Chapter 4: Magnetic Field - YSL Physicsyslphysics.weebly.com/uploads/4/8/2/6/48261583/slides_4.pdf · Chapter 4: Magnetic Field In this chapter, ... form a resultant magnetic field

Chapter 4: Magnetic Field In this chapter, you will learn about the force of a

magnetic field on a moving charge and on a current-

carrying conductor. Magnetic are also produced by

currents flowing through conductors.

Overview Magnetic Field

Magnetic Field

Produced by

Current-Carrying

Conductor

Magnetic Force Torque Sources and

Magnetic Field

Lines

Motion of

Charge in

Magnetic and

Electric Field

Moving

Charge Current-

Carrying

Conductor

Two

Parallel

Conductors

Straight

Wire

Circular

Loop Solenoid

4.1 Magnetic Field

Define magnetic field.

Identify magnetic field sources and sketch

their magnetic field lines.

Learning Objectives

Magnetic Field Magnetic field is defined as the region around a magnet

where a magnetic force can be experienced.

Magnetic field has two poles, called north (N) and south

(S). These magnetic poles are always found in pairs

whereas a single magnetic pole has never been found.

Like poles (N-N or S-S) repel each other.

Opposite poles (N-S) attract each other.

Magnetic Field Sources and

Magnetic Field Lines

Magnetic field lines are used to represent a magnetic field.

The characteristics of magnetic field lines:

The lines do not intersect one another

The lines form a closed loop: magnetic field lines leave the

North-pole and enter the South-pole.

The lines are closer together at the poles. (The number of

lines per unit cross-sectional area is proportional to the

magnitude of the magnetic field.)

Two sets of magnetic field lines can be superimposed to

form a resultant magnetic field line.


Magnetic Field Lines

Magnetic field can be represented by crosses or by

dotted circles as shown in figures below:

Magnetic field lines

enter the page

perpendicularly

Magnetic field lines

leave the page

perpendicularly


Magnetic Field Lines The pattern of the magnetic field lines can be

determined by using two methods:

Using compass needles

Using sprinkling iron filings on paper

Bar Magnet

One bar magnet

Horseshoe or U magnet

Two bar magnets (unlike pole)

Two bar magnets (like pole)

Current-carrying conductor

A stationary electric charge is surrounded by an electric

field only.

When an electric charge moves, it is surrounded by an

electric field and a magnetic field. The motion of the

electric charge produces the magnetic field.

Earth magnetic field

Note that the Earth’s

“North Pole” is really

a south magnetic pole,

as the north ends of

magnets are attracted

to it.

Example 1

a)

b)

N

N

S

S

N

N

S

S

4.2 Magnetic Field Produced by

Current-Carrying Conductor Use magnetic field:

for a long straight wire.

at the centre of a circular coil.

at the centre of a solenoid.

at the end of a solenoid.

Learning Objectives

r

IB

2

0

r

IB

2

0

nIB 0

nIB 02

1

Types of Current-Carrying Conductors

1. A long straight wire

2. A circular coil

3. A solenoid

A Long Straight Wire

Direction: Right hand grip rule

A Long Straight Wire

Magnetic field at any point from the conductor (wire):

where

B = magnetic field strength / flux density (T)

I = current in the wire (A)

r = perpendicularly distance of B from

the wire (m)

μo = the permeability of free space (vacuum)

= 4π×10-7 Henry per metre (H m-1)

r

IB

2

0

B • r

A Circular Coil

Direction:

Right hand

grip rule

A Circular Coil

Magnetic field at the centre of the circular coil :

where

R = radius of the circular coil.

µ0 = permeability of free space

4π × 10-7 H m-1

I = current

N = number of coils (loops)

R

IB

2

0

R

NIB

2

0

At the centre of

ONE circular coil

At the centre of N

circular coils

B

R

A Solenoid

A Solenoid

Direction:

Right hand grip rule

To determine the north pole

of the solenoid (/magnet)

A Solenoid The magnitude of magnetic field intensity at the centre

(mid-point/ inside) of N turn solenoid is given by

The magnitude of magnetic field intensity at the end of N

turn solenoid is given by:

l

NIB 0

nIB 0

nl

Nand

nIB 02

1

Example 2

Determine the magnetic field strength at point X

and Y from a long, straight wire carrying a current

of 5 A as shown below.

Example 2 – Solution

Example 3

A circular coil having 400 turns of wire in air has

a radius of 6 cm and is in the plane of the paper.

What is the value of current must exist in the coil

to produce a flux density of 2 mT at its center?


Example 4

An air-core solenoid with 2000 loops is 60 cm

long and has a diameter of 2.0 cm. If a current of

5.0 A is sent through it, what will be the flux

density within it?


Example 5

A solenoid is constructed by winding 400 turns of

wire on a 20 cm iron core. The relative

permeability of the iron is 13000. What current

is required to produce a magnetic induction of 0.5

T in the center of the solenoid?

Example 5 – Solution N = 400, l = 0.2 m, B = 0.5 T, µr = 13000

1-2

7

0

m H 1063.1

10413000

r

A 015.0

2.0

4001063.15.0

2

I

l

NIB

Additional knowledge:

Iron-core solenoid has

stronger magnetic field

strength than air-core

solenoid because

the soft iron inside

the solenoid becomes

a magnet itself when

the current is flowing.

Bl

NIB ,

Example 6

Two straight parallel wires are 30 cm apart and

each carries a current of 20 A. Find the magnitude

and direction of the magnetic field at a point in the

plane of the wires that is 10 cm from one wire and

20 cm from the other if the currents are

a) in the same direction,

b) in the opposite direction.


a)

IX IY

0.3 m

0.2 m 0.1 m

A

IX IY

BX

BY

View from top

rX rY

A 20 III YX



b)

IX IY

0.3 m

0.2 m 0.1 m

A

x

IX IY

BX

BY

View from top

rX rY

A 20 III YX


4.3 Force on a moving charged particle

in a uniform magnetic field

Use magnetic force,

Describe circular motion of a charge in a

uniform magnetic field.

Use relationship

BvqF

CB FF

Learning Objectives

Force on a Moving Charged

Particle A stationary electric charge in a magnetic field will not

experience any force. But if the charge is moving with a

velocity, v in a magnetic field, B then it will experience

a force. This force known as magnetic force.

The magnitude of the magnetic force can be calculated by

using the equation below:

BvqF

sinqvBF

where

q : magnitude of the charge

θ : angle between v and B

Right Hand Rule

C

B

A +

IMPORTANT!

Thumb indicates the

direction of the

magnetic force exerted

on a positive charge.

If the charge is

negative, direction of

the force is opposite. Direction of the force

if the charge is

electron (negative)

BAC

First vector Second vector

Force on a Moving Charged

Particle The direction of the magnetic force can be determined by

using right hand rule:

F

B

v

IMPORTANT!

Thumb indicates the

direction of the

magnetic force exerted

on a positive charge.

If the charge is

negative, direction of

the force is opposite.

BvqF

Alternatively,

Fleming’s Hand

Rule

Right Hand

- Negative charge

- Electron

Left Hand

- Positive charge

- Proton

Thumb – direction of Force

First finger – direction of Field

Second finger – direction of Velocity.

Example 7

Determine the direction of the magnetic force, exerted on a

charge in each problems below:

B

vF

(into the page)

B

v

F

(to the left)

Example 7


charge in each problems below:

B v

X X X X

X X X X

X X X X

F

(to the left)

v

I

F

(upwards)

B

X X

X

X

X

X

X X

Circular Motion of a Charge in a

Uniform Magnetic Field


Uniform Magnetic Field Consider a charged particle moving in a uniform

magnetic field with its velocity perpendicular to the

magnetic field.

As the particle enters the region, it will experience a

magnetic force which the force is perpendicular to the

velocity of the particle. Hence the direction of its

velocity changes but the magnetic force remains

perpendicular to the velocity.

Since the path is circle therefore the magnetic force FB

contributes the centripetal force Fc (net force) in this

motion. Thus

CB FF


Uniform Magnetic Field

CB FF

r

mvBqv

2

sin and θ = 90°

Bq

mvr

where m : mass of the charged particle

v : magnitude of the velocity

r : radius of the circular path

q : magnitude of the charged particle


Uniform Magnetic Field The period of the circular motion, T makes by the particle

is given by

The frequency of the circular motion makes by the

particle is given by

v

rT

2

Bq

mT

2

Tf

1

m

Bqf

2

Note!

Circular motion if v

is perpendicular to

B

90 90

Helical motion if v is

not perpendicular to B

Spiral if v is not

constant

Bq

mvr

vr

CB FF

Example 8

A charge q1 = 25.0 μC moves with a speed of 4.5×103 m s-1

perpendicularly to a uniform magnetic field. The charge

experiences a magnetic force of 7.31×10-3 N. A second

charge q2 = 5.00 μC travels at an angle of 40.0 o with

respect to the same magnetic field and experiences a

1.90×10-3 N force. Determine

a. The magnitude of the magnetic field

b. The speed of q2.


Example 9

An electron at point A in figure above has a speed v of

2.50 106 m s-1. Determine the magnitude and direction of

the magnetic field that will cause the electron to follow the

semicircular path from A to B.

(Given e = 1.601019 C and me= 9.111031 kg)



4.4 Force on a current carrying

conductor in a uniform magnetic field

Use magnetic force,

BlIF

Learning Objectives

Force on a Current-Carrying

Conductor When a current-carrying conductor is placed in a

magnetic field B, thus a magnetic force will acts on that

conductor.

The magnitude of the magnetic force exerts on the

current-carrying conductor is given by

BlIF

sinIlBF

where

l : length of the conductor

θ : angle between l and B

Force on a Current-Carrying

Conductor

F

B

l

IMPORTANT!

vector l is the same as

the direction of the

current flows I through

the conductor.

BlIF

Example 10


conductor carrying current, I in each problems below.

B I

X X X X

X X X X

X X X X

F

(to the left)

B I

X X X X

X X X X

X X X X

F

(to the right)

Alternatively,

Direction of F can also be determined by using

Fleming’s Left Hand Rule

Thumb – direction of Force

First finger – direction of Field

Second finger – direction of Current.

The magnetic force on the conductor

has its maximum value when the

conductor (and therefore the current)

and the magnetic field are

perpendicular (at right angles) to

each other then = 90

The magnetic force on the conductor

is zero when the conductor (and

therefore the current) is parallel to the

magnetic field then = 0

Note!

90sinmax BIlF

BIlF max

0sinmin BIlF

0min F

Example 11

A wire of 20 cm long is placed perpendicular to the

magnetic field of 0.40 Wb m-2.

a. Calculate the magnitude of the force on the wire when

a current 12 A is flowing.

b. For the same current in (a), determine the magnitude of

the force on the wire when its length is extended to 30

cm.

c. If the force on the 20 cm wire above is

60×10-2 N and the current flows is12 A, find the

magnitude of magnetic field was supplied.



Example 12

A square coil of wire containing a single turn is placed in a

uniform 0.25 T magnetic field. Each side has a length of

0.32 m, and the current in the coil is 12 A. Determine the

magnitude of the magnetic force on each of the four sides.

90o B

I


4.5 Forces between two parallel

current carrying conductors

Derive magnetic force per unit length of

two parallel current carrying conductors.

Use magnetic force per unit length,

Define one ampere.

d

II

l

F

2

210

Learning Objectives

Forces Between Two Parallel

Current Carrying Conductors

Consider two identical straight conductors X and Y

carrying currents I1 and I2 with length l are placed parallel

to each other as shown in figure below. The conductors are

in vacuum and their separation is d.

View from the top

Problem Solving Strategy

Step 1

Determine the direction of the

magnetic flux density due to

the neighbouring current-

carrying conductor

Step 2

Determine the direction of the

magnetic force acting on the

current-carrying conductor

(object)

d

2I

2I

1I

1I

X Y

Q

d

2I

2I

1I

1I

X Y

Q F

B B

Force exerted on conductor X

The magnitude of the magnetic flux

density, B2 at point Q on conductor

X due to the current in conductor Y

is given by:

d

IB

2

202

Direction: out of the page

Conductor X carries a current I1 and in the magnetic field B2

then conductor X will experiences a magnetic force, F21.

The magnitude of F21 is given by

sin1221 lIBF

90sin2

120

21 lId

IF

lI

d

IF 1

2021

2

Direction:

towards

conductor Y

Force exerted on conductor Y

The magnitude of the magnetic flux

density, B1 at point P on conductor

Y due to the current in conductor X

is given by

d

IB

2

101

Direction: into the page

Conductor Y carries a current I2 and in the magnetic field B1

then conductor Y will experiences a magnetic force, F12.

The magnitude of F12 is given by

sin2112 lIBF

90sin2

210

12 lId

IF

lId

IF 2

1012

2

Direction:

towards

conductor X

Conclusion

From equation above, thus the force per unit length is

given by

d

LIIFFF

2

2102112

The properties of this force : Attractive force

d

II

L

F

2

210

Forces Between Two Parallel

Current Carrying Conductors If the direction of current in conductor Y is change to

upside down as shown in figure

The magnitude of F12 and F21 can be determined by using

equations above and its direction by applying right hand

rule.

View from the top

Conclusion: Type of the force is repulsive force

One Ampere If two long, straight, parallel conductors, 1.0 m apart in vacuum carry

equal 1.0 A currents hence the force per unit length that each

conductor exerts on the other is

The ampere (one ampere) is defined as the current that flowing in

each of two parallel conductors which are 1.0 metre apart in vacuum,

would produce a force per unit length between the conductors of

2.0 10-7 N m-1.

d

II

L

F

2

210

17 m N 100.2 L

F

)01(2

)01)(01)(104( 7

.π

..πx

L

F

Example 13

Two long straight parallel wires are placed 0.25 m apart in

a vacuum. Each wire carries a current of 2.4 A in the same

direction.

a. Sketch a labelled diagram to show clearly the direction

of the force on each wire.

b. Calculate the force per unit length between the wires.

c. If the current in one of the wires is reduced to 0.64 A,

calculate the current needed in the second wire to

maintain the same force per unit length between the

wires as in (b).

(Given µ0 = 4π × 10-7 T m A-1)



4.6 Torque on a Coil

Use torque,

where N = number of turns

Explain the working principle of a moving

coil galvanometer.

BANI

Learning Objectives

Torque on a Coil

Top view

Side view

Torque is the tendency of

a force to rotate an object

about an axis.

Torque on a Coil

For a coil of N turns, the magnitude of the torque is given

by

BANI

sinNIAB

where

τ : torque on the coil

A : area of the coil

B : magnetic flux density

I : current flows in the coil

θ : angle between vector area, A (the normal to plane of the coil)

and B

N : number of turns (coils)

Note!

The torque is zero when θ = 0˚ and is maximum when θ = 90˚ as

shown in figures below.

0sinNIABτ

0τ

90sinNIABτ

NIABτ

The Working Principles of a

Moving Coil Galvanometer



The galvanometer is the main component in analog

meters for measuring current and voltage.

It consists of a magnet, a coil of wire, a spring, a pointer

and a calibrated scale.

The coil of wire contains many turns and is wrapped

around a soft iron cylinder.

N S



The coil is pivoted in a radial magnetic field, so that no

matter what position it turns, the plane of the coil is

always parallel to the magnetic field.



The basic operation of the galvanometer uses the fact

that a torque acts on a current loop in the presence of a

magnetic field.

When there is a current in the coil, the coil rotates in

response to the torque (τ = NIAB ) applied by the

magnet.

This causes the pointer (attached to the coil) to move in

relation to the scale.



The torque experienced by the coil is proportional to the

current in it; the larger the current, the greater the torque

and the more the coil rotates before the spring tightens

enough to stop the rotation.

Hence, the deflection of the pointer attached to the coil

is proportional to the current.

The coil stops rotating when this torque is balanced by

the restoring torque of the spring.



From this equation, the current I can be calculated by

measuring the angle θ.

constant) (torsional spring theofconstant stiffness theis where

torquerestoringforce magnetic todue Torque s

k

NAB

kI

kNIAB

coil theof anglerotation :θ radianin

Example 14

A 20 turns rectangular coil with sides 6.0 cm x 4.0 cm is

placed vertically in a uniform horizontal magnetic field of

magnitude 1.0 T. If the current flows in the coil is 5.0 A,

determine the torque acting on the coil when the plane of

the coil is

a. perpendicular to the field,

b. parallel to the field,

c. at 60 to the field.



4.7 Motion of charged particle in

magnetic field and electric field

Explain the motion of a charged particle

in both magnetic field and electric field.

Derive and use velocity, in a

velocity selector.

B

Ev

Learning Objectives

Velocity Selector

Velocity selector consists of a pair of parallel metal plates

across which an electric field can be applied in uniform

magnetic field.

Consider a positive charged particle with mass m, charge q

and speed v enters a region of space where the electric and

magnetic fields are perpendicular to the particle’s velocity

and to each other as shown in figure above.

Velocity Selector

The charged particle will experiences the electric force FE is

downwards with magnitude qE and the magnetic force FB is

upwards with magnitude Bqv as shown in figure above.

If the particle travels in a straight line with constant velocity

hence the electric and magnetic forces are equal in

magnitude (dynamic equilibrium). Therefore

EB FF

qEqvB 90sin

B

Ev

Only particles with velocity

equal to E/B can pass through

without being deflected by the

fields. Equation above also

works for electron or other

negative charged particles.

Mass Spectrometer When the charged particle entering

the region consists of magnetic field

only, the particle will make a

semicircular path of radius r.

and

if B = B’ ,

CB FF

r

mvqvB

2'

B

Ev

BrB

E

m

q'

A mass spectrometer is a device used for separating atoms or molecules according to their mass.

2rB

E

m

q

B

B’

(Outside velocity

selector)

(Inside velocity

selector)

Example 15

A velocity selector is to be constructed to select ions

(positive) moving to the right at 6.0 km s-1. The electric

field is 300 V m-1 upwards. What should be the magnitude

and direction of the magnetic field?


Example 16

An ion enters a uniform magnetic field B. The path of the

ion is a spiral as shown in the figure.

a. Determine whether the ion is positively or negatively

charged.

b. Give reasons for the shape of the path.


Summary Magnetic Field

Magnetic Field

Produced by

Current-Carrying

Conductor

Magnetic Force Torque Sources and

Magnetic Field

Lines

Motion of

Charge in

Magnetic and

Electric Field

Moving

Charge Current-

Carrying

Conductor

Two

Parallel

Conductors

Straight

Wire

Circular

Loop Solenoid

Documents

Chapter 4: Magnetic Field - YSL Physicsyslphysics.weebly.com/uploads/4/8/2/6/48261583/slides_4.pdf · Chapter 4: Magnetic Field In this chapter, ... form a resultant magnetic field