Chapter 4

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Chapter 4: Magnetic Field

4.1 Magnetic Field

L.O 4.1.1 Define 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.

L.O 4.1.2 Identify magnetic field sources and sketch their 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 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

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The pattern of the magnetic field lines can be determined by using two methods:

Using compass needles

Using sprinkling iron filings on paper

Magnetic field sources:

i. Bar magnet

One bar magnet Horseshoe or U magnet

Two bar magnets (unlike pole)

Two bar magnets (like pole)

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ii. 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.

A circular coil A long straightwire A solenoid

iii. Earth magnetic field

Example

Note that the Earth’s “North

Pole” is really a south magnetic

pole, as the north ends of

magnets are attracted to it.

N S

N S

N S

S N

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4.2 Magnetic Field Produced by Current-Carrying Conductor

L.O 4.2.1 Use magnetic field equations

a) Long Straight Wire

View (a) from the top:

Magnitude of the magnetic field at any point

from the conductor (wire):

where

B = magnetic field strength / magnetic flux

density (T)

I = current in the wire (A)

r = perpendicularly distance of B from

the wire (m)

μo = constant of proportionality

known as the permeability of

free space (vacuum)

= 4π x 10-7

Henry per metre (H m-1

)

The direction of magnetic field around the

wire or coil can be determined by using the

right hand grip rule as shown in Figure(b).

Thumb – direction of current

Other fingers – direction of magnetic

field (clockwise OR anticlockwise)

r

IB

2

0

r B •

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b) Circular Coil

View from the top:

(a)

(b)

Magnitude of the magnetic field at the centre

of the circular coil:

For ONE circular coil

For N circular coils

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)

The direction of magnetic field around the

wire or coil can be determined by using the

right hand grip rule as shown in Figure(a).

Example of multiple circular loops:

B

B

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c) Solenoid

View from the top:

(a)

The magnitude of magnetic field intensity at

the centre (mid-point/ inside) of N turn

solenoid is given by

Since , therefore it can be written as

The magnitude of magnetic field intensity at

the end of N turn solenoid is given by:

where

n = numbers of turns per unit length

µ0 = permeability of free space

4π × 10-7

H m-1

I = current

N = number of coils (loops)

The directions of the fields can be found by

viewing the current flows in the solenoid from

both end or applying the right hand grip rule

as shown in Figure (a).

Thumb – north pole

Other fingers – direction of current in

solenoid.

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Example

Question Solution

Determine the magnetic field strength at

point X and Y from a long, straight wire

carrying a current of 5 A as shown below.

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?

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?

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?

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Question Solution

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

i. in the same direction,

ii. in the opposite direction.

Exercise

Question

Two long straight wires are oriented perpendicular to the page as shown in Figure below.

The current in one wire is I1 = 3.0 A pointing into the page and the current in the other wire is

I2 = 4.0 A pointing out of the page. Determine the magnitude and direction of the nett

magnetic field intensity at point P. (Given 0=4 x 10-7

H m-1

)

Answer: 8.93×10-6

T, 63.1° below +ve x-axis

A 2000 turns solenoid of length 40 cm and resistance 16 is connected to a 20 V supply.

Find the magnetic flux density at the end of the axis of the solenoid.

Answer: 3.95×10-3

T

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4.3 Force on a Moving Charged Particle in a Uniform Magnetic Field

L.O 4.3.1 Use magnetic force equation

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

The direction of the magnetic force can be determined by using right hand rule:

Example

Determine the direction of the magnetic force, exerted on a charge in each problems below:

where

q : magnitude of the charge

: angle between ⃗ and ⃗⃗

F

v

B

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.

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L.O 4.3.2 Describe circular motion of a charge in a uniform magnetic field

L.O 4.3.3 Use relationship FB = FC

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

r

mvBqv

2

sin and θ = 90°

Bq

mvr

The period of the circular motion, T makes by the particle is given by

v

rT

2

Bq

mT

2

The frequency of the circular motion makes by the particle is given by

T

f1

m

Bqf

2

where m : mass of the charged particle

v : magnitude of the velocity

r : radius of the circular path q : magnitude of the charged particle

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Example

Question Solution

A charge q1 = 25.0 μC moves with a speed of

4.5 x 103 m s

-1 perpendicularly to a uniform

magnetic field. The charge experiences a

magnetic force of 7.31 x 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 x 10 -3

N force. Determine

a. The magnitude of the magnetic field

b. The speed of q2.

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)

Exercise

Question

Calculate the magnitude of the force on a proton travelling 3.0 x107 m s

-1 in the uniform

magnetic flux density of 1.5 Wb m-2

, if

a. the velocity of the proton is perpendicular to the magnetic field.

b. the velocity of the proton makes an angle 50 with the magnetic field.

(Given the charge of the proton is 1.60 x 10-19

C)

Answer: 7.2×10-12

N, 5.5×10-12

N

An electron is moving in a magnetic field. At a particular instant, the speed of the electron is

3.0 x 106 m s

-1. The magnitude of the magnetic force on the electron is 5.0 x 10

-13 N and the

angle between the velocity of the electron and the magnetic force is 30. Calculate the

magnitude of the magnetic field. Given e = 1.601019

C.

Answer: 1.2 T

A proton is moving with velocity 3 x 10 5 m s

-1 vertically across a magnetic field 0.02 T.

Calculate

a. kinetic energy of the proton

b. the magnetic force exerted on the proton

c. the radius of the circular path of the proton. (mp = 1.67 x 10 -27

kg)

Answer: 7.52 x 10-17

J , 9.6 x 10 -16

N, 0.16 m

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4.4 Force on a Current Carrying Conductor in a Uniform Magnetic Field

L.O 4.4.1 Use magnetic force equation

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

The direction of the magnetic force can be determined by using right hand rule:

One tesla is defined as the magnetic flux density of a field in which a force of 1 newton

acts on a 1 metre length of a conductor which carrying a current of 1 ampere and is

perpendicular to the field.

Example

Determine the direction of the magnetic force, exerted on a conductor carrying current, I in

each problems below.

where l : length of the conductor

: angle between l and ⃗⃗

F

l

B

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Example

Question Solution

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 x 10-2

N and the current flows is12 A,

find the magnitude of magnetic field was

supplied.

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.

Exercise

Question

A straight wire with a length of 0.65 m and mass of 75 g is placed in a uniform magnetic field

of 1.62 T. If the current flowing in the wire is perpendicular to the magnetic field, calculate

the current required to balance the wire? (g = 9.81 ms-2

)

Answer: 0.70 A

A wire of length 0.655 m carries a current of 21.0 A. In the presence of a 0.470 T magnetic

field, the wire experiences a force of 5.46 N. What is the angle (less than 90o) between the

wire and the magnetic field?

Answer: 57.62°

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4.5 Forces between Two Parallel Current-Carrying Conductors

L.O 4.5.1 Derive magnetic force per unit length of two parallel current-carrying

conductors

L.O 4.5.2 Use magnetic force per unit length equation

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.

Force exerted on conductor X Force exerted on conductor Y

The magnitude of the magnetic flux density,

B2 at point Q on conductor X due to the

current in conductor Y is given by:

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

The magnitude of the magnetic flux density,

B1 at point P on conductor Y due to the

current in conductor X is given by

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

d

IB

2

101

d

IB

2

202

View from the top

sin1221 lIBF and 90

90sin2

120

21 lId

IF

lId

IF 1

2021

2

Direction: out of the page

Direction: towards

conductor Y

Direction: into the page

sin2112 lIBF and 90

90sin2

210

12 lId

IF

lId

IF 2

1012

2

Direction: towards

conductor X

d

LIIFFF

2

2102112

Conclusion: Type of the force is attractive force

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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.

Conclusion: Type of the force is repulsive force

L.O 4.5.3 Define 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 x 10-7

N m-1

.

Example

Question Solution

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

)

View from the top

d

II

L

F

2

210).π(

).)(.)(πx(

L

F

012

0101104 7

17 100.2 mNxL

F

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4.6 Torque on a Coil

L.O 4.6.1 Use torque equation

Torque is the tendency of a force to rotate an object about an axis.

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

BANI

sinNIAB

Top view Side view

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)

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L.O 4.6.2 Explain 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.

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.

This working principles of a moving coil galvanometer also used in voltmeter (multiplier),

ammeter (shunt), ohmmeter and multimeter.

Structure of a moving-coil galvanometer

coil theof anglerotation :θ radianin

constant) (torsional spring theofconstant stiffness theis where

torquerestoringforce magnetic todue Torque s

k

NAB

kI

kNIAB

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Example

Question Solution

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.

Exercise

Question

A rectangular coil of 10 cm x 4.0 cm in a galvanometer has 50 turns and a magnetic flux

density of 5.0 x 10-2

T. The resistance of the coil is 40 and a potential difference of 12 V is

applied across the galvanometer, calculate the maximum torque on the coil.

Answer: 3.0 x 10-3

N m.

A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is held in a radial

magnetic field of flux density 0.15 T and its suspension has a torsional constant of 3.0 × 10-6

N m rad-1

. Determine the current is required to give a deflection of 0.5 rad.

Answer: 1.0 × 10-3

A

Calculate the magnetic flux density required to give a coil of 100 turns a torque of 0.5 Nm

when its plane is parallel to the field. The dimension of each turn is 84 cm2 , and the current

is 9.0 A.

Answer: 66.1 mT

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4.7 Motion of Charged Particle in Magnetic Field and Electric Field

L.O 4.7.1 Explain the motion of a charged particle in both magnetic field and electric

field

L.O 4.7.2 Derive and use velocity equation in a velocity selector

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.

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

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

EB FF

qEqvB 90sin

B

Ev

When the charged particle entering

the region consists of magnetic field

only, the particle will make a

semicircular path of radius r.

CB FF

r

mvBqv

2

and B

Ev

2rB

E

m

q

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Example

Question Solution

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 Vm-1

upwards. What should be the magnitude and

direction of the magnetic field?

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.

Exercise

Question

An electron moving at a steady speed of 0.50106 m s

1 passes between two flat, parallel

metal plates 2.0 cm apart with a potential difference of 100 V between them. The electron is

kept travelling in a straight line perpendicular to the electric field between the plates by

applying a magnetic field perpendicular to the electron’s path and to the electric field.

Calculate :

a. the intensity of the electric field.

b. the magnetic flux density needed.

Answer : 0.50104 V m

1; 0.010 T

An electron with kinetic energy of 8.01016

J passes perpendicular through a uniform

magnetic field of 0.40103

T. It is found to follow a circular path. Calculate

a. the radius of the circular path.

b. the time required for the electron to complete one revolution.

(Given e = 1.761011

C kg-1

, me = 9.111031

kg)

Answer: 0.595 m, 8.92×10-8

s