# FORM SIX: PHYSICS STUDY NOTES-TOPIC 1: ELECTRO-MAGNETISM

TOPIC 1: ELECTRO-MAGNETISM

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 ðœƒ  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/m2

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 S-pole.

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   = 0o or  1800

F = BQV

F=   BQVSin00 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   =900

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= 900   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 – 4T. 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 -19C.

Solution

F = BQV

B= 0.1T

V= m/s
F=0.1 X 1.6 X 10-19 X 5 X 106X Sin 90

Q = 1.6 X 10-19C

= 900

1. An electron experiences the greatest force as it travel   at 3.9 x105 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 -13N

Q= 1.6 X10 -19C

V=3.9X105m/s

B = 13.14T

By right hand rule per cross product, the direction of the magnetic field is towards northward.

1. An  α  – particle of mass 6.65 x 10-27 kg is  travelling at right angles to a magnetic field with a speed of 6×105m/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 X10-27Kg

V = 6 x 105m/s

B = 0.2T

= 900

F = BQV

= (0.2 x 2x 1.6×10-19) x x Sin90Ëš

Acceleration of α – particle

F= mÉ‘

É‘= =

1.  A  copper  wire  has  1.0 x 1029 free  electrons per  cubic meter, a  cross sectional  area  of  2mm2 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= 1×1029m-3   e = 1.6×10-19c   A= 2mm2 = 2×10-6m2

I = 5A

Force on each electron F= BQ  Sin

Q= 1.6 x 10-19c

B= 0.15T

Q=900

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  = 00 or 1800

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  = 900

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 ðœƒ 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 =900 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 Iron-cored 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 ðœƒ 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

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 I1 and I2 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   = ByI1L

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

I1 = Current through short conductor

l = Length of short conductor

I2 = 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

1. 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)

1. A wire placed along the South-North direction carries currents of 5A from South to North. Find the magnetic field  due to a 1cm piece of wire at a point 200cm North-East from the place.

Solution

By RHGR, The field is vertically vertical downwards

1. 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 106 m/s.  Calculate the magnetic field produced by electron at the centre ( e= 1.6 x 10-9c)

Number the revolution made by the electron in 1 second is

Current   =

I =   1.27 X 1016 X 1.6 X10-19

1S

I =     2.04 X 10-3A

Magnetic field produced by the electron at the centre is

1. 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 = 10×10-2m

B = 4  x 10-7x 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

1. 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  1/8th 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

1. 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

1. A long straight wire carries a current of 50A. An  electron  moving at  107ms 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 V1  is towards the wire. The  angle between  VI  and  B is  900 force on  electron

F= BQV

F = 2x 10-4 x 1.6×10-19x107x Sin 900
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
3.2×10-16N

(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

1. 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

1.  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.

1.  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

1.   (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 IA and IB  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 r1 is the radius of n-turn coil. Then

N

First case                                  Second case

Solution
C.  Magnetic field at the centre of circular coil

1.  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 X10-19c

Current produced I =

I = 2 x 1.6 x 10-19

2

I =1.6 x10-19A

Magnetic field at the centre of the circle orbit of the helium is,

B=

B= 1.256 x 10-25T

1.  A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5×10-4m2 it is uniformly wound with 2000turns carrying a current of 2A and the magnetic  flux in the iron is 8x 10-3Wb. 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 μr

1.  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

1. 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 x0.182

2(0.182 + 0.182) 3/2

B= 4.4 x 10-4T

1. 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

= 900
F= 0.02 X 5 X 0.06Sin900
F = 0.006N

By Fleming’s Left hand rules the force is directed towards West

1. 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

1.   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

10-4I2 = 0.49

1.  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  900

Force on the conductor per unit

(b)      When current  is  flowing from south to  north    =  00

Force on the conductor per unit length

1. 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 10-3V

1.  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

II  = 2A

I2 = 10A

r = 100 x 10-3m

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.

1.  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  F1 and  F2 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 (μr ) and magnetic susceptibility (Xm).

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

1. A diamagnetic substance is feebly repelled by a strong magnet.
2. The magnetic susceptibility ( ) of a diamagnetic substance has a small negative value.
3. The relative permeability ( ) of a diamagnetic substances is slightly less than 1
4. 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

1. 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

1. The magnetic susceptibility of the paramagnetic substance has small positive value

It is because  and

1. 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

1. 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
2. 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
• Fe2O3

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

1. 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.

1. A ferromagnetic substance is strongly attracted by a magnet
2. When a rod of ferromagnetic substance is suspended in a uniform magnetic field, it quickly aligns itself in the direction of the field.
1. They retain their magnetization even when their magnetizing force is removed.
2. 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 B-H 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 B-H 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-2T-I or Jm-Iwb-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/m2

1. (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

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 0.1m2 area of x-section 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 10-4TA-4M

(d)    Steel it is because has high coactivity. This ensures the stay of magnetism in steel for a longer period

1. (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

1. (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 I-H 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

1. (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 -1Tm

(ii) Susceptibility of the material

= (1 + )

= 1328.62 Am -1

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  4×103 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 1022Am2  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

1. (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

1. (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 7700C 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

1. (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/m3.

(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/m3 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/cm2   assume  that  each Iron atom has  a dipole moment of  9.27 x 10-24 Am2

Solution

H = 2700A/m

H

I = 1.35 X 106A/m

One Kilo mole (55.85Kg) of Iron has 6.2 x1026 atoms. Therefore, number of atoms in 1 m3 of Iron

Average magnetic moments per iron atoms

=1.59 x 10-23 Am2

Solution

f =50 HZ

cycle

Length of cubic domain l

=1 M = 10-6M

Volume of Domain

V = 3

V = (10-6)3

V = 10-18M3

Mass of domain

=Volume X Density

=10-12cm x7.9g

= 7.9 x10 -12g/cm

It is given that 55g of Iron contain 6.023 x1023 Iron atoms (Avogadro’s no)

Number of atoms in the domain is

N = 6.023 X 1023 X 7.9 X10-12

55

N = 8.65 X 1010atoms

The maximum possible dipole moment   is achieved per the case when all the atomic domains are perfect aligned (This condition is   unrealistic)

= (8.65 x1010) x (9.27 x10-24)

= 8 x 10-13 AM2

Maximum intensity of magnetization of the Domain is

=     =

=   8 x105 Am-1

NUMERICAL PROBLEMS

1. (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

1. (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 10-4Wb what  is the  permeability  of the  material  of the  rod?

μ = 0.625 x10-3 Wb A -1 m-1

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

1. i)   1500AT/m
2. ii)   9.43T

iii)  7.5 X106A/m

1. (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/m2, find the  relative  permeability and  susceptibility

Xm = 317.5

Mr = 318.5

1.  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

1.  An  Iron  of  volume  10-4m3 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 = 25Am2

1. 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
2. i)                     The  relative  permeability
3. ii)                  Absolute  permeability  of  the  material

= 1. 4

μ= 5.6 x10-7H/m

2. a)       Why does a paramagnetic sample display greater magnetization (per the same magnetizing field) when cooled?
3. b)      Why is diamagnetism, in contrast almost independent of temperature?
4. 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 Fm on the charged particle is given by

Fm = 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

= 00

From

Fm = 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    =  1800

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  =900

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

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.

1. 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

1. i)                     V1 =  V Acting in the  direction of  the  field
2. ii)                  V2 =  Vsin acting perpendicular  to the  direction

The  perpendicular  component V2 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 V2 while  moving in  the  direction of the  field with  a speed  VI

PARAMETERS OF MOTION
The perpendicular  component of  velocity  V2 determines the parameters of  the  circular  motions  while  the  horizontal component of  velocity VI decides the  pitch of  helix

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  900

(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

1.  An electron and a proton moving in a circular path at  3×106 Ms-1 in a uniform magnetic field  of  magnitude  2 x 10-4T. Find the radius  of  the  path

Solution

1.  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.

1. (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 = 00  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.

1.  A  become  of  α-particles 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

=

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

1. A proton with charge – mass ratio of 108 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    .  108

1. (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 105 M5-1 enters a  uniform field  of  0.3T  at  an angle  of  600 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 -19C
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 x10-2M

(ii)  Pitch of helix of d

d = 4.37 x10-2M

1. (a) A particle  of  charge ðœƒ 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

1. 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 Fc

Fm = Fc

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                            ðœƒ = 900

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 = 10-3M

(ii)    When electrons enters  the magnetic field  making an  angle ðœƒ = 300 with  the  field

r     = 0.5 x10-3M

(C)   Solution

Let 1, 2 and 3 be the suffix force proton, deuteron and α – particle respectively

K .E, = K.E2 = K.E3

1 MIV1 = 1 M2V2= 1 M3V3

2               2                  2

If MI then M2 = 2M     and   M3 = 4M

MV2I = 2MV22 = 4MV23

VI =   2 = 2V3

V2 =   VI

Also

r =   MV

BQ

If QI = Q, Then Q2 = Q    and Q3 = 2Q

rI = MIVI       =       MVI

BQI                BQ

r2 = M2V2   =   2M.   V1

BQ2         BQ

rI:  r:  r3  =  MVI   :     MVI  :   MVI

BQ               BQ         BQ

rI : r2 : r3 =  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 10-16C) moving at a speed of 3 x107m/s in a magnitude field of 6 x 10-4T 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

1. 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 10-3m

1. An α-particle is describing a circle of radius of 0.45m in a field of magnetic Induction 1.2Wb/m2. 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 α-particle is 6.8 x10-27Kg and its charge is 3.2 x 10-19 C

V=2.6 X107 m/s
f= 9.2 x106 sec-1

MAGNETIC TORQUE ON RECTANGULAR COIL IN UNIFORM FIELD
Consider a rectangular conductor ABCD having length L and N-turns carrying a current I and placed in a magnetic field between N and S-pole 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 = cross-sectional 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 ðœƒ = 900

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)      60to the field

Solution

The area of the coil

A= (5 x 10 -2) x (2 x10-2)

A= 10-3 m2

(a)      From

A=10-3

I = 2A

N=10

B=0.1T

τ= 0.1 X10-3 X10 X2

τ = 2 X10-3NM

(b)

τ=BANI

τ= 0

(c)

τ = BANICos ðœƒ

τ= 0.1 x10-3 x10x2xCos 600

τ= 10 -3NM

1. 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 = πr2

= πx72

= 154cm2
=      154×10-4m2

Magnetic toque τ

τ = BANI

τ = (154 X10-4) X 100 X100

τ = 150 T

1. 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 = πr2 = π (0.05) M2

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.

1.  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 cross-sectional area 10-5m2 and  force of electron  density in the  wire  is  1029m-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 = BeVd

F = Be.

F =

F= 10-24N

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 2.5cm2. If the coil is in the radial field of 2.0 10-9Nm per degree what current is needed to give a deflection of 600?

Example 2

A moving coil galvanometer with a coil of 15 turns and an area of 0.02m2 is suspended by a torsion wire which has restoring constant of 9.00 10-6Nm 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 =200020

=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 0-200v has resistance R of 1500 per volt.(fsd) what resistance should be converted in series with it to give a range of 0-2000v.

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 10-4â„¦

Total resistance for the new scale is

=15002000 = 30 105 â„¦

Extra resistance required is

(30-3) 105 =27 105 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 VH called Hall voltage

MAGNITUDE HALL VOLTAGE

Suppose  VH 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  Fv  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 = Fv

BeV = eE

BV = E

BV=
VH=BVd…………………..(i)

Where

E = The strength of the uniform electric field between X and Y due to the Hall voltage

VH = 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 VH 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

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 BY

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 = Eo sinwt where Eo = NABw

A gain w =

Therefore

1. Eo =

Exercise 1

The magnetic flux QB 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 QB = 8t2 +5t +5 where QB is measured in millimebers and t in seconds

1. What is the magnitude of induced in the loop when
2. 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 r1 and an axle at the Centre of radius r2 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 self-induction and produced is called self-induced

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 self-inductance 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:

1. Mutual inductance
2. 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 self-inductance 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

(Unit for L is Henry)

ENERGY STORED IN AN INDUCTOR

Because of of the self-induction 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 self-inductance 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 Io 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 self-inductance

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

...

Thanks for reading FORM SIX: PHYSICS STUDY NOTES-TOPIC 1: ELECTRO-MAGNETISM

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