Chapter 4courses.minia.edu.eg/attach/2edadyLec2.pdf · CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES 81 The magnitude of the magnetic force is F q B v sinI, where I is the angle measured

CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES

79

Chapter 4

MAGNETIC FIELDS AND MAGNETIC FORCES

Properties of Magnet

1- A magnet attracts iron pieces or iron fillings. The filings cling

near the ends, at the (poles}) of the magnet.

2- The magnetization is zero at the middle of the magnet and

maximum at the poles.

3- When a magnet is freely suspended so that it can swing in a

horizontal plane, it comes to rest in N-S direction. N-pole points

towards the north, and S-pole points towards the south.

4- First law of magnetism: '' like poles repel, unlike poles attract”.

5- When a magnet is broken into pieces, each piece is found to be a

magnet with two poles, i.e. no monopole magnet.


80

What causes magnetism? It is the spinning of the electrons in the atoms

of materials, and the behavior of these electrons, which give a material

its magnetic properties.

Magnetic field: ''It is the region around a magnet in which a magnetic

force is exerted''. We can describe magnetic interactions in two steps:

1- A moving charge or a current creates a magnetic field in the

surrounding space (in addition to its electric field)

2- The magnetic field exerts a force F on any other moving charge

or current that is present in the field.

Magnetic field is a vector quantity and denoted by B . At any position

the direction of B is defined as the direction in which the north pole

of a compass needle tends to point.

The Magnetic Force on a Moving Charge:

The magnetic force, BF , on a test charge, oq , moving with velocity v in

a magnetic field, is

qv B F


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The magnitude of the magnetic force is v sinF q B , where is the

angle measured from the direction of v to the direction of B . The force

is perpendicular both to the velocity and to the magnetic field. Since the

force is proportional to the velocity, charges at rest do not experience

magnetic forces. [B] = T (Tesla) = N

A.m. Also, 1T =10

4 Gauss.


82

EXTRA NOTICES:

1- The charge should be inmotion

(i.e. current).

2- Positive and negative charges

experience forces in opposite

directions.

3- The force is greatest when the

charge moves perpendicular to the

magnetic field and zero when the

charges move parallel to the field.

4- The size of the force also depends

on the magnitudes of the magnetic

field and the electric charge


83

(current) and on the speed of the moving charge.

5- The magnetic force does not accelerate the charge, but deflect it.

Magnetic Field Lines and Magnetic Flux

The magnetic flux lines or lines of force represent the magnitude and

direction of the magnetic field, i.e. it is vector quantities.

``Note that:

1- `The density of lines is

proportional to the

strength of the field (i.e.

the lines are close together

where the field is strong

and vice versa).

2- The lines go from North

Pole to the South Pole.

3- The lines never cross each

other.

Magnetic Flux and Gauss's Law for Magnetism

We can divide any surface into elements of area dA. For each element we

determine B , the component of B normal to the surface at the position

of that element. We define the magnetic flux Bd through this area as

cos .Bd B dA B dA B d A


84

Where is the angle between the direction of B and the line

perpendicular to the surface.

The total magnetic flux through the surface is the sum of the

contributions from the individual area elements:

cos .B B dA B dA B d A

Magnetic flux is a scalar quantity. In the special case in which B is

uniform over a plane surface with total area A, B and are the same at

all points on the surface, and


85

cosB B A BA

The SI unit of magnetic flux is called weber (Wb).

21 1 . ,1 1Wb T m T N A

Gauss's law for magnetism

The total magnetic flux through a closed surface is always zero.

Symbolically

. 0B d A

If the element of area dA is at right angles to the field lines, then B B ;

calling the area dA , we have

BdB

dA

That is, the magnitude of magnetic field is equal to flux per unit area

across an area at right angles to the magnetic field. For this reason,

magnetic field B is sometimes called magnetic flux density.

A Beam of Charged Particles in a Magnetic Field:

Consider a beam of positively charged particles with velocity v .If this

beam enters a magnetic field at right angle to its direction of motion, it

will experience a force perpendicular to both velocity and magnetic

field, i.e. it will be deflected.


86

Circular motion: The beam will move in circular path if the velocity is

small and the magnetic field is strong. In that case:

Magnetic force = Centripetal force

2v vv

mq B m B

R qR

where R is the radius of the circle. Also, the angular frequency ( ) and

the periodic time (T ) of a rotating charged particle are:

v,

2 2

v

qB

R m

mT

q B

The beam will move in helical (spiral) path if the velocity vector has

an angel with the magnetic field.

The beam of negative charges will be deflected in the reverse

direction.


87

Application of motion of charged particles

Velocity selector: In a region of

crossed or perpendicular

magnetic field B and electric

field E perpendicular to v, the

forces cancel when

v vE

q B qEB

Only particles with speeds equal

to E/B can pass through without

being deflected by the fields. By

adjusting E and B appropriately,

we can select particles having a

particular speed for use in other

experiments. Because v does not

depend on the charge, a velocity

selector works also for electrons or other negatively charged particles.

Thomson's e/m experiment

When a particle is accelerated through a potential difference, V ,

energy conservation requires that the kinetic energy equals the loss

of electric potential energy

221 v

v2 2

qqV m

m V


88

In this experiment, a beam of electron is used, so

221 v 2

v2 2

e eVeV m v

m V m

The electron passes between the plates and strike the screen at the end of

the tube, which is coated with a

material that fluoresces at the

point of impact. The electrons

pass straight through the plates

when

2

2

2 e so

m 2

E eV E

B m VB

All the quantities on the right

side can be measured, so the

ratio e/m of charge to mass can

be determined.


89

The Magnetic Force on a Current Carrying Wire:

The magnetic force BF on a segment of wire, L , carrying current, I, in a

magnetic field, B , is given by

vt

oB o o

qq q I

t

LF B B L B L B

The total force on the wire is the vector sum of the forces on the

segments.

Examples

Example1: An electron is projected into a uniform magnetic field given

by ˆ ˆ1.4 i + 2.1 j T.B . Find the vector expression for the force on the

electron when its velocity is 5 ˆ3.7 10 j m/s v .

Answer:

5 5 14

x y z

ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k

ˆ ˆv v v v 0 3.7 10 0 (1.4)(3.7 10 )( k) (8.3 10 k) N

1.4 2.1 0

B e e

x y z

q q

B B B

F B

Example 2: A proton is moving with a velocity of 6 ˆ6.0 10 i m/s v at a

point where the magnetic field is given by ˆ ˆ ˆ3.0 i 1.5 j 2.0 k T B .

What is the magnitude of the acceleration of the proton at this point?

Answer:


90

2 2 2 15 21.44 10 m/sx y za a a a

Example 3: An electric field and a magnetic field normal to each other.

The electric field is 4.0 kV/m and the magnetic field strength is 2.0 mT.

They are act on a moving electron to produce no force,

a- Calculate the electron speed.

b- Calculate the kinetic energy of the electron.

Answer:

a- 3

6

3

4.0 10v 2.0 10 m/s

2.0 10

E

B

b-

2 31 6 2 181 1v (9.11 10 )(2.0 10 ) 1.82 10 J 11.4 eV

2 2k m

H.W. A proton with velocity 6 ˆ2.0 10 i v (m/s) moves horizontally

into a region of space in which there is an electric field

3 ˆ5.0 10 j N/C E and a magnetic field B. Find the smallest magnetic

field such that the proton will continue to move horizontally, i.e. un-

deflected. [Ans: 3 ˆ2.5 10 k T B ]


91

Example 4: A proton with a velocity of 6.0×106 m/s Travels at right

angles to magnetic field of 0.5 Tesla. What is the frequency of the

proton's orbit?

Answer:

2 196

27

1.6 10 0.57.62 10 Hz

2 2 2 2 1.67 10

v v qB v qBqvB m f

r r m r m

Example 5: Alpha particles (m = 273.3 10 kg , 2q e ) are accelerated

from rest through a potential difference of 1.0 kV. They then enter a

region of magnetic field B = 0.2 T perpendicular to their direction of

motion. What is the radius of the bath?

Answer:

21 2v v

2

q Vq V m

m

27

2

19

2 3.3 10 kg 1000 Vv 1 2 V 12.2 10 m

0.2 T 2 1.6 10 C

m mR

qB B q

Example 6:

A wire 72 cm in length has a mass of 15 g. It is suspended by a pair of

flexible leads in a magnetic field B = 0.54 T pointing out of the page, as


92

shown in the following figure. What current must exist in the wire for the

tension in the supporting leads to be zero?

Answer: At the balance, tension Magnetic force = Gravitational force

(i.e. B GF F ), then we have:

0.015 9.80.38 A from Q to P

0.72 0.54

B

in

L B mgI

L L L

mgI

LB

F

Example 7: A wire bent into a semicircle of radius R = 2.0 m forms a

closed circuit and carries a current of 1.5

A. The circuit lies in the xy-plane, and a

uniform magnetic field B = 3.0 T is

present along the y axis, as shown in the

figure. Find the magnitude of the

magnetic force on the curved portion of

the wire.

Answer:

Magnetic Dipoles: A current loop, of area A, behaves like a magnet

and has a magnetic dipole moment

ˆIAM n

0

(2 )3 18 N

circle straight circle straight

straight

F F F F

F i l B i R


93

Where n̂ is a unit vector normal to the plane of the loop. Its direction is

found by placing the fingers of the right hand along the current; the

right thumb then indicates the direction of n̂ and M . For N -current

loops

ˆNIAM n .

Example 8: The current loop, in the

following figure, consists of one loop with

two semicircles of different radii. If the

current in the circuit is 2 A, a = 3.0 cm and

b = 5.0 cm, then the magnetic dipole

moment of the current loop is:

Answer:

In a uniform magnetic field, a magnetic dipole behaves much like an

electric dipole in uniform electric field, and the net force on the dipole

will be zero. The torque, , on the dipole and the potential energy, U, are

given by:

, cosU MB M B

The magnetic torque on a flat current-carrying loop of wire by a uniform

magnetic field B is maximum when the plane of the loop is parallel to B .

2 2

2

(1)(2) (1.05) (0.3)2

0.01 A.m , into the page.

NI

M A


94

Example 9: The plane of area 4.0 cm2 rectangular loop of wire is

parallel to a 2.0 T magnetic field. The loop carries a current of 6.0 A.

Calculate the magnitude of the torque acts on the loop.

Answer:

4 3

sin

(1)(6)(4 10 )(2)sin90 4.8 10 N.m

NI NIAB

M B A B

Example 10: A 100 turns coil, lies in

xz-plane, has an area of 2.0 m2 and

carries a current I = 0.3 A in the

direction indicated in the figure. The

coil lies in a magnetic field directed

along the x-axis and has a magnitude

of 1.5 T. What is magnitude and

direction of the torque on the coil?

Answer:

ˆ ˆ100 0.3 2j 1.5 i

ˆ ˆ ˆ90 j i 90 k, 90 N.m along the positive z axis

NI

M B A B

Example 11: A 100-turn circular coil of wire with radius 1.0 cm carries

a current of 0.5 A. What torque will be exerted on the coil when it is

placed in a magnetic field of 5.0 mT which makes an angle of 60o with

the plane of the coil?

Answer:

x

y

z

I


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2 2 ˆ(100)(0.5) (1.0 10 ) k

ˆ0.0157 k.

NI

M A

3 50.0157 5.0 10 sin30 3.93 10 N.m M B

Example12: A square loop (L=1.00 m)

consists of 100 closely wrapped turns of

0.20 A. The loop is oriented as shown in

figure (5) in a uniform magnetic field of

0.10 T directed in the positive x- direction.

What is the torque (in N.m) on the loop? (j

is a unit vector in the +y-direction.)

Answer:

ˆ ˆk i sin150

ˆ(100)(0.2)(1.0)(0.1)(0.5) 1.0 j

NI NI A B

M B A B

Example 13: A current of 17 mA is maintained in a circular loop of 2 m

circumference which is parallel to the y-z

plane (see Figure 4). A magnetic field B =

(- 0.8 k) T is applied. Calculate the torque

exerted on the loop by the magnetic field.

(i, j and k are the unit vectors in x, y and z

directions, respectively).

Answer:


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2 2

3

3

1 1 1 ˆ2 2m m m i

1 ˆ ˆ1 17 10 ( 0.8) i k

ˆ4.33 10 j

r r A r

NI A B

A

M B A B

Example 14: A current of 16 mA is maintained in a single circular loop

of radius 0.32 m. An external magnetic field of 0.8 T is directed parallel

to the plane of the loop.

a- Calculate the magnetic moment of the current loop.

Answer:

23

3 2

16 10 0.32

5.1 10 C m

IA

M

b- What is the magnitude of the torque exerted on the loop by the

magnetic field?

3

3 2

sin 90 5.1 10 0.8

4.1 10 C m T

M B M B

Sources Of Magnetic Field

1. Magnetic field of a moving charge

To find the magnetic field B of a point charge q moving with constant

velocity v at the point P, let's call the location of the moving charge at a

given instant the source point and the point P where we want to find the

field the field point. Experiments show that the magnitude of B is


97

proportional to q , the particle

velocity v , 21 r and to the sin .

But the direction of B is

perpendicular to the plane

containing the line from source

point to field point and the

particle velocity v .

0

2

sin

4

q vB

r

where 0

4

is proportionality

constant? B has the greatest

magnitude when sin 1 or

90 . If the charge q is

negative, the direction of B is

opposite.

The unit of B is tesla (T).

2. Magnetic Field of a Current

Element

There is a principle of

superposition of magnetic fields:

The total magnetic field caused

0

2

ˆ

4

qv rB

r


98

by several moving charges is the vector sum of the fields caused by the

individual charges.

We begin by calculating

the magnetic field caused

by a short segment dl of a

current carrying a

conductor, the total

moving charge dQ in the

segment is

dQ nqAdl

Where n is the number of

the moving charged

particles per unit volume,

A is the cross sectional

area, and Adl is the

volume.

The moving charges in

this segment are

equivalent to a single

charge dQ , travelling with

a velocity equal to a drift

velocity v . The magnitude

of the resulting field dB at any point P is

0 0

2 2

sin sin

4 4

d ddQ v dq nv AdldB

r r

,


99

but

dn dq v A I ,

Then 0

2

sin

4

IdldB

r

In the vector form, using the unit vector r̂ , we have

This is the Biot and Savart law.

We can use this law to find the total magnetic field at any point in space

due to the current in a complete circuit.

0 0

2 3

ˆ OR

4 4

Idl r Idl rB B

r r

3. Magnetic Field of a Straight Current-Carrying Conductor

An important application of the Biot and

Savart law is finding the magnetic field

produced by a straight current carrying

conductor. From the figure

2 2 2 2 and sin sin( )r x y x x y

0

2 2 3 24 ( )

a

a

xdyB

x y

The final result is

0

2 2

2

4

I aB

x x a

When the length 2a of the conductor is

very great in comparison to its distance x

0

2

ˆ

4

Idl rd B

r


100

from the point P, we can consider it to be infinitely long. When a is

much larger than x , 2 2x a is approximately equal to a , hence in the

limit a

0

2

IB

x

Thus, at all points on a circle of radius r around the conductor, the

magnitude of B is

0 (near a long, straight, current-carrying conductor)2

IB

r

Note:

The total magnetic flux

through any closed surface is

always zero. This implies that

there are no isolated magnetic

charges or magnetic

monopoles. Any magnetic

field lines that enters a closed

surface must also emerge from

that surface.

. 0B d A


101

4. Force Between Parallel Conductor

The lower conductor produces a B field that, at the position of the upper

conductor, has magnitude

0 2

IB

r

The force that this field exerts on a length L of the upper conductor is

'F I L B

Where the vector L is in the direction of the current 'I and has the

magnitude L. Since B is perpendicular to the length of the conductor and

hence to L , the magnitude of this force is

'' 0

2

II LF I LB

r

And the force per unit length is

'

0

2

IIF

L r

Applying the right-hand rule to 'F I L B shows that the force on the

upper conductor is directed downward.

Note:

1- The two conductors carrying current in the same direction attract

each other. If the direction of either current is reversed, the forces

also reverse.

2- Parallel conductors carrying currents in opposite directions repel

each other.


102

Defining the Ampere:

The attraction or repulsion between two straight, parallel, current-

carrying conductors is the basis of the official SI definition of the

ampere:

One ampere is that unvarying current that, if present in each of two

parallel conductors of infinite length and one meter apart in empty space,

causes each conductor to experience a force of exactly 72 10 Newton's

per meter of length.

5. Magnetic Field of a Circular Current Loop

We can use the law of Biot and Savart

to find the magnetic field at a point P

on the axis of the loop, at a distance x

from the center. As the figure shows,

d l and r are perpendicular, and the

direction of the field dB lies in the x-y

plane. Since

``2 2 2r x a

`0

2 24 ( )

I dldB

x a

0

2 2 2 2 1 2cos

4 ( ) ( )x

I dl adB dB

x a x a

0

2 2 2 2 1 2sin

4 ( ) ( )y

I dl xdB dB

x a x a


103

The situation has rotational symmetry about the x-axis, so there can’t be

a component of the total field B perpendicular to this axis. For every

element d l there is a corresponding element on the opposite side of the

loop, with opposite direction. These two elements give equal

contributions to the x-component of d B , but opposite components

perpendicular to the x-axis. Thus all the perpendicular components

cancel and only the x-component survive.

0 0

2 2 3 2 2 2 3 24 ( ) 4 ( )x

I IaadlB dl

x a x a

The integral of dl is just the circumference of the circle 2dl a , then

2

0

2 2 3 22( )x

IaB

x a

The direction of the magnetic field is given by a right-hand rule.

Magnetic field on the axis of a coil

2

0

2 2 3 2 (on the axis of circular loops with radius )

2( )x

NIrB N r

x r

0 (at the center of circular loops with radius )2

x

NIB N r

r

Magnetic field due to a current in a circular arc wire

, is the radius and is the angle of the arc4

o

NIB R

R


104

Ampere's Law

It states that: the line integral of B around any closed path equal to 0

times the net current through the area enclosed by the path.

0. enclBdl I

The positive sense of current is determined by a right-hand rule


105

Applications of Ampere’s Law

Example 1: Magnetic field of a long, straight, current-carrying

conductor:

.

2

2

B d s

o enc

enco

B ds B r i

iB

r

Example 2: Magnetic field inside a long straight wire with current (a

long cylindrical conductor): a cylindrical conductor with radius R

carries a current. The current is uniformly distributed over the cross

sectional area of the conductor.

.

2

2

2

2

2

B d s

o enc o

o

rB ds B r i i

R

iB r r R

R

Current density (current per unit area) 2

iJ

R , so

22

2 ( )enc

i ri J r

R

Magnetic field of a solenoid:

a coil of wire that acts as a magnet when an electric current flows

through it

http://www.ninjawords.com/coil

http://www.ninjawords.com/wire

http://www.ninjawords.com/magnet

http://www.ninjawords.com/electric+current


106

inside solenoid, near centres o o

NB nI I

L

Where B is the magnetic field anywhere within the solenoid, its unit is

Tesla (T), o is the permeability of free space, 74 10 T.m/Ao , n is

the number of coils per meter in the solenoid, L is the length of the

solenoid, and I is the current passing through the solenoid.

0 outside solenoid B


107

Electromagnetic Induction

1. Electromagnetic Experiments

Here is what we observe: When there is

no current in the electromagnet, so that

0B , the galvanometer shows no current.

1- When the electromagnet is turned

on, there is a momentary current

through the meter as B increases.

2- When B levels off at a steady

value, the current drops to zero, no

matter how large B is.

3- With a coil in a horizontal plane,

we squeeze it so as to decrease the

cross-sectional area of the coil. The meter detects current only


108

during the deformation, not before or after. When we increase the

area to return the coil to its original shape, there is current in the

opposite direction, but only while the area of the coil is changing.

4- If we rotate the coil a few degrees about the horizontal axis, the

meter detects current during the rotation, in the same direction as

when we decrease the area. When we rotate the coil back, there is

a current in the opposite direction during this rotation.

5- If we jerk the coil out of the magnetic field, there is a current

during motion, in the same direction as when we decreased the

area.

6- If we decrease the number of turns of the coil by unwinding one

or more turns, there is a current during unwinding, in the same

direction as when we decrease the area. If we wind more turns

onto the coil, there is a current in the opposite direction during the

winding.

7- When the magnet is turned off, there is a momentary current in

the direction opposite to the current when it was turned on.

8- The faster we carry out any of these changes, the greater the

current.

9- If all these experiments are repeated with a coil that has the same

shape but different material and different resistance, the current in

each case is inversely proportional to the total circuit resistance.

This shows that the induced emfs that are causing the current do

not depend on the material of the coil but only on its shape and

the magnetic field.


109

Note: The common element in all these experiments is changing

magnetic flux through the coil connected to the galvanometer.

2. Faraday's Law

The common element in all

induction effects is changing

magnetic flux through the circuit.

Remember that

. cosB B A BA

If B is uniform over a flat area A .


110

Faraday's law of induction states:

The induced emf in a closed loop equals the negative of the time rate

of change of magnetic flux through the loop.

In symbols, Faraday's law is

Bd

dt

If we have a coil with N identical turns, and if the flux varies at the same

rate through each turn, the total rate of change through each turn is

BdN

dt

3. Lenz's Law

Lenz's Law is a convenient method for determining the direction of an

induced current or emf. Lenz's law is not an independent principle: it can

be derived from Faraday's law.

Lenz's law states: The direction of any magnetic induction effect is such

as to oppose the cause of the effect.


111

4. Motional Electromotive Force

Figure shows the rod moves to the right at a constant velocity v in a

uniform magnetic field B directed

into the page. A charged particle q

in the rod experiences a magnetic

force with magnitude F q vB .

This magnetic force causes the free

charges in the rod to move,

creating a positive charge at the

upper end and a negative charge at

the lower end. This in turn creates

an electric field E within the rod.

There is a downward electric force

with magnitude qE to cancel

exactly the upward magnetic force

with magnitude qvB . Then

qE qvB and the charges are in

equilibrium. The magnitude of

potential difference

ab a bV V V EL vBL

Now suppose the moving rod slides along a stationary U-shaped

conductor, forming a complete circuit. No magnetic force acts in the


112

charges, but the charge redistributes itself, creating an electric field

within it. This field establishes a current in the direction shown. The

moving rod has become a source of electromotive force. We call this emf

a motional electromotive force, denoted by .

vBL

Note: The motional emf; length and velocity perpendicular to uniform B .

Examples

Faraday’s Law of Electromagnetic Induction and Lenz’s Law

1. For the following scenarios, determine whether the magnetic flux

changes or stays the same. If the flux changes: indicate whether it is

increasing or decreasing (and in which direction). Explain your answer.


113

2. Find the direction of the induced current for the solenoid in the figure

below, when the magnet is _____.

3. A circular loop (radius of 10 cm or 0.10 m) is placed in a uniform

magnetic field of magnitude, B = 2.0 T, where the face of the loop is

perpendicular to the direction of the magnetic field.


114

a. Determine the magnetic flux through the loop.

2

B = B dA = BAcos = 0.628 T m

b. The loop is then rotated 90o in 3.0 seconds. What is the magnetic flux

through the loop at the end of the 3.0 seconds?

2

B = B dA = BAcos = 0 T m

c. What is the induced emf in the loop during the rotation?

2

2B T m

s

d 0.628 T m = = = 0.0209 or V

dt 3 s

4. A person moves a 2-m rod at a constant velocity of 3 m/s in a

magnetic field, B= 2.0 T. The rod is perpendicular to the direction of

the B field.

a. What is the direction of induced current in the rod?

`Ans. in the +z direction

b. Determine the induced emf in the

rod.

Ans. =vBL= 12 V

c. The resistance in the rod (and

connecting wires) is 2 . What is the current in the rod?

Ans.

12Vi = = = 6A

R 2


115

d. Determine the magnitude of the magnetic force acting on the rod.

BF =iL B =24N

5. Consider a 1-m conducting rod attached at each end by conducting

rails. The rails are connected at the top and the total loop has a

resistance of 5- . (see figure below). The rod falls to the ground at a

constant velocity, v. The apparatus is inside a constant magnetic field, B

= 3.0 T (directed out of the page). The mass of the rod is 0.5kg.

`

a) What is the magnetic force on the falling rod, due to the magnetic

field?

ˆ ˆB netF = F - mg = mg j = 4.9N j

b) What is the induced current in the rod?


116

BB

F 4.9NF = iLBsin = 4.9N i = = = 1.63A

LB (1m)(3T)

c) What is the induced electromotive force?

= iR = (1.63A)(5 ) = 8.15 V

d) What is the equation for the rate of change of magnetic flux for this

problem?

Bd dA d(Lh) dh

= B = B = BL = BLv = -dt dt dt dt

e) How fast is the rod falling?

ˆ ˆ ˆ

ms

8.15 Vv = - j = - j = -2.72 j

BL 3 T m

f) When the rail falls for 1 sec, verify that energy is conserved.

mmg ε s

εiP =P mgv=εi v = = 2.71 check with (e)

mg

Or

mg εP =P mgv=εi mgv=4.9 2.72 13.3, εi 13.3

Generator

6. A water powered generator, shown below, to convert mechanical

energy into electrical energy. A rotating wheel receives falling

water forcing a wire loop (N=500), located within a constant

magnetic field B=0.01 T (as shown), to rotate counter-clockwise at

a rate of 150 rpm. The length of the segment normal to the B field

(side a) are 0.20 m and the length of the segment parallel to the

field (side b) is 0.15 m.


117

a. What is the area of the

region of the coil within

the magnetic field?

Answer.

2

A = ab

A = (0.15m)(.20m)

A = 0.030 m

b. Determine the general

equation for the magnetic

flux through the coil in terms of area A, B, and angular velocity

.

. B= B dA=B A cos t

c. What is the angular velocity of the rotating coil?

Answer.

rot radmin s

1 min 2 rad = 150 = 15.7

60 s 1 rot

d. Calculate the induced electromotive force around the loop.

Answer.

T2

T2

B

0avg T 0

2

avg

d d =-N =-N B A cos t

dt dt

NBA sin t dt2NBA 4NBA

= = - cos t = -T T

= -1.5 V

The instantaneous emf is: (t)= NBA sin( t+ )


118

e. What direction does the induced current flow around the

coil? Explain.

Answer.

The current will flow clockwise (looking down on the armature),

accordance with RHR.

Self Inductance:

7. A solenoid, r=0.01 m, l=0.03 m (length) and N=100, is in series with

a 10 resistor, both of which are

in parallel with a 10 resistor, all

of these are in series with a 5 V

power supply.

a. Determine the inductance, L,

of the solenoid.

Answer.

2 -6oL = N A = 3.96x10 H

b. When the power supply is initially connected. What is the

current across the solenoid?

Answer.

io-solenoid = 0A

c. What is the initial current drawn from the power supply?

Answe.

o

Vi = = 0.50 A

R


119

d. After 1 minute, what is the current through the solenoid?

Ans. 60sRt

-73.96x10 sL--

solenoid max

5Vi = i 1-e = 1-e = 0.50 A

10

e. What is the total current drawn from the power supply?

Answer.

itot= 1.00A

f. How much energy is stored in the inductor after 1 minute?

Answer.

U = ½Li2 = 4.95x10

-7 J

Examples

Example 1: A cylindrical conductor of radius 2.5 cmR carries a

current 2.5 AI along its length; this current is uniformly distributed

through the cross section of the conductor. Calculate the magnetic field

midway along the radius of the wire (that is, at / 2r R ).

Answer:

2

7 5

2

, 2

2.5( / 2) 4 10 1.0 10 T

2 4 (0.05)

o

o

IB r r R

R

IB R

R

------------------------------------------------------------------------

Example 2: What current in a solenoid 15-cm long wound with 100

turns would produce a magnetic field equal to that of the earth, which is

5.1×10-5

T?


120

Answer:

52

7

5.1 10 0.156.1 10 A

4 10 100

s o o

s

o

NB nI I

L

B NI

L

--------------------------------------------------------------------------

Example 3: A solenoid is formed by tightly winding a single layer of

wire. The wire is 1.0 mm in diameter. What is the magnitude of the

magnetic field inside the solenoid when there is a current of 0.081 A in

the windings?

Answer:

7 4

3

14 10 0.081 1.02 10 T

10

s o o

NB nI I

L

------------------------------------------------------

--------------------

Example 4: Three long wires parallel to the

x-axis carry currents as shown in the

following figure. If I = 20 A, what is the

magnitude of the magnetic field at the origin?

31 21 2 3

1 2 3

5

,2

34 2 31.2 10 T.

2 2 1 3 2

o

o o

II IB B B B

a a a

I I II


121

Example 5: A segment of wire is formed into

the shape shown in the following figure and

carries a current I. What is the resulting

magnetic field at the point P?

Answer:

1 2 3 2

2

0 0,

3 3 into the page

4 4 2 8o o o

B B B B B

I I IB B

R R R

Example 6: Figure 10 shows two concentric, circular wire loops, of radii

r1 = 15 cm and r2 = 30 cm, are located

in the xy plane. The inner loop

carries a current of 8.0 A in the

clockwise direction, and the outer

loop carries a current of 10.0 A in the

counter clockwise direction. Find the

net magnetic field at the center.

Answer:

.

1

2

7

1

7

2

1 2

4 10 833.5 T into the page.

2 2(0.15)

4 10 1020.9 T out of the page.

2 2(0.3)

33.5 20.9 12.6 T into the page.

or

or

IB

r

IB

r

B B B

-------------------------------------------------------------------


122

Example 7: A circular loop of radius 0.1 m has a resistance of 6 Ohms.

If it is attached to a 12 V battery, how large a magnetic field is produced

at the center of the loop?

Answer:

7 5

122 A,

6

24 10 2 1.3 10 T

4 4 0.1o

IV

R

IB

R

----------------------------------------------------------------------

Example 8: How many turns should be in a flat circular coil of radius

0.1 m in order for a current of 10 A to produce a magnetic field of

33.0 10 T at its center?

Answer:

3

7

2 2 3.0 10 0.148 Turns

2 4 10o

o

NI BRB N

R I

--------------------------------------------------------------------

Example 9: Two long parallel wires, a distance d apart, carry currents of

I and 5I in the same direction. Locate the point r, from I, at which their

magnetic fields cancel each other.

Answer:

1 2at equilibrium 2 2 ( )

Solve for r, one can find 6

o oI IF F

r d r

dr

---------------------------------------------------------------------------


123

Example 10: Two long parallel conductors, separated by a distance

10 cma , carry currents in the same direction. If 1 5.0 AI and

2 8.0 AI , what is the force per unit length exerted on each conductor by

the other?

Answer:

1 2

7 5

2

5.0 8.04 10 8.0 10 N/m attractive

2 0.1

o

I IF

l a

MAGNETIC PROPERTIES OF MATERIALS

All matter is composed of atoms and atoms are composed of protons,

neutrons and electrons. The protons and neutrons are located in the

atom's nucleus and the electrons are in “constant motion” around the

nucleus. Electrons carry a negative electrical charge and produce a

magnetic field as they move through space. A magnetic field is produced

whenever an electrical charge is in motion.

This may be hard to visualize on a subatomic scale but consider an

electric current flowing through a conductor. When electrons (electric

current) are flowing through the conductor, a magnetic field forms

around the conductor. The magnetic field can be detected using a

compass.


124

Since all matter is comprised of atoms, all materials are affected in some

way by a magnetic field. However, not all materials react the same way.

At the atomic level, the motion of an electron gives rise to current loop

magnetic dipole moment magnetic field

Magnetic dipole moment m [A.m2]

um m N i A n

Where N is the number of electrons, i is the current, and A is the area.

It is a vector quantity and its direction can be found by the right hand

screw rule.

The magnetic moments associated with atoms have three origins:

1 The electron orbital motion.

2 The change in orbital motion caused by an external magnetic

field.

3 The spin of the electrons.

Magnetization M [A.m-1]

It is equal to the magnetic dipole moment per unit volume

magnetic momentM

volume

Permeability

Permeability is a material property that describes the ease with which a

magnetic flux is established in the component. It is the ratio of the


125

magnetic flux density to the magnetic intensity and, therefore,

represented by the following equation: B

H

Magnetic susceptibility ( m )

It represents the response of a system to the external magnetic field.

When a material is placed within a magnetic field, the material's

electrons will be affected. However, materials can react quite

differently to the presence of an external magnetic field. This reaction is

dependent on a number of factors such as the atomic and molecular

structure of the material, and the net magnetic field associated with the

atoms. In most atoms, electrons occur in pairs. Each electron in a pair

spins in the opposite direction, so when electrons are paired together,

their opposite spins cause their magnetic fields to cancel each other.

Therefore, no net magnetic field exists. Alternately, materials with some

unpaired electrons will have a net magnetic field and will react more to

an external field.

Most materials can be classified as ferromagnetic, diamagnetic or

paramagnetic.

Relation between the magnetic field (B), external magnetic field (H)

and magnetization (M):

m o m(1 )B H M H

where


126

m is the susceptibility, is the permeability, and o is the permeability

of free space (air).

Diamagnetic materials 0m

Small and negative susceptibility.

Slightly repelled by a magnetic field.

Do not retain the magnetic properties when the external field is

removed.

Magnetic moment – opposite direction to applied magnetic field.

Solids with all electrons in pairs - no permanent magnetic

moment per atom.

Properties arise from the alignment of the electron orbits under

the influence of an external magnetic field.

Most elements in the periodic table, including copper, silver, and

gold, are diamagnetic.

m(argon) ~ -1.010-8 m(copper) ~ -1.010-5

H

B

Diamagnetic material

m < 0 (small)

B = o (1+ m ) H

Permeability

= o (1+ m ) =slope of B-H line


127

Paramagnetic materials 0m small

Small and positive susceptibility.

Slightly attracted by a magnetic field.

Material does not retain the magnetic properties when the

external field is removed.

Properties are due to the presence of some unpaired electrons

and from the alignment of the electron orbits caused by the

external magnetic field.

Examples - magnesium, molybdenum, lithium, and tantalum.

m(oxygen) ~ 2.010-6 m(aluminium) ~ 2.110-5

Ferromagnetic materials

Large and positive susceptibility.

Strong attraction to magnetic fields.

Retain their magnetic properties after the external field has been

removed.

Ideal magnetic material

or paramagnetic material

m > 0 (small)

B = o r H = H

= constant = slope of B-H curve

B

H


128

Some unpaired electrons so their atoms have a net magnetic

moment.

Strong magnetic properties due to the presence of magnetic

domains. In these domains, large numbers of atomic moments

(1012 to 1015) are aligned parallel so that the magnetic force

within the domain is strong. When a ferromagnetic material is in

the demagnetized state, the domains are nearly randomly

organized and the net magnetic field for the part as a whole is

zero. When a magnetizing force is applied, the domains become

aligned to produce a strong magnetic field within the part.

Iron, nickel, and cobalt are examples of ferromagnetic materials.

o ( )B H M Magnetization is not proportional to the applied

field.

m(ferrite) ~ 100 m(iron) ~ 1000

Magnetic Domains

Ferromagnetic materials get their magnetic properties not only because

their atoms carry a magnetic moment but also because the material is

made up of small regions known as magnetic domains. In each domain,

all of the atomic dipoles are coupled together in a preferential direction.

This alignment develops as the material develops its crystalline

structure during solidification from the molten state. Magnetic domains


129

can be detected using Magnetic Force Microscopy (MFM) and images of

the domains like the one shown below can be constructed.

Magnetic Force

Microscopy (MFM) image

showing the magnetic

domains in a piece of heat

treated carbon steel.

During solidification a trillion or more atom moments are aligned

parallel so that the magnetic force within the domain is strong in one

direction. Ferromagnetic materials are said to be characterized by

"spontaneous magnetization" since they obtain saturation

magnetization in each of the domains without an external magnetic

field being applied. Even though the domains are magnetically

saturated, the bulk material may not show any signs of magnetism

because the domains develop themselves are randomly oriented

relative to each other. Ferromagnetic materials become magnetized

when the magnetic domains within the material are aligned. This can be

done by placing the material in a strong external magnetic field or by

passes electrical current through the material. Some or all of the

domains can become aligned. The more domains are aligned, the


130

stronger the magnetic field in the material. When all of the domains are

aligned, the material is said to be magnetically saturated. When a

material is magnetically saturated, no additional amount of external

magnetization force will cause an increase in its internal level of

magnetization.

demagnetized Material Magnetized Material

The Hysteresis Loop and Magnetic Properties

A great deal of information can be learned about the magnetic properties

of a material by studying its hysteresis loop. A hysteresis loop shows the

relationship between magnetization and external magnetic field. This

relationship, for many ferromagnetic materials, is different when external

field


131

is increasing from when it is decreasing. An example hysteresis loop is

shown below.

The materials of both (a) and (b) remain strongly magnetized when

external field is reduced to zero. Since (a) is also hard to demagnetized, it

would be good for permanent magnets. Since (b) magnetizes and

demagnetizes more easily, it could be used as a computer memory

material. The material of (c) would be useful for transformers and other

alternating devices when zero hysteresis would be optimal.


132

Documents

Chapter 4courses.minia.edu.eg/attach/2edadyLec2.pdf · CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES 81 The magnitude of the magnetic force is F q B v sinI, where I is the angle measured