers of windings of 850 turns each and carries a current of 5.57A. what is the magnitude of the magnetic field at the centre of the solenoid
Solution
The magnitude of the magnetic field at the centre of a solenoid is given by
But
 A to void has a core ( non – ferromagnetic) of inner radius 20cm and over radius 25cm around which 1500 turns of a wire are wound. If current in the wire is 2A
Calculate the magnetic field
(i) Inside the to void
(ii) Outside the to void
Solution
( i) The magnitude of the magnetic field inside the toroid is given by
2
B = 0.003T
(ii)The magnetic field outside the toroid is Zero. It is all inside the toroid.
 A solenoid 1.5m long and 4cm in diameter possess 10 turns cm. A current of 5A is flowing through it. Calculate the magnetic induction
(i) Inside and
(ii) At one end on the axis of the solenoid
Solution
n = = =
(i) Inside the solenoid , the magnetic induction is given by
B =
B = 4
B =
(ii) At the end of the solenoid the magnetic induction is given by
 (a) How will the magnetic field intensity at the centre of a circular loop carrying current change, if the current through the coil is doubled and the radius of the coil is halved?
(b) A long wire first bent in to a circular coil of one turn and then into a circular
coil of smaller radius having n turns, if the same current passes in both the cases, find the ratio of magnetic fields produced at the centers in the two cases.
(c) A and B are two concentric coils of centre O and carry currents I_{A} and I_{B} as shown in figure
If the ratio of their radii is 1:2 and ratio of flux densities at O due to A and B is 1:3, find the value of
Solution
(a) Magnetic field at the centre of circular coil
( b)Suppose r is the radius of one turn coil and the r^{1 }is the radius of nturn coil. Then
N
First case Second case
Solution
C. Magnetic field at the centre of circular coil
 A helium nucleus makes a full rotation in a circle of radius 0.8m in two seconds. Find the value of magnetic field at the centre of the circle.
Solution
The charge on helium nucleus
Q= e
Q= 1.6 X10^{19}c
Current produced I =
I = 2 x 1.6 x 10^{19}
2
I =1.6 x10^{19}A
Magnetic field at the centre of the circle orbit of the helium is,
B=
B= 1.256 x 10^{25}T
 A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5×10^{4}m^{2} it is uniformly wound with 2000turns carrying a current of 2A and the magnetic flux in the iron is 8x 10^{3}Wb. What is the relative permeability of iron?
Solution
Length of ring l
l = 2
l = 2 x 0.10m
Number of turns per unit length n
n = =
If M is the absolute permeability of iron, then magnetic flux density of iron ring is
B =
B =
Magnetic flux
Magnetic flux = BA
Relative permeability of Iron μ_{r}
 Two flat circular coils are made of two identical wires each of length 20cm one coil has number of turns 4 and the other 2. If the some current flows though the wire in which will magnetic field at the centre will be greater?
Solution
For the first coil
For second coil
Therefore, magnetic field will be greater in coil with 4 turns
 A plat circular coil of 120 turns has a radius of 18cm and carries currents of 3A. What is the magnitude of magnetic field at a point on the axis of the coil at a distance from the centre equal to the radius of the coil?
Solution
Number of turns n = 120
Radius of the coil r = 0.18 m
Axial distance x = 0.18m
Current in coil I = 3A
^{ }B = (4 x 10^{7)} x 120 x3 x0.18^{2}
2(0.18^{2 +} 0.18^{2}) ^{3/2}
B= 4.4 x 104T
 A current of 5A is flowing upward in a long vertical wire. This wire is placed in a uniform northward magnetic field of 0.02T. How much force and in which direction will this field exert on 0.06 length of the wire?
Solution
B = 0.02T
I = 5A
L = 0.06
= 90^{0 F= 0.02 X 5 X 0.06Sin900 F = 0.006N}
By Fleming’s Left hand rules the force is directed towards West
 A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspend in mind air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field?solution
M = 200 X 10^{3} kg
I = 2A
l = 1.5m
B =?
F=BIL
Mg = BIL
B = Mg = 200 x 10^{3 }x 9.8
IL 1.5 X 2
B = 0.65T
 Two long horizontal wires are kept parallel at a distance of 0.2cm apart in a vertical plane . both the wires have equal currents in the same direction the lower wire has a mass of 0.05kg/m if the lower wire appears weightless what is the current in each wire ?
Solution
Let I amperes be the current in each wire the lower wire is acted upon by two forces.
Since the lower wire appears weightless the two forces were equal over 1m length of the wire
10^{4}I^{2 }= 0.49
 The horizontal component of the earth magnetic field at a certain place is 3 x 10^{5} and the direction of the field is from the geographic south to the geographic North A very long straight conductor is carrying a steady current of 1A. what is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) East to West
(b) South to North
Solution
(a) When current is flowing from east to west 90^{0}
Force on the conductor per unit
(b) When current is flowing from south to north = 0^{0}
Force on the conductor per unit length
 A horizontal straight wire 5cm long of mass 1.2gm^{1} placed perpendicular to a uniform magnetic field of 0.6T if resistance of the wire is 3.85cm^{1} calculate the P.d that has to be applied between the ends of the wire to make it just self supporting
Solution
The current (i) in the wire is to be in such a direction that magnetic force acts on it vertically upward. To make the wire self supporting its weight should be equal to the upward magnetic force.
Resistance of the wire
R = 0.05 x 3.8
= 0.19
Required P. (I) V = IR
V = 19.6X10^{3} X 0.19
V = 3.7 X 10^{3}V
 A conductor of length 2m carrying current of 2A is held parallel to an infinitely long conductor carrying current of 10A at a distance of 100mm. find the force on small conductor
Solution
I_{I} = 2A
I_{2} = 10A
r = 100 x 10^{3}m
l = 2m
Force on unit length of short conductor by the long conductor is give by
Force on length l = 2m of short conductor by the long conductor is
The force will be attractive if the direction of current is the same in two conduction and it will be repulsive if the conductors carry current in the opposite directions.
 In the figure below, determine the position between two wire which experience zero resultant force due to charge Q placed at that point
Solution
The force unit length acting in each wire of the parallel wire is given by
Let be the force per unit length in the wire carrying a current of 14A
Since F_{1} and F_{2} have the same magnitude but they are acting in opposite direction for resultant force to be zero
Assume that the charge Q is placed at a distance X from the wire carrying the
The charge Q is placed 4m from the either wire.
CLASSIFICATION OF MAGNETIC MATERIALS
All substances are affected by magnetic field some attain weak
magnetic properties and some acquire strong magnetic properties
and some acquire strong magnetic properties.
The magnetic properties of the substances are explained on the basis of modern atomic theory.
The atoms that make up any substance contain electrons that orbit around the central nucleus.
Since the electrons are charged they constitute an electric current and therefore produce magnetic field .
Thus an atom behave as a magnetic dipole and possesses magnetic dipole moment.
The magnetic properties of a substance depend upon the magnetic moments of its atoms.
IMPORTANT TERMS USED IN MAGNETISM
The following terms are used in describing the magnetic properties of the materials:
(i) Magnetic flux density (B)
Is a measure of the number of magnetic field lines passing per unit area of the material.
The greater the number of magnetic field lines passing per unit are of the material
(ii) Magnetic permeability
Is a measure of its conductivity for magnetic field lines
The greater the permeability of the material the greater is its conductivity for the magnetic field line and vice versa
Since magnetic field strength B is the magnetic field lines passing per unit area of the material, it is a measure of magnetic permeability of the material.
Suppose magnetic flux density in air or vacuum is . If vacuum/air is replaced by a material, suppose the magnetic flux density in the material becomes B
Then ratio B/ called the relative permeability . of the material.
(i) Relative permeability .
Is the ratio of magnetic flux density B in that material to the
magnetic flux density that would be if the material were
replaced by vacuum/ air .
Clearly is a pure number and its value per vacuum/air is 1
Relative permeability of a material may also be defined as the ratio of absolute permeability of the material to absolute permeability of vacuum/air.
(ii) Magnetizing force/ Magnetic intensity
Is the number of ampere – turns flowing per unit length of the toroid.
The SI Unit of magnetizing force H is Ampere – turns per meter (AT/m)
Consider a toroid with n turns per unit length carrying a current I. if the absolute permeability of toroid material is M, then magnetic flux density B in the material is
The quantity is called magnetizing force or magnetic intensity
Therefore, the ratio in a material I is from
; B=
Thus if the some magnetizing force is applied to two identical air cored and iron cored toroid, then magnetic flux density produced inside the toroid is
(iii) Intensity of magnetization ( ) is the magnetic moment developed per unit volume of the material.
When a magnetic material is subjected to a magnetizing force , the material is magnetized
Intensity of magnetization is the measure of the extent to which the material is a magnetized and depends upon the nature of the material
where:
= magnetic moment developed in the material
V= volume of the material
If m is the pole strength developed,
is the area of X – section of the material and 2l is the magnetic length. Then
Hence Intensity of magnetization of a material may be defined as the pole strength developed per unit area of cross – section of the material.
Thus the SI unit of I is Am^{1} which is the same as the SI unit of H
Magnetic susceptibility is the ratio of intensity of magnetic on I developed in the material to the applied magnetizing force H
The magnetic susceptibility of a material indicates how easily the material can be magnetized.
The unit of I is the same as that of H so that is a number
Since I is magnetic moment per volume is also called volume susceptibility of the material .
Consider a current carrying toroid having core material of relative permeability
The total magnetic flux density B in the material is given by
Where
= magnetic flux density due to current in the coils.
= magnetic flux density due to the material (Magnetization of the material)
…………………(i)
…………………(ii)
Here I is the intensity of magnetization induced in the toroid material
B = +
Now,
Equation (iii) give the relation between relative permeability (μ_{r} ) and magnetic susceptibility (X_{m}).
CLASSIFICATION OF MAGNETIC MATERIALS
All materials or substances are affected by the external magnetic field. Some attain weak magnetic properties and acquire strong magnetic properties.
On the basis of their behavior in external magnetic field , the various substance classified into the following three categories
(i)Diamagnetic materials
(ii)Paramagnetic materials
(iii)Ferromagnetic materials
(i) DIAMAGNETIC MATERIAL
When a diamagnetic substance is placed in a magnetic field in the
magnetic field lines prefer to passs through the surrounding air rather
than through the substance.
Diamagnetic materials are materials which can not be affected by the magnetic field.
They are repelled by magnetic field e.g. lead, silver, copper, zinc, water, gold bismuth etc.
These substances when placed in a magnetic field are weakly magnetized in a direction opposite to that of the applied field.
PROPERTIES OF DIAMAGNETIC MATERIALS
 A diamagnetic substance is feebly repelled by a strong magnet.
 The magnetic susceptibility ( ) of a diamagnetic substance has a small negative value.
 The relative permeability ( ) of a diamagnetic substances is slightly less than 1
 When a rod of diamagnetic substances is suspended freely in a uniform magnetic field, the rod comes to rest with its axis perpendicular to the direction of the applied field.See figure below
This gives the relation between relative permeability and magnetic susceptibility of the material.
(ii)PARAMAGNETIC MATERIALS
Are materials which when placed in a magnetic field are weakly magnetized in the direction of the applied field
The paramagnetic substances include the Aluminum antimony , copper sulphate, Crown grass etc
Since the weak induced magnetic field is in the direction of the applied field, the resultant magnetic field in the paramagnetic substance is slightly more than the external field
Hence the magnetic susceptibility of a paramagnetic substance is positive having
It clear that the relative permeability for such substances will be slightly more than 1
= 1 +
Paramagnetic substance loses its magnetism as soon as the external magnetic field is removed
BEHAVIOR OF PARAMAGNETIC SUBSTANCES IN AN EXTERNAL MAGNETIC FIELD
When a paramagnetic substance is placed in an external magnetic field the dipoles are partially aligned in the direction of the applied field.
Therefore the substance is feebly magnetized in the direction of the applied magnetic field. This result into a weak attractive force on the substances.
In the absence of the external magnetic field the dipoles of the paramagnetic substances are randomly oriented and therefore the net magnetic moment of the substance is zero.
Hence the substance does not exhibit Para – magnetism
PROPERTIES OF PARAMAGNETIC SUBSTANCES
 The relative permeability of a paramagnetic substance is always more than 1
The result field B inside a paramagnetic substance is more than the external field Bo
 The magnetic susceptibility of the paramagnetic substance has small positive value
It is because and
 The magnetic susceptibility of a paramagnetic material varies inversely as the absolute temperature
Paramagnetism is quite sensitive to temperature. The lower the temperature the stronger is the paramagnetism and vice versa
 A paramagnetic substance is feebly attracted by the strong magnet. It is because a paramagnetic substance develops weak magnetization in the direction of the applied external magnetic field
 When a paramagnetic substance is placed in a magnetic field, the magnetic field lines of force prefer to pass through the substance rather than through air.
Therefore the resultant field B inside the substance is more than the external field Bo
FERROMAGNETIC MATERIALS
Are the materials which when placed in a magnetic field are strongly magnetic in the direction of the applied field.
Ferromagnetic substances includes
 Iron
 Cobalt
 Nickel
 Fe_{2}O_{3}
 Gadolinium
Since the strong induced magnetic field is in the direction of the applied magnetic field, the resultant magnetic field inside the ferromagnetic substance is very large compared to external field
It is clear that ferromagnetism is very stronger form of magnetism. When external field (magnetic field) is removed some ferromagnetic substances retain magnetism
PROPERTIES OF FERROMAGNETIC SUBSTANCES
1. The relative permeability ( ) of the ferromagnetic substance is very large
Now
The resultant field B inside a ferromagnetic substance is very large as compared to the external filed Bo
 The magnetic susceptibility ( ) of a ferromagnetic substance is positive has a very high value.
It is because = 1 and 1 for this reason, ferromagnetic substance can be magnetized easily and strongly.
 A ferromagnetic substance is strongly attracted by a magnet
 When a rod of ferromagnetic substance is suspended in a uniform magnetic field, it quickly aligns itself in the direction of the field.
 They retain their magnetization even when their magnetizing force is removed.
 When a ferromagnetic substance is placed in a magnetic field the magnetic field lines tend to crowd into the substance
DOMAIN
Is the region of the space over which the magnetic dipole movements of the atoms are aligned in the same direction.
(i) In the absence of the external magnetic field the domain of the ferromagnetic materials are randomly oriented as shown below.
In other words, within the domain all the magnetic moments are aligned in the same direction but different domains are oriented randomly in different direction.
The result is that one domain cancels the effect of the other so that the net magnetic moment in the material is zero.
Therefore a ferromagnetic material does not exhibit magnetism in the normal state
(ii) When a ferromagnetic substance is placed in an external magnetic field a net magnetic moment develops the substance.
This can occur in two ways
(a) By displacement of boundaries of the domains i.e. the domains that already happen to be aligned with the applied field may grow in size whereas those oriented opposite to the external field reduce in size.
(b) By the rotation of the domains i.e. the domains may rotate so that their magnetic moments are more or less aligned in the direction of the magnetic field.
The result is that there is net magnetic moment in the material in the direction of the applied field.
Since the degree of alignment is very large even for a small external magnetic field the magnetic field produced in ferromagnetic material is often much greater than the external field.
CURIE TEMPERATURE
Is the temperature at which the ferromagnetic substance becomes paramagnetic
It is also known as Curie point of the substance
Ferromagnetism decreases with the increases in temperature
When a ferromagnetic substance is heated magnetization decreases because random thermal motions tend to destroy the alignment of the domains
At sufficiently high temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic.
In a ferromagnetic substance the atom appear to be grouped magnetically into what are called domains.
This occurs because the magnetic dipole moments of atoms of a paramagnetic substance exert strong force on their neighbor so that over a small region of space the moments are aligned with each other even with no external field.
Above Curie temperature these forces disappear and ferromagnetic substances become paramagnetic.
HYSTERESIS
Is the phenomenon of lagging of flux density (B) behind the magnetic force (H) in ferromagnetic materials subjected to cycles of magnetization.
When a ferromagnetic substance e.g. iron is subjected to cycle of magnetization (i.e. it is magnetized first in one direction and then in the other) it is found that flux density B in the materials lags behind the applied magnetizing force H.
This phenomenon is known as Hysteresis.
If a piece of ferromagnetic material is subjected to one cycle of magnetization the result BH curve is a closed loop a b c d e f a called Hysteresis loop.
B always lags behind H, Thus at point b, H is zero but flux density B has a finite positive value ob similarly at point e, H is zero but flux density B has a finite negative value X .
HYSTERESIS LOOP
Consider an Iron cored toroid carrying current I
If N is the total number of turn and l the length of toroid, then magnetizing force is
The value of H can be changed by varying current in the coil
Consider that when the Iron cored toroid is subjected to a cycle of magnetization the resultant B H curve traces a loop a b c d e f a called hysteresis loop
(i) To start with the toroid is unmagnetised and its situation is represented by point O in graph
As H is increased ( by increasing current I),B increases along and reaches its saturation value at a this stage
(i) all the domains are aligned
(ii) If now H is gradually reduced by decreasing current in the toroid it is found that curve follows the path instead of
At point b, H = O but flux density in the material has a finite value of +Br called residual flux density
REMANENCE
Is the flux density left behind in the sample after the removal of the magnetizing force (H). It is also called Residual magnetism or retentively.
B lags behind H. This effect is called Hysteresis
(iii) In order to reduce flux density in the material to zero, it is necessary to apply H in the reverse direction
This can be done by reversing the current in the toroid
When H is gradually increased in the reverse direction the curve follows the path
At point C, B =O and H = HC, the value of H needed to wipe out residual magnetism is called coercive force it.
COERCITIVITY OF THE SAMPLE
Is the value of reverse magnetizing force required to wipe out the residual magnetism in the sample
Now H is further increased in the reversed direction until point d is reached where the sample is saturated in the revision direction ( ).
If the H is now reduced to zero, point e is reached and the sample again retain magnetic flux density ( )
The remaining part of the loop is obtained by increasing current to produce H in the original direction .
The hysteresis loop results became the domains do not become completely unaligned when H is made zero.
The area enclosed by the hysteresis loop represents loss in energy
This energy appears in the material as heat.
HYSTERESIS LOSS
This is the loss of energy in the form of heat when a ferromagnetic material is subjected to cycles of magnetization.
Hysteresis loss is present in all those electrical machines whose iron parts are subjected to cycles of magnetization.
The obvious effect of hysteresis loss is the rise in temperature of the machine.
HYSTERESIS LOOP
Is the loop traced by the resultant BH curve when the Iron – cored toroid is subjected to a cycle of magnetization.
The shape and size of hysteresis loop largely depends upon the nature of the material
The choice of a ferromagnetic material per a particular application often depends upon the shape and size of the hysteresis loop
(i) The smaller the hysteresis loop area of a ferromagnetic material the smaller is the hysteresis loss
The hysteresis loop per silicon steel has a very small area.
For this reason, silicon steel is widely used per making transformer cores and rotating machines which are subjected to rapid reversals of magnetization
(ii) The hysteresis loop per hard steel indicates that this material has high retentivity and Coercivity.
Therefore hard steel is quite suitable for making permanent magnets.
But due to the large area of the loop there is a greater hysteresis loss
For this reason, hard steel is not suitable for the construction of electrical machines
(iii)The hysteresis loop for wrought iron shows that this material has fairly good residual magnetism and Coercivity.
Hence it is suitable for marking cores of electromagnets
APPLICATIONS OF FERROMAGNETIC MATERIALS
Ferromagnetic material (E.g. iron, steel nickel, cobalt etc) are widely used in a number of applications
The choice of ferromagnetic material for a particular for a particular application depends upon its magnetic properties such as
(i) Retentivity
(ii)Coercivity
(iii) Area of the hysteresis loop
Ferromagnetic materials are classified as being either
(i) Soft (soft iron)
(ii) Hard (steel)
Figure below shows the hysteresis loop for soft and hard ferromagnetic materials
The table below gives the magnetic properties of hard and soft ferromagnetic materials
Magnetic property  Soft  Hard 
Hysteresis loop  narrow  Large area 
Retentivity  High  High 
Coercivity  low  high 
(i) PERMANENT MAGNETS
The permanent magnets are made for hard ferromagnetic materials (steel, cobalt, carbon steel)
Since these materials have high relativity the magnet is quite strong
Due to their high Coercivity, they are unlikely to be demagnetized by stray magnetic field
(ii) TEMPORARY MAGNETS / ELECTROMAGNETIC
The Electromagnets are made from soft ferromagnetic materials e.g. soft iron.
Since these materials have low coercively they can be easily demagnetized.
Due to high saturation flux density they make strong magnets
(iii) TRANSFORMER CORES
The transformer cores are made from soft ferromagnetic materials
When a transformer is in use, its core is taken through many cycles of magnetization
Energy is dissipated in the core in the form of heat during each cycle. The energy dissipated is known as hysteresis loss. And is proportional to the area of hysteresis loop
Since the soft Ferromagnetic materials have narrow hysteresis loop (smaller loop areas) they are used for making transformer cores.
WORKED EXAMPLE
1. (a) How does a permanent magnet attract an unmagnetised iron object?
(b) Show that the unit of magnetizing force is Nm^{2}T^{I} or Jm^{I}wb^{I}
(C) Why is electromagnets made of soft iron?
(d) An Iron ring has a cross – sectional area of 400 and a mean diameter of 25cm. It is wound with 500trns. If the relative permeability of iron is 5000 find;
(i) The magnetizing force
(ii) The magnetic flux density set up in the ring.
The coil resistance is 474 and the supply voltage is 240V
Solution
( a) The magnet’s field cause a slight alignment of the domains in the unmagnetised iron object so that the object becomes a temporary magnet with its north pole facing the south pole of the permanent magnet and viceversa. Therefore attraction results.
(b) The SI units of magnetizing (H) are Ampere/M(Am^{I} )
Now
Also
( c) The soft Iron has very small residual magnetism and coercive force. Therefore the material loses magnetism as soon as the magnetizing force is removed for this reason electromagnets are made of soft iron
(d) Current through the coil I
Mean length of the ring l
l = 2
l = 2 x (12.5 x 10^{2)}
^{ } l = 0.785m
(i) Magnetizing force H
(ii) Magnetic flux density B
From
B = 2.02wb/m^{2}
(a) Diamagnetic is a property of a material. Discuss
(b) What is the magnetic susceptibility and permeability of a perfectly diamagnetic Substance?
(c) Why is Diamagnetism independent of temperature?
(d) The core of a toroid having 3000 turns has inner and outer radii of 11cm and 12cm respectively. The magnetic flux density in the core per a current of 0.70A is 2.5T
What is the relative permeability of the core?
Solution
(a) Diamagnetism is a natural reaction to the applied magnetic field. Therefore it is present in all materials but is weaker even than paramagnetism. As the result, diamagnetism is overwhelmed by paramagnetic and ferromagnetic effects in materials that display these other forms of magnetism
Solution
B = H + I
B = (H+I)
For perfectly diamagnetic substance
B = O
(H+I) = O
H +I = O
I = H
Susceptibility
=
Also
= 1 +
= 1 – 1
= O
(c) The induced magnetic moment in atoms of a diamagnetic substance is not affected by the thermal motion of the atoms. For this reason, diamagnetic is independent of temperature.
(d) Solution
Mean radius of toroid
The magnetic flux density in the toroid is
3 .(a) (i) What is a non Magnetic material?
(ii) Can there be a material which is non magnetic?
(b) What do you mean by the greater susceptibility of a material?
(c) An iron rod of 0.1m^{2} area of xsection is subjected to magnetic field of 1000
Calculate its magnetic permeability. Given susceptibility of iron are 599.
(d) Which material is used to make permanent magnets and why?
Solution
(a) (i) A non –magnetic material is that which is not affected even by strong magnetic fields.
(ii) No, every material is at least diamagnetic.
(b) From
For a given H, α Thus the greater value of the susceptibility of the material the greater will be its intensity of magnetization i.e. more easily can be magnetized .Thus greater value of its susceptibility for iron means that it can be easily magnetized.
(c) Solution
= 1 +
But
=
= (1+ )
=4 ^{7}(1+599)
=7.54 x 10^{4}TA^{4}M
(d) Steel it is because has high coactivity. This ensures the stay of magnetism in steel for a longer period
 (a) A toroid of mean circumference 50cm has 500 turns and carries a current of 0.15A
(i) Determine the magnetizing force and magnetic flux density if the toroid has an air core
(ii) Determine the magnetic flux density and intensity of magnetization if the core is filled with iron of relative permeability 5000
(b) Why do magnetic lines of force prefer to
(c) What is the SI unit of magnetic susceptibility?
Solution
(a) (i) given
L = 50cm = 50 x 10^{2} m
N = 500 turns
I = 0.15A
Magnetizing force H
Magnetic flux density
(b) It is because permeability of Iron (Ferromagnetic material ) is very high as compared to that of air .
Therefore, have no Units
 (i) Figure below shows the variation of intensity of magnetization (I) versus the applied magnetic field intensity (H) for two magnetic material A and B
(a) Identify the materials A and B
(b) For the material A, plot the variation of I with temperature.
(ii) The relative permeability of a material is
(i) 0.999
(ii) 1.001
Solution
The slope IH graph gives the magnetic susceptibility of the material
For material A the slope is positive and has a small value. Therefore, material A is paramagnetic.
For material B, the slope is position and has a large value. Therefore , material B is ferromagnetic
(a) The intensity of magnetization of a paramagnetic material A is inversely proportional to the absolute temperature therefore I – T graph for material A will be as shown in figure below .
(ii) (i) Diamagnetic material
(ii) Paramagnetic material
 (a) Graph below shows the variation of intensity of magnetization(I) versus the applied
Magnetic field intensity (H) for two magnetic material A and B
(i) identify the materials A and B
(ii) draw the variation of susceptibility with temperature for material B
(b)A magnetizing for of 360 produce a magnetic flux density of 0.6T in a ferromagnetic material. Calculate
(i) Permeability
(ii) susceptibility of the material
Solutions
(a) for material A the susceptibility ( = slope of I – H graph) is small and positive therefore material A is paramagnetic and for material B, Susceptibility is small and Negative. Therefore , material B is Diamagnetic
iii) The susceptibility of a diamagnetic material B is independent of temperature therefore – T graph for material B will be shown in graph below.
(b) (i) Permeability of material
= 1.67 X 10^{3} A ^{1}Tm
(ii) Susceptibility of the material
= (1 + )
= 1328.62 Am ^{1}
 (a) What is Magnetic solution?
(b) Name two materials which have
(i) Position susceptibility
(ii) Negative susceptibility
(c) Obtain the earth’s magnetization , assuming that the earth’s field can be approximated by a giant bar magnet of magnetic moment 8.0 x 10 ^{22} A ? Radius of earth = 6400km
(d) A bar magnet has Coercivity of 4×10^{3} A/m it is desired to demagnetize it by inserting it inside a solenoid 12cm long and having 60turns. With current should be sent through the solenoid
Solution
(a) Magnetic saturation
Is the maximum magnetization that can be obtained in the material when all the domains of a ferromagnetic material are in the direction of the applied magnetic filed
(b) (i) paramagnetic material e.g. Aluminum and Antimony
(ii) Diamagnetic materials e.g. Copper and Zinc
(c) magnetic moment M = 8 x 10^{22}Am^{2} radius of the earth Re = 6400km
Magnetization of the earth is given by
The bar magnetic has a Coercivity of 4 i.e. it needs a magnetic intensity H = 4 to get magnetized
 (a) Define hysteresis loop(b)What does the area of hysteresis loop indicate?
(c) What is the use of hysteresis loop?
(d) Why is soft iron preferred in making the core of a transformer?
Solution
(a) hysteresis loop
Is the resulting B – H curve (Closed loop) obtained when a ferromagnetic material is subjected to one circle of the magnetization
(b) the area of hysteresis loop is a measure of energy wasted in a sample when it is taken through s complete of magnetization
(c) The hysteresis loop of a material tells us about hysteresis loss retentively and Coercivity. This knowledge helps us in selecting materials for making electromagnetic permanent magnets cores of transformer
(d) The area of hysteresis loop for soft iron is small. Therefore energy dissipated in the core for cycle magnetization is small. For this reason, the core of a transformer is made of soft iron
 (a) state curie law
(b) Give the graph between I and B/T
(c) What happens if an Iron magnet is melted?
(d) Copper Sulphate is paramagnetic with a susceptibility of 1.68×10^{4} at 293K. What is the susceptibility of copper at 77.4K if it fellows curie law?
Solution
(a) Curie law
States that intensity of magnetization (I) if a paramagnetic substance is directly proportional to the external magnetic field (B) and inversely proportional to the absolute temperature(T) of the substance.
I
I
Combining these two factors, we have
where
C is a constant of proportionality and is called curie constant
This law is physically reasonable As B increase the alignment of magnetic moments increases and therefore I increases
If the temperature is increased the thermal motions will make alignment difficult thus decreasing I
The curve law is found to hold good so long as does not become too large
Since
(b) The temperature of molten iron 770^{0}C is above the Curie temperature i.e. On malting the iron becomes paramagnetic. Therefore it loses its magnetism
Solution
According to curie law the susceptibility depends inversely on the temperature
 (a) A solenoid 0.6m long is wound with 1800 turns of copper wire. An iron rod having a relative permeability of 500 is placed along the axis of the solenoid. What are the magnetic intensity H and field B when a current of 0.9A flows through the wire? What is the intensity of magnetization I in the iron? Find the average magnetic moment per iron atom. Density of iron is 7850 Kg/m^{3}.
(b) An Iron sample having mass 8.4Kg is repeadly taken over cycle of magnetization at a frequency of 50cyles per second it is found that energy equal to 3.2 x J is dissipated as heat in the sample in 30Minutes if the density of the iron is 7200kg/m^{3} find the energy dissipated per unit volume per cycle in the iron sample.
(c) A domain in ferromagnetic iron is in the form of a cube of side length 1μM. Estimate the number of Iron atom in the domain and the maximum possibility dipole moment and magnetization of the domain and the maximum possible dipole moment and magnetization of the domain. The molecular mass of Iron is 55g/mole and its density is 7.9g/cm^{2} assume that each Iron atom has a dipole moment of 9.27 x 10^{24} Am^{2 }
Solution
H = 2700A/m
H
I = 1.35 X 10^{6}A/m
One Kilo mole (55.85Kg) of Iron has 6.2 x10^{26} atoms. Therefore, number of atoms in 1 m^{3} of Iron
Average magnetic moments per iron atoms
=1.59 x 10^{23} Am^{2}
Solution
f =50 HZ
cycle
Length of cubic domain l
=1 M = 10^{6}M
Volume of Domain
V = ^{3}
V = (10^{6})^{3}
V = 10^{18}M^{3}
Mass of domain
=Volume X Density
=10^{12}cm x7.9g
= 7.9 x10^{ 12}g/cm
It is given that 55g of Iron contain 6.023 x10^{23} Iron atoms (Avogadro’s no)
Number of atoms in the domain is
N = 6.023 X 10^{23} X 7.9 X10^{12}
55
N = 8.65 X 10^{10}atoms
The maximum possible dipole moment is achieved per the case when all the atomic domains are perfect aligned (This condition is unrealistic)
_{ }= (8.65 x10^{10}) x (9.27 x10^{24})
= 8 x 10^{13} AM^{2}
Maximum intensity of magnetization of the Domain is
= =
= 8 x10^{5} Am^{1}
NUMERICAL PROBLEMS
 (1)The magnetic moment of a magnet (10cm x 2cm x 1cm) is 1AM? What is the intensity of magnetization?
I = 5 X A/m
 (2)An iron rod of cross sectional area 4 is placed with its length parallel to a magnetic flied of intensity 1600 A/M the flux through the rod is 4 x 10^{4}Wb what is the permeability of the material of the rod?
μ = 0.625 x10^{3} Wb A ^{1} m^{1}
 (3) A toroid winding carrying a current of 5A is wound with 300turns/miter of core. The core is Iron which has a magnetic permeability of 5000Mo under the given conditions
Find (i) the magnetic intensity H
(ii) Flux density B
(iii) Intensity of magnetization I
 i) 1500AT/m
 ii) 9.43T
iii) 7.5 X10^{6}A/m
 (4)A specimen of Iron is uniformly magnetized by a magnetizing field of 500 A/m. if magnetic induction in the specimen is 0.2Wb/m^{2}, find the relative permeability and susceptibility
X_{m} = 317.5
M_{r} = 318.5
 Consider a toroid of 1000 turns and mean radius 25cm. what is the B field in the toroid if there is a current of 2A?
What will be the field when the toroid is filled with Iron per which μ = 100H/m?
= 1 .6 x 10 ^{3}
B = 0.16T
 An Iron of volume 10^{4}m^{3} and relative permeability 1000 is placed inside a long solenoid wound with storms/cm. if a current of 0.5A is passed through the solenoid, find the magnetic moment of the rod.
M = 25Am^{2}
 The flux through a certain toroid clangs from 0.65m Wb to 0.91M Wb when Air core is replaced by another material. What are
 i) The relative permeability
 ii) Absolute permeability of the material
= 1. 4
μ= 5.6 x10^{7}H/m
 Answer the following Questions
 a) Why does a paramagnetic sample display greater magnetization (per the same magnetizing field) when cooled?
 b) Why is diamagnetism, in contrast almost independent of temperature?
 c) Distinguish between a soft and a hard magnetic material, giving an example of soft magnetic materials are those which can easily be magnetized but do not retain their magnetism (retentively )
An example of soft magnetic material
Is soft Iron i.e. Iron in a reasonably pure state. It is otherwise known as wrought iron
Hard magnetic material
Are those which are difficultly to magnetic but once magnetized, can retain the magnetism per long
These are usually used making permanent magnetic
An example of hand magnetic material is steel which consists of iron and a small % of carbon
MOTION OF CHARGED PARTICLE IN UNFORM MAGNETIC FILED
Consider a charged particle of charge +Q and mass M moving with a velocity V in the plane of the paper.
Suppose this charged particle enters a uniform magnetic filed B which is perpendicular to the plane of the paper and directed outward
Clearly the entry of the charged particle is at right angles to the magnetic field
The force i.e. magnetic force F_{m} on the charged particle is given by
F_{m} = BQV
The magnetic force Fm acts at right angle to the plane containing V and B
On entering the magnetic field at M the charged particle experiences a force of magnitude and is deflected in the direction shown
This force is at right angle to the direction of motion of the charge particle and therefore, cannot change the speed of charge particle it only charge its direction of motion
A moment later, then the particle reaches point N the magnitude of force Fm acting on it is the same as it was at M but the direction of force is different (Fm is still perpendicular to V )
Thus the force is perpendicular to the direction of motion of the charged particle at all times and has a constant magnitude
The magnetic force does not change the speed or kinetic energy of the charge particle it only charges the direction of the charged particle
When the moving charged particle is inside the uniform magnetic field, it moves along a circular path.
When the initial velocity of the particle is parallel to the magnetic field
= 0^{0}
From
F_{m }= BQV
Fm = BQV
Fm = 0
Thus in this case the magnetic field does not exert any force on the charge particle
Therefore the charged particle will continue to move parallel to the magnetic field then = 180^{0}
Therefore, the particle will continue to the move in the original direction.
When the initial velocity of the particle is perpendicular to the magnetic field =90^{0}
From
Fm = BQV
Max. Value Fm = BQV
PARAMETERS OF MOTION
A force of constant magnitude always acts perpendicular to the direction of motion of the charged particle.
Therefore , provides the necessary centripetal force to more the charged particle in a circular path in the circle of radius r perpendicular to the field
 i) RADIUS OF PATH
The acceleration of a particle moving along a circular path of radius r is given by
For a given charge mass and magnetic field r V. this means that fast particles move in large circles and slow ones in small circles.
 ii) TIME PERIOD
The time taken by the charged particle to complete one circular revolution in the magnetic field is its Time period T
From
Thus Time period of the charged particle is independent of the speed (V) and the radius of the path
It only Depends on the magnitude of B and charge to mass ratio of the particle .
FREQUENCY
The number of circular revolutions made by the charged particle in one second is its frequency f
f =
f = 1
There Frequency of the charged particle is also independent of speed (V) and radius (r) of the path
ANGULAR FREQUENCY
From
=2πf
But
Then
Again Angular frequency of the charged particle is independent of the speed (V) and radius (r) of the path..
Since T, f and of a charged particle moving in a magnetic field are independent of its speed (V) and the radius (r) of the path.
In fact all the charged particles with same Q/M and moving in a uniform magnetic field B will have the same value of T, f and w
MOTION OF CHARGED PARTICLE ENTERING UNIFORM MAGNETIC FIELD AT AN ANGLE
Suppose the charged particle moving with velocity V enters a uniform magnetic field B making an angle to the direction of the field
Diagram
The velocity V can be resolved into two rectangular components
 i) V_{1} = V Acting in the direction of the field
 ii) V_{2} = Vsin acting perpendicular to the direction
The perpendicular component V_{2} moves the charged particle in a circular party while the horizontal component moves it in the direction of the magnetic field
In other words, the charged particle corers circular path as well as linear path. Consequently the charged particle will follow a helical path.
The charged particle rotates in a circle at speed V_{2 }while moving in the direction of the field with a speed V_{I}
PARAMETERS OF MOTION
The perpendicular component of velocity V_{2 }determines the parameters of the circular motions while the horizontal component of velocity V_{I} decides the pitch of helix
(i) Radius of path
From
(ii) T, f and w
Since time period (T), frequency (f) and Angular frequency(ω) of a charged particle moving in a uniform magnetic field are independent of speed V and radius (r) of the path, these values remain the same
(iii) Pitch of helix (d)
It is the linear distance covered by charged particle when it completes
one circular revolution Or It is the linear distance covered by charged
particle during time T
d = T
d = V T.
d = V
The following points may be noted about the behavior of charged particle in a Uniform magnetic field
(i) If a charged particle is at rest V=0 in a magnetic field, it experiences no force
From
= BQV
= BQ
= 0
(ii) If a moving charged particle enters a uniform magnetic field at right angles to the field it describes a circular path
(iii) If a moving charged particle enters a uniform magnetic field. Making an angle to the direction of the field it describes a helical path
(iv) A moving charged particle in a magnetic field experience maximum force when angle between V and B is 90^{0}
(v) Since magnetic force does not change the speed of a charged particle it means that K.E of the charged particle remains constant in the magnetic field.
(vi) Since magnetic force (Fm) is perpendicular to V, it does not work. Therefore work done by the magnetic force on the charged particle is zero
WORKED EXAMPLE
 An electron and a proton moving in a circular path at 3×10^{6} Ms^{1} in a uniform magnetic field of magnitude 2 x 10^{4}T. Find the radius of the path
Solution
 An electron and a proton moving with the same speed enter the same magnetic field region at right angles to the direct of the field. For which of the two particles will the radius of circular path be smaller?
Solution
From
Since the mass of electron is less than that of the proton the radius of the circular path of electron will be smaller.
 (a) What will be the path of a charged particle moving along the direction of a uniform magnetic field?
(b) A moving charged particle enters a magnetic
Solution
(a) When a charged particle moves along the direction of a uniform magnetic field ,it experiences no force = 0^{0} therefore the charged particle will more along its original straight path
(b) Helical path since the velocity of the charged particle can be resolved into two rectangular components one along the field, and the other perpendicular to the field. The velocity component perpendicular to the field causes the charged particle to more in a circular path while the velocity component along the field cause it to more it in the direction of the field. The combination of these two motions course the charged particle to move in a helical path.
 A become of αparticles and of proton of the same velocity V, entries a uniform magnetic field at right angles to the field lines. The particle describes circular paths. What is the ratio of radii of the two paths?
Solution
Radius of αparticle path
=
Of proton path
= ……………………….
Take equation (i) equation (ii)
Therefore, radius of x – particle path is twice that of proton’s path
Note
α – particle
Charge = 2e
Mass = mass of helium nucleus
 A proton with charge – mass ratio of 10^{8 }CKg ^{1 }is moving in a circular orbit in a uniform magnetic field of 0.5T. calculate the frequency of revolution
f = BE
2πM
F = B. e
2π M
F = 0.5 . 10^{8}
2π
 (a) What happens when a charged particle is projected perpendicular to a uniform magnetic field ?
(b) A beam of protons moving with a velocity of 4 x 10^{5} M5^{1} enters a uniform field of 0.3T at an angle of 60^{0} to the direction of a magnetic field find
(i) The radius of helical path of proton beam
(ii) Pitch of helix
Given mass of proton = 1.67 x 10 ^{27} Kg charge on proton = 1.6 x 10 ^{19}C
Solution
(a) When a charged particle is projected perpendicular to a uniform magnetic field
(i) Its path is circular in plane perpendicular to B and V
(ii) Its speed and Kinetic energy remain the same
(iii) The magnitude of force remains the same Fm = BQV. Only the direction of velocity of the particle charges.
(iv) The force acting on the particle is independent of the radius of the circular path
(v) The time period of revolution of the particle is independent if V and r
(b) (i) Solution
r = 1.2 x10^{2}M
(ii) Pitch of helix of d
d = 4.37 x10^{2}M
 (a) A particle of charge ðœƒ moves in a circular path of radius r in a uniform is P = BQr
(b) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0KV enters a region of magnetic field of 0.15 determine the trajectory of the electron if the field
(i) is traverse to its initial velocity
(ii) makes an angle of 30 with the initial velocity
 c) A proton a deuteron and an α – particle whose kinetic energies are same enter perpendicularly to a uniform magnetic field. Compare the radii of their circular paths
(a) Solution
The magnetic force Fm provides the necessary centripetal force F_{c}
F_{m} = F_{c}
BQ =
Momentum P = BQr
(b) Solution
When an electron (e) is accelerated through a p.d of V, it acquires energy eV. IF Vis the velocity gained by electron, then
V =
V=9X10^{31}
(i) Force on the electron due to transverse field is
Fm = BQr ðœƒ = 90^{0}
Since magnitude of Fm is constant and Fm is perpendicular to both V and B the electron will move in a circle of radius r. The necessary centripetal force is provided by Fm
r = 10^{3}M
(ii) When electrons enters the magnetic field making an angle ðœƒ = 30^{0} with the field
r = 0.5 x10^{3}M
(C) Solution
Let 1, 2 and 3 be the suffix force proton, deuteron and α – particle respectively
K .E, = K.E_{2 }= K.E_{3}
1 M_{I}V_{1} = 1 M_{2}V_{2}= 1 M_{3}V_{3}
2 2 2
If MI then M_{2} = 2M_{ }and M_{3} = 4M
MV^{2}_{I} = 2MV^{2}_{2} = 4MV^{2}_{3}
V_{I }= ^{2} = 2V_{3}
V_{2 }= VI
Also
Radius of the path r
r = MV
BQ
If QI = Q, Then Q2 = Q and Q3 = 2Q
r_{I} = M_{I}V_{I } = MV_{I}
BQ_{I} BQ
r_{2} = M_{2}V_{2} = 2M. V_{1}
BQ_{2} BQ
r_{I:} r_{2 }: r_{3} = MV_{I} : MV_{I} : MV_{I}
BQ BQ BQ
r_{I} : r_{2} : r_{3} = 1 : : 1
NUMERICAL PROBLEMS
1. What is the radius of the path of an electron (mass 9 x 10^{31})Kg and charge(1.6 x 10^{16}C) moving at a speed of 3 x10^{7}m/s in a magnitude field of 6 x 10^{4}T perpendicular to it? What its frequency? Calculate its energy in KeV(1Ev=1.6×10^{19 }J)
r= 0.28m
f= 1.7 x10^{7 }Hz
E = 2.53KeV
 An electron after being accelerated through a p.d of 100V enters a uniform magnetic field of 0.004T perpendicular to its direction of motion. Find the radius of the path described by the electron
r = 8.4 10^{3}m
 An αparticle is describing a circle of radius of 0.45m in a field of magnetic Induction 1.2Wb/m^{2}. Find its speed, frequency of rotation and kinetic energy. What potential difference will be required which will accelerate the particle so as to this much energy to it? The mass of αparticle is 6.8 x10^{27}Kg and its charge is 3.2 x 10^{19 }C
V=2.6 X10^{7 }m/s
f= 9.2 x10^{6 }sec^{1}
MAGNETIC TORQUE ON RECTANGULAR COIL IN UNIFORM FIELD
Consider a rectangular conductor ABCD having length L and Nturns
carrying a current I and placed in a magnetic field between N and Spole
of bar magnets.
Since the velocities of the electrons in the sides side AB and BC are perpendicular to the magnetic induction B then these sides will experience the maximum magnetic forces equal to F.
F= BIL
The direction of the magnetic force in these two sides is given by Fleming’s Left Hand Rule as shown in the figure below.
The two parallel and equal forces will constitute the turning of the rectangular called the magnetic torque.
Mathematically
The magnetic torque or couple is given by
= Total force x Perpendicular distance
τ=BIL x b
τ=BI (L x b)
Lb = crosssectional area=A
For the rectangular coil of N turns
ELECTROMAGNETIC MOMENT(M)
This is the magnetic torque acting on the coil when it is parallel to a
uniform field whose flux density is one tesla.It is the property of the
coil is defined as the couple required to hold the coil at right angles
to
field.
Thus, in equation τ=m when B=1I
from
Consider a rectangular coil ABCD placed at an angle to the magnetic field of flux density B
The perpendicular distance between the parallel force is and not b. Then the magnetic force or compile is given
Special cases
(i) When
Plane of the loop is parallel to the direction of magnetic field
Thus, the Torque on a current loop is maximum when the plane of the loop is parallel to the direction of magnetic field is given by
(ii) when ðœƒ = 90^{0}
Plane of the loop is perpendicular to the direction of magnetic field
Thus , the torque on a current loop is minimum (zero) when the plane of the loop is perpendicular
(iii) When B = 1T
when B = 1T, then the magnetic torque is numerical equal to magnetic moment
From
B = 1T
MAGNETIC TORQUE AT AN ANGLE α BETWEEN THE AXIS OF THE COIL AND NORMAL TO THE PLANE OF THE COIL
We can also express torque in another useful form. If normal to the plane of coil makes an angle α with the direction of the magnetic field
From
τ = BANI cos (90 – α)
τ = ANIBSin α
Since M = IAN
When a current carrying coil is placed in a uniform magnetic field, torque acts on it which tends to rotate the coil so that the plane of the coil is perpendicular to the direction of magnetic field.
WORK DONE BY TORQUE
If the magnetic torque displaces the coil through the small angular displacement
The work done by the torque is given by
The total work done is obtained by integration the above equation within limits
If the magnetic field displaces the coil through small angular displacement the work done by torque is given by
dw = τdα
dw = mbSinαdα
WORKED EXAMPLES
1. A vertical rectangular coil of sides 5cm by 2cm has 10 turns
and carries a current of 2A. Calculate the torque on the coil when it is
placed in a uniform horizontal magnetic field of 0.1T with its plane.
(a) Parallel to the field
(b) Perpendicular to the field
(c) 60^{0 }to the field
Solution
The area of the coil
A= (5 x 10 ^{2}) x (2 x10^{2)}
A= 10^{3 }m^{2 }
(a) From
A=10^{3}
I = 2A
N=10
B=0.1T
τ= 0.1 X10^{3} X10 X2
τ = 2 X10^{3}NM
(b)
τ=BANI
τ= 0
(c)
τ = BANICos ðœƒ
τ= 0.1 x10^{3} x10x2xCos 60^{0}
τ= 10 ^{3}NM
 2.Given a uniform magnetic field of 100T in East to West direction and a 44cm long wire with a current carrying capacity of at most 10A. what is the shape and orientation of the loop made of this wire which yields maximum turning effect on the loop?
Solution
A current carrying planar loop will experience maximum together if its area to the direction of the magnetic field for a given perimeter , a circle has the maximum area.
If 44cm wire is bent into a circular
2πr = 44
πr = 22
22r = 22
7
r = 1
7
R = 7cm
Area of loop = πr^{2}
= πx7^{2}
= 154cm^{2}
= 154×10^{4}m^{2}
Magnetic toque τ
τ = BANI
τ = (154 X10^{4}) X 100 X100
τ = 150 T
 A circular coil of wire of 50 turns and radius 0.05 carries current of 1A the wire is suspended vertically in a uniform magnetic field of 1.5T. the direction of magnetic field is parallel to the plane of the coil
(a) Calculate the Toque on the coil
(b) Would your answer charged if the circular coil is replaced by a plane coil of some irregular shape that has the same area (all other particulars are unaltered? )
(a) Solution
B = 1.5T
A = πr^{2} = π (0.05) M^{2}
A = 7.85 X 10^{3}
N = 50
I = 1A
τ = 1.5 X (7.85 X10^{3}) X50X1
τ = 0.589NM
(b) Since torque on the loop is independent of its shape
provide area (A) remains the same the magnitude of the torque
will remain unaltered.
 A circular coil of 20turns and radius 10 cm is placed in a uniform magnetic field of 0.2T normal to the coil. If current in the coil is 5A find.
(i) Total torque on the coil
(ii) Total force on the coil
^{(iii) }Average force on each electron in the coil due to the magnetic field. The coil is made of copper wire of crosssectional area 10^{5}m^{2} and force of electron density in the wire is 10^{29}m^{3}
Solution
(i) The toque on the coil is given by
Since
τ = 0
(ii) The net force on a planar current loop in a uniform magnetic field is always zero
(iii) Magnetic force on each electron
F = BeV_{d}
F = Be.
F =
F= 10^{24}N
MOVING COIL METERS
A galvanometer deflects or measures small amount of current passing through it and it gives the direction to which that current is flowing.
In these instruments a rectangular of fine insulated copper wire is suspended in an strong magnetic field as shown in the figure below. The field is set up between soft iron poler pierces Ns attacked to a powerful permanent magnet.
(i) Millimeter
The magnet field is radial to the core and pole pieces over the region which the coil can swing. In this case the deflected coil always comes to rest with the plane parallel to the field in which it then situated.
The moving coil galvanometer ha s has hair spring and jewel bearings. The coil is around in the rigid but light aluminium frame which also comes to carries a pivot. The current is led in and out of the springs.
Aluminum pointer P shows the deflection of coil, it is balanced by counter weight Q.
THEORY OF MOVING COIL GALVANOMETERS
The rectangular coil is situated in the radia field B when the current is passed into it the coil rotate to an angle Q which depend on the length of the spring.
No matter where the coil comes to rest, the field B in which it is situated always along the plane of the coil because the field is radial.
It is shown that the torque T of the coil is given by T = B A N I. ie.
In equilibrium the deflecting torque T is equal to the opposing torque due to the elastic forces in the springs. The opposing torque
Torque = CQ
Where C is the constant of the spring.
Example 1
A galvanometer coil has a coil has 100 turns which each turn having an area of 2.5cm^{2}. If the coil is in the radial field of 2.0 10^{9}Nm per degree what current is needed to give a deflection of 60^{0}?
Example 2
A moving coil galvanometer with a coil of 15 turns and an area of 0.02m^{2} is suspended by a torsion wire which has restoring constant of 9.00 10^{6}Nm per degree of twist if the current of â„¦MA is passed through the coil whose plane is parallel to a uniform magnetic field of 0.03T. What will be the deflection of the coil?
CURRENT SENSITIVITY.
Therefore the greater the sensitivity is obtained with a stronger field B atom value of c that is a week springs and greater value of N and the value of A. However the size and number of turns of a mound increase the resistance of the meter which is not desirable
VOLTAGE SENSITIVITY
If the resistance of a moving coil meter is R the p.d.v across its terminals where a current I flows it. It is given by V = IR————— (1)
But we have BAIN = C ———— (2)
Hence the voltage sensitivity depends on the resistance R of the meter.
Unlike the current sensitivity.
CONVERTING A MILLIAMETER TO AN AMMETER.
Moving coil meters give full scale deflection for current smaller than those generally met in laboratory. In order to measure the current of the order of an ampere or more we connect a low resistance s called the shunt across. The terminals of moving coil meter
The shunt turns most of the current to be measured I away from the coil.
Suppose the coil of meter has a resistance r of 20â„¦ and full deflected by the current of 5MA. If we want to convert it so that its full scale deflection is 5A; Then the shunt s must be connected which will extra current that is (5 0.005) A or 4.995A.
Potential difference across the shunt p.d across the coil ie.
Example
A milli ammeter has a full scale reading of 0.8MA and resistance of 75 ohms. What is the value of a single resistor which mould a current it into an ameter capable of reading 15amps at full scale
SOLUTION:
CONVERTING A MILLIAMETER TO VOLTIAMETER
Suppose we have a moving coil meter which requires 5MA for full scale deflection and also let suppose that the resistance of its coil r is 20 â„¦
When this milliameter is full deflected the p.d across it is given by u = rI
= (20â„¦) (510^{3})
=0.1u
If the coil resistance is constant the instrument can be used as a voltimeter giving a full scale deflection for p.d of 0.1 or 100mv so this milliameter can possess two scales for current and for voltage as shown in (1) above.
The p.d to be measured in the laboratory is usually greater than 100mv. Therefore to measure such a p.d we put resistor R in series of with the coil as shown in fig (2) above:
For example if we wish to measure to measure up 10v is applied between terminals CD then the scale current of 5MA flows through the moving coil. That is
V = (R + r) I
10 =(R + 20) 5 10^{3}
Or
R =2000–20
=1980â„¦
The resistance r is called a multiplier.
Many voltmeters contain a series of multiplier contains a series of multiplier of different resistances which can be chosen by a switch…. and socket arrangement shown in fig (3) above
Example
A voltmeter whose range is 0200v has resistance R of 1500 per volt.(fsd) what resistance should be converted in series with it to give a range of 02000v.
Solution
Given
But when the resistance in series they have the same current through it
New solution:
Total resistance for this voltmeter is
=1500 200 =30 10^{4}â„¦
Total resistance for the new scale is
=15002000 = 30 10^{5} â„¦
Extra resistance required is
(303) 10^{5} =27 10^{5} â„¦
MAGNITUDES FOR CURRENT CARRYING CONDUCTORS
Laws of Biot and Savant.
It state that the flux density dB= at point P due to a small element dl of a conductor carrying
Where r is the distance from the point P to the element is the angle formed it to P.
B: FLUX DENSITY = INDUCTION OR MAGNETIC INDUCTION
MAGNETIC FIELD
Equation (1) can be written as
dB=KIdlsinx
Where K is the constant of proportionality and it depends on the medium in which the
B AT THE CENTRE OF A NARROW CIRCULAR COIL
Suppose the coil is in air has a radius of r carries a steady current I and it is considered to consist of current element of length dl. Each element is at the distance r from the centre o and it is at the
Example
A coil of wire with 15 turns of radius 6.0cm, has a current of 3.5A flowing through it. What is the magnetic flux density at the center of the coil?
Example
What is the magnitude of the flux density produced the center of a coil of radius 5cm carrying current of 4A in air.
Example
A circular coil of radius 6cm consisting of 5 turns carries a current supplied from 2v accumulator of negligible internal resistance. If the coil has a total resistance of 2â„¦. Calculate the magnetic field induced at the centre
B DUE TO A LONG STRAIGHT WIRE AT A DISTANCE d SIDE THE WIRE
Consider a very long wire YN carrying a current I. Take P to be a point outside the wire but also this point is considered to be very near to this wire.
THE HALL EFFECT
Is the phenomenon where by e .m. f or voltage is set up
transversely or across a current carrying conductor when a
perpendicular magnetic field is applied
Consider a piece of conducting material in a magnetic field of flux density B
Suppose that the field is directed (perpendicularly) into the paper and that there is a current flowing from right to left.If the material is a metal the current is carried by electrons moving from left to right
Consider the situation of one of these electrons and suppose that it has a velocity V
The electron feels a force F which by Fleming’s left hand rule, is directed downwards .
Thus in addition to the electron flow from left to right electrons are urged away from face Y and towards face X. Anegative charged builds up on X, leaving a positive charged on Y so that a potential difference is established between X and Y. The buildup of charge continues until the potential difference becomes so large that it prevents any further increase . This maximum, potential difference is called the Hall voltage
Hall voltage
Is the potential difference created across a current carrying metal strip when the strip is placed in a magnetic field perpendicular to the current flow in the strip.
Actually, the magnetic field does not have to be totally perpendicular to the strip the magnetic field only needs to have a component that is perpendicular
The flow ceases when the e .m .f reaches a particular V_{H} called Hall voltage
MAGNITUDE HALL VOLTAGE
Suppose V_{H} is the magnitude of the Hall voltage and d is the width of the slab (the separation of x and y). Then the Electric field strength E set up across the slab is numerical equal to the potential gradient.
E =
let F_{v} be the force exerted on an electron by the P.d between X and Y. Therefore when the buildup of charged on X and Y has ceased
F = F_{v}
BeV = eE
BV = E
BV=
V_{H}=BVd…………………..(i)
Where
E = The strength of the uniform electric field between X and Y due to the Hall voltage
V_{H} = Hall voltage
d = The separation of X and Y
HALL VOLTAGE
Is the potential difference created across a current
carrying metal strip when the strip is placed in a magnetic
field perpendicular to the current flow in the strip
Actually the magnetic field does not have to be totally
perpendicular to the strip the magnetic field only needs to have
a component that is perpendicular
The flow ceases when the e .m. f reaches a particular value V_{H }called Hall voltage. It has been shown that the current I in a material is given by I = neAV
Where
n = the number of electron per unit value
e = the charge on each electrons
v = the drift velocity of the electrons
A = the cross – sectional area of the material
V = —————– (ii)
Sub equation (ii) into equation (i)
From
In figure A = d t and therefore
ELECTROMAGNETIC INDUCTION
An electric current create magnetic field, the reverse effects of
producing electricity by magnetism was discovered by Faraday and is
called electromagnetic induction
Induced can be generated in two ways
(a) By relative moment ( The generator effect )
if the bar magnet is moved in and out of a stationary galvanometer or
small current is recorded during the motion but not at other time
movement of the coil towards or away from the stationary magnet has the
same results (figure above)
relative motion between the magnet and coil is necessary, the direction
of the of induced current depends on the direction of the relative
motion. And magnitude of current is produced increase with
i. the speed of the motion
ii. The number of turns in the coil
iii. The strength of the magnet used
(b) By changing a magnetic field (transformer effects)
In this case two coils are arranged one inside the other (Figure below) to galvanometer
Rheostat the other called secondary is connected to galvanometer,
switching the current on or off in the primary causes impulse of and
current to be induced in the secondary. Varying the primary current
quietly altering the value of rheostat has the same effect.
Electromagnetic induction thus they occurs only when there is only
change in primary current and also in magnetic fields it induces
LAWS OF ELECTROMAGNETIC INDUCTION
While the magnitude of the induced EMF is given by Faraday law. Its direction can be predicted by Lenz’s Law
LENZ’S LAW
The direction of induced is such that it tends to oppose the flux change
which causing it and does oppose it if induced current flows
Faraday or Newman’s laws
The induced is directly proportional to the rate of change of the flux through the the coil.
If E = induced then
NOTE
I). The minus sign express Lenz’s Law
II). NÉ¸ is the flux linkage in the coil
INDUCED EMF IN A MOVING ROD
Area swept in 1 second
AB is a wire which can be moved by a force F in a contact with a
smooth metal rails PQ and RS. A magnetic field of flux density B acts
downwards perpendicular to the plane of the system.
As the wire AB cuts the flux density the is produced by the current I and is in opposition to the motion
Therefore
F= BIL ……………………………………………………..i
Where l is the distance between two rails
And I = ……………………………………………………………ii
Where is the resistance of the wire
If the wire is moving with a speed V then F’ = F ……………….iii
F’ = ……………………………4
Power = = = Force x velocity
= ……………………… 5
Also power = = = …………….6
Equating equation 5 and 6
=
I.e. E = BLV (This is the induced in a moving coil)
INDUCED EMF IN A ROTATING COIL
Consider a coil of an area A and its normal makes an angle of with the magnetic field B_{Y }
The flux linkage with the coil of n turns is expressed as
N = ………………………………………………………1
The induced emf is given by
E = = – = =
E = since
If the maximum value of emf is denoted by _{o }
Then
E = E_{o }sinwt where E_{o }= NABw
A gain w =
Therefore
 E_{o }=
Exercise 1
The magnetic flux Q_{B }through the loop perpendicular to the plane of the coil and directed into the paper as shown in the diagram is varying according to the equation Q_{B }= 8t^{2} +5t +5 where Q_{B }is measured in millimebers and t in seconds
 What is the magnitude of induced in the loop when
 What is the direction of the current through R?
Solution
E =
E = 16t + 5
E = 53Mv
Exercise 2
What is the maximum induced in a coil of 500turns, each with an area of , which makes 50reflections per second in a uniform magnetic field of flux density 0.04T?
Solution
B = 0.04T
2.5Volts
INDUCED EMF IN ROTATING DISC – DYNAMO
Consider a copper disc which rotates between poles of magnets. Connections are made to its circle and the circumference. An induced emf is obtained between the Centre of the disc and one edge. We assume that magnetic field is uniform over the radius xy
The radius continuously cuts the magnetic flux between the poles of the magnet. For this straight conductor, the velocity at the end of x is zero and that at the other end y where w is the angular velocity of the disk
Average velocity of is
An induced in straight conductor is given by
In this case
^{…………………………………………………….i)}
Since …………………………………………ii)
If the disc has the radius r_{1 }and an axle at the Centre of radius r_{2} the area swept out by a rotating radius of the metal disc is – = – in this case the induced would be
– f
The direction of the E is given by Fleming’s right hand rule
As the disc rotates clockwise the radius moves to the left at the same time as the radius moves to right
If the magnetic field covers the whole disk, induced in the two radii would be in opposite direction. So the resultant emf between yz would be zero. The emf between the Centre and the rim of the disc is the maximum which can be obtained
Qn.
A circular metal disc with a radius of 10cm rotates at 10revolutions per seconds. If the disc is in a uniform magnetic field of 0.02T at a right angle to the plane of the disc. What will be the induced between the Centre and the rim of the disc?
Solution
B = 0.02T
SELF INDUCTANCE (L)
An induced emf appear in the coil if the current in that coil is changed is called selfinduction and produced is called selfinduced
For a given coil produced no magnetic materials nearly the flux linkage proportional to the current I
Or
Where L is a constant proportionality which is called selfinductance of a coil
From Faraday’s law in such a coil the induced
Substitute i) in ii)
or
Hence the unit of inductance . A special name the Henry has been given to this combination of units
Two coils A and B have 200 and 800turns respectively. A current 2Amperes in A produces a magnetic flux of in each turn of A, compute:
 Mutual inductance
 Magnetic flux through A when there is a current of 4.0 Ampere in B
iii. The induced when the current in A changes 3A to 1A in 0.2seconds
SELF INDUCTANCE (L) FOR THE COIL
The induced ,
(By integrating the equation we have)
Therefore
The selfinductance may be defined as the flux linkage per unit current, when is in wabers and I is in amperes then L is in henry:
Magnetic flux density for a long coil is given by with an iron core with a relative permeability of
The flux density is given by since
Thus the flux linkage
(Unit for L is Henry)
ENERGY STORED IN AN INDUCTOR
Because of of the selfinduction that act when the current in the coil change, electrical energy must be supplied in setting up the current against the .
If L is the selfinductance of the inductor then the back across it is given by
…………………………i)
Hence rate at which work is done against the backward emf.
Power = EI…………………………ii)
Substitute equation i) into ii)
Then equation ii) becomes
The work done to bring the current from zero to a steady state value I_{o} is
Therefore
MUTUAL INDUCTANCE (M)
The may be induced by in one circuit by changing current in another. This phenomenon is often called mutual induction and the pairs of circuits which shows it are said to have mutual inductance
The mutual inductance m between the two circuits is defined by the following equation
Induced in B by changing = M (rate of change of current in A) i.e.
The unit of mutual inductance is Henry the same as that of selfinductance
MUTUAL INDUCTION
Since the rate of change in flux in B then
QUANTITY OF ELECTRICITY INDUCED
Consider a close circuit of total resistance R Ohms which has a total flux linkage with magnetic field B. if the flux linkage starts to change
Induced , but current
Flux linkage will not change at a steady rate and a current will not be constant. But throughout it changes. Its charge is being carried round the circuit. If a time t seconds is taken to reach a new constant value the charge carried round the circuit in that time is
Where is the number of linkage at t=o and is the number of linkage time t
Thus
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