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Chapter 25 Electromagnetic waves

25.1Electromagnetic vibrations


Electric vibration E = E0 sin (wt – kx)

Magnetic vibration B = B0 sin (wt- kx)

E, B and the direction of propagation of the electromagnetic wave is always perpendicular to one another

The speed of electromagnetic waves


Electromagnetic waves such as light are transverse waves. However, it is not particles which oscillate, but an electric field (E) coupled with a magnetic field B , as shown above. The speed of the waves through space, called the speed of light,c depends on the permittivity,e0 and permeability m0, of a vacuum :

speed of light (in vacuum) clip_image006 = 3.0 x 108 m s-1

25.2 Relationship between e0, m0 and c

The relationship of e0 , m0 and c

Velocity of electromagnetic wave, c = clip_image008


e0 = 8.85 x 10-12 F m-1 (Free space permittivity) and

m0 = 4p x 10-7 H m-1 (Free space permeability)

c = 2.998 x108ms-1

Example 1

The velocity of an electromagnetic waves in vacuum is given by the equation clip_image010

(a) Calculate this velocity.

(b) What deduction can be made from this information?

(c) What is it that oscillates in the electromagnetic waves?

25.3 Electromagnetic wave spectrum

The electromagnetic spectrum

Light Is one member of a whole family of transverse waves called the electromagnetic spectrum. In empty space, these waves all travel at the same speed: 300 000 km s-1 . Electromagnetic waves are emitted whenever electrons or other charged particles oscillate or lose energy. The greater the energy change, the lower the wavelength.


A table comparing electromagnetic wave and mechanical wave (e.g. sound wave)


E.M Wave

Mechanical Wave

Propagates through vacuum



Requires a medium for propagation




3.0 x 108 m s-1

(speed of light)

330 m s-1 (for sound)

Type of wave



(Sound waves)

and transverse

(Water waves)

4 Electromagnetic Wave Spectrum




Radio wave

> 1

< 108


10-4 – 1

108 – 1012


10-6 – 10-4

1012 – 1014

Visible light

4 x 10-7 – 7 x 10-7

75 x 1014 – 4.3 x 1014


10-9 – 10-7

1015 – 1017


10-11 – 10-9

1017 – 1019

Y- rays

< 10-10

> 1018

Example 2

The wavelength of visible light ranges from 4.0 x 10-7 m to 7.5 x 10-7 m. What is the frequency range of visible light?

Chapter 24 Electronics

Electronic circuits handle small, changing electric currents. The changes are called signals. The diagram below shows the main parts of a basic electronic system.



•Devices which change signals from one form to another (e.g. sound to electrical) are called transducers(a device that converts one type of energy to another).

•Input signals can be very weak. But by means of electronic circuits, they can control output signals which can be much stronger. The output power is provided by the supply (usually low voltage DC).


Two possible functions of electronic circuits are described below.

1. Switching Depending on the input signals received, the output voltage is either HIGH (close to the supply voltage) – or LOW (zero), so the output device is either ON or OFF.

2. Amplification The output signals are an amplified (magnified) version of the input signals. For example, very low voltage AC from a microphone causes a higher voltage AC output for a loudspeaker.

The voltagegain of an amplifier is defined like this:



•In electronics, the term ‘voltage’ is commonly used for potential, PD, and EMF.

The key component in electronics is the transistor. It can amplify or act as a switch. An integrated circuit (IC), in a package as on the right, may have thousands of transistors and other components formed on a single chip of silicon.

Input devices

Input transducers (e.g. microphones) are called sensors. They must be linked to the circuit in such a way that any change they detect (e.g. a pulse of sound) causes a change in input voltage.

Sensors generating a voltage These include some microphones, and those light sensors that work like solar cells . Ideally, the voltage is in proportion to the external change causing it.

Sensors with varying resistance These include:

• light-dependent resistors (LDRs) ,

•thermistors (temperature-dependent),

•strain gauges (resistance changed by stretching).

To produce the necessary voltage change, the sensor forms one part of a potential divider . For example, when light falls on the LDR below, its resistance falls. The LDR therefore takes a smaller share of the supply voltage, so the voltage V falls.


Output devices


These include loudspeakers, buzzers, LEDs, and relays. Light-emitting diodes (LEDs) These emit light when a small current passes through. They can be used as indicator lamps to show the presence of an output voltage. Like all diodes, they only conduct in one direction (see D1 and D2).

To avoid damage, a resistor must be placed in series with an LED to limit the current through it.

The LED on the right can take a current of 0.01 A, which produces a 2 V drop across it. If a circuit’s output voltage is 9 V, there needs to be a 7 V voltage drop across the resistor, so its resistance should be 7/0.01 = 700 S2.

Relays These are electromagnetic switches. A small current activates an electromagnet. This closes (or opens) a switch which can control the flow of a larger current in a separate circuit, e.g. a circuit with a mains heater or motor in it.


Impedances and matching

Resistance is one form of impedance (see D4). The microphone and loudspeaker below each have impedance. The amp[ifer has input impedance and output impedance rather as a battery has internal resistance .


In the diagram, the various impedances are represented by resistors in the input and output circuits.

In each circuit, it is the higher impedance that has the higher voltage drop across it. So

For maximum input voltage, the microphone’s impedance should be as low as possible.

For maximum output voltage, the loudspeaker’s impedance should be as high as possible. However, this gives a very low output current and almost no sound!

For maximum output power, the loudspeaker’s impedance should be the same as the output impedance. (Output voltage x output current is then at its highest.)

A transformer can be used to match impedances and give maximum power transfer. A transformer with an impedance Z in its primary circuit is equivalent to an impedance of Z(N2/ N1)2.

24.1 Operational amplifiers (Open Loop)

Op Amp beroperasi asas untuk menggandakan voltan.

Open loop bermaksud bila di bekalkan Vin di dalam op amp ianya akan menghasilkan Vo, voltan output.

Di bawah di tunjukkan gambarajah op amp


This IC has one output and two inputs. It amplifies the different( between the two input voltages. It is a differential amplifer.

• If V+ > V the output voltage is positive (+).

• If V+ < V the output voltage is negative (-).

• If V+ = V the output voltage is zero.

If only one input is in use (and the other is at zero voltage),

• a positive (+) voltage on the inverting input causes a negative (-) output voltage,

• a positive (+) voltage on the non-inverting input causes a positive (+) output voltage.


• The + and – signs on the inputs show whether they invert of not. They do not indicate + or – voltage.

Power supply A three-terminal DC supply is required, giving 0 V (earth) and typically, ±9 V. The supply connections are often excluded from diagrams.

Op amp features These include:

• an extremely high voltage gain, typically around 105, though less at high frequencies,

• a very high input impedance (e.g. 1012 W),

• a very low output impedance (e.g. 102 W).

The above op amp is also known as a differential Amplifier- it amplifies the voltage between Vin (non-inverting and inverting voltage), so the Vo given by


A0, ialah open loop gain (pemalar gandaan)


Graf Hubungan diantara Vo vs Vin (V+ – V)


Daripada graf didapati

1. Gradient ,m = A0

2. Op amp tetap juga beroperasi walaupun A0 mengganda sehingga 105 kali, tetapi inya bergantung kepada voltan yang di bekalkan.

3. voltan yang di gandakan tidak boleh melebihi voltan yang dibekalkan, pada kaadaan ini ianya dinamakan voltan tepu.

Contoh 1


The circuit diagram shows an op-amp with an open loop gain of 105 connected to a ±9V voltage supply. Sketch graph to show the variation of the input voltage Vin and output voltage Vout with time t for each of the following input voltages.

(a) sinusoidal a.c voltage of peak value 20mV

(b) sinusoidal a.c voltage of peak value 120 mV

(c) sinusoidal a.c voltage of peak value 2.0 V

24.2 Op amp (close loop)

Differential amplifier – it amplifies the voltage between inverting and non inverting, so V output given by


Open loop gain (gandaan voltan gelung-buka), clip_image027

In the case of op amp, A0 is depend on the frequency.

To reduce gain(actually to reduce frequency), op amp are used with a resistor or wire linking the output to one input.


This kind of circuit is called a close loop.

Close loop bermaksud sebahagian V0 disambung secara selari kepada voltan Vin (contohnya kepada V.

When connecting output to the inverting input. The connection is called with another name as

Negatif feed close loop back in short it just called a Negatif feedback.

Negative feedback


The amplifier circuit on the right uses a closed loop to feed back a set fraction of the output voltage to the inverting input. This is called negative feedback because the signal being fed back partly cancels the input signal.

Advantages using negatif feeback

  1. voltan output tidak sefasa dengan voltan input
  2. mengurangkan voltan input setelah ianya disuap balik
  3. reducing gain (voltage gain)

Advantages of reducing gain (Using closed loop op amp) These include the following:

• Input signals can be stronger without causing saturation and, therefore, distortion.

• The gain is less affected by temperature changes.

• The bandwidth is greater, i.e. the gain is constant over a wider range of input frequences (see below).

Note : Bandwidth = frequency which the gain is constant

Graf A0 vs frekuensi


24.3 Inverting and non-inverting amplifiers

V0 = A0 (V2 – V1) where A0= open loop gain

1 Inverting amplifier

Gain clip_image035



Op amp as an inverting amplifier


The amplifier circuit on the right uses a closed loop to feed back a set fraction of the output voltage to the inverting input. This is called negative feedback because the signal being fed back partly cancels the input signal.

To calculate the amplifier’s closed-loop voltage gain clip_image040 the following assumptions are made:

• The op amp is not saturated. It has such a high open-loop gain that the voltage difference between its inverting and non-inverting inputs must be negligible. Therefore P is effectively at 0 V. (P is called a virtual earth.)

• The current through P is negligible. Therefore, the current through Rf is the same as through Ri (see D3).

If I is the current through Rf (feedback resistor) and Ri,

voltage drop across Rf = IRf

voltage drop across Ri = IRi

But if P is at 0V,

voltage drop across Rf = 0 – Vout = -Vout

voltage drop across Ri = Vin – 0 = Vin

From the above, it follows that clip_image042

So closed-loop voltage gain = clip_image044

From the closed-loop voltage gain equation below:

• An inverting amplifier has a negative gain.

• The gain does not depend on the characteristics of the op amp. It is set by the values of Rf and Ri. For example, if Rf = 200 kW and Ri = 20 kW, the gain = -200/20 = -10.

Contoh 2

The circuit diagram shows an inverting amplifier the power supply is ± 9V.


(a) what is the voltage gain of the circuit ?

(b) what is the output voltage V0.

(i) x = 200 kW and Y = 50 kW

(ii) x = 50 kW and Y = 150 kW

2 Non-inverting amplifier

Gain clip_image048



Op amp as a non-inverting amplifier


In the circuit on the top, the input signal goes to the non-­inverting input, and the potential divider provides negative feedback. With the same assumptions as above,

closed-loop voltage gain clip_image054

A non-inverting amplifier has a much higher input impedance than the inverting type. This is useful, for example, where a high-impedance microphone is connected to the input (see Impedances and matching in H16).

Voltage follower This is a non-inverting amplifier in which Rf is zero (i.e. a direct connection) and Ri is infinite (i.e. removed). In this case, the gain is 1, so the output voltage is the same as the input voltage. However, the output impedance is much less than the input impedance, which makes the circuit useful for impedance matching.

Contoh 3

In the circuit diagram shown, the input signal is Vin and the output signal is V0


(a) state and explain the type of amplifier

(b) what is the output voltage V0 if Vin

(i) +1.5 V

(ii) +2.5 V

(c) sketch a graph to show the variation of the output voltage V0 with time t, if


24.4 Use of operational amplifiers

3 Uses of op-amps

(a) Changing sinusoidal wave forms to rectangular waveforms


(b) Automatic electrical switch/ Comparator


On the right, the LED is switched on automatically when darkness falls. The switching is done by an op amp which compares the voltages from two potential dividers:

V is fixed by the ratio of R2 to R1,

V+ varies with the resistance of the LDR.

There is no feedback, so a small difference between V and V+ saturates the op amp, causing maximum output voltage.

In bright light, the LDR has a low resistance, so V+ is low and less than V . The op amp is saturated, but its output voltage is negative, so the LED cannot conduct.

As the light intensity falls, the resistance of the LDR rises, and so does V+. When V+ > V the op amp saturates in the opposite direction. The output voltage is now positive, so the LED conducts and lights up.

(c) Voltage follower


V0=Vi, Gain ACL = 1

(d) Summing amplifier



(e) Differential Op-Amp



(f) AC Amplifier


(g) The Schmitt Trigger


24.5 Oscillators


The Wein oscillator – used both positive and negative feedback

1.positive feedback through R1 and C1

2.Negative feedback through resistor in series

3. R1=R2 = R, C1 = C2 = C

4. Frequency clip_image080

Chapter 23 Alternating currents

AC terms

The graph below shows how the current from an AC supply varies with time. Here are some of the terms used to describe AC. Note the similarities with those used for circular motion and SHM.

Continue reading

Chapter 22 Electromagnetic induction



Above, a magnet is moved into a coil. If the flux through the coil changes (at a steady rate) by DF in time Dt, then an EMF of clip_image004 is induced in each turn. But there are N turns in series. So, the total induced EMF E is as follows:


For example, if the flux changes by 6 Wb in 2 s, and the coil has 100 turns, then the total induced EMF is 300 V.

Continue reading

Chapter 21 Magnetic fields

21.1 Magnetic field B

Magnetic Field B is a field of force that exist around the magnetic body or current carrying conductor[1].

Magnetic field B, also known as magnetic flux density or magnetic field Intensity[2].


we can summarize the result with the following equation:

(Direction) clip_image004

(Magnitude) clip_image006

21.2 Force on a moving charge


If the charge q moving in a magnetic field B with a velocity v. The angle between B and v is q the charge will experience magnetic force as clip_image006[1].

Unit bagi magnetic flux density adalah Tesla,T atau Wbm-2

To find out the direction of the force also can be shown using Fleming’s left-hand rules.

Contoh 1

Satu eletron bergerak dalam medan magnet. Pada satu ketika laju elektron ialah 3.0 ´ 106 ms-1. Magnitud daya magnet yang bertindak pada elektron ialah 5.0 ´ 1013 N. Sudut di antara arah halaju elektron dan arah daya magnet ialah 30°. Kira ketumpatan fluks magnet pada kedudukan elektron medan itu.

[cas elektron = -1.6 ´ 10-19 C]

21.3 Force on a current-carrying conductor


Above, a current-carrying wire is at right-angles to a uniform magnetic field. The field exerts a force on the wire. The direction of the force is given by Fleming’s left-hand rule . The size of the force depends on the current I, the length l in the field, and the strength of the field. This effect can be used to define the magnetic field strength, known as the magnetic flux density, B:

FB = BIl (1)

B is a vector. The SI unit of B is the testa (T). For example, if the magnetic flux density is 2 T, then the force on 2 m of wire carrying a current of 3 A is 2 x 2 x 3 = 12 N.

If a wire is not at right angles to the field, then the above equation becomes

FB=BIl sinq

where q is the angle between the field and the wire. As q becomes less, the force becomes less. When the wire is parallel to the field, sin q = 0, so the force is zero.

Contoh 2


Satu konduktor lurus panjang 2m membawa arus 1.5A dan di letakkan dalam medan magnet. Medan itu mempunyai ketumpatan fluks magnet bermagnitud 3.0 ´ 10-3 T. Sudut antara arah medan dan konduktor ialah 30°. Kirakan magnitud daya magnet yang bertindak pada konduktor dan nyatakan arah daya itu.

21.4 Magnetic fields due to currents


The Biot-Savart law,

On the top, a short length dl of thin wire, carrying a current I, causes a magnetic flux density dB at P. According to the Biot-Savart law


With a suitable constant, this can be turned into an equation:


For a vacuum (and effectively for air), the value of k is 10-7 T m A-1. However, in practice, another constant, m0, is used, and the above equation is rewritten as follows:

clip_image020 (2)

m0clip_image022 is called the permeability of free space. Its value is 4p x 10-7 T m A-1. This is not found by experiment. It is a defined value, linked with the definition of the ampere .

Calculating magnetic flux density Using equation clip_image020[1] and calculus, it

is possible to derive equations for B near wires and inside coils carrying a current I.


B near an infinitely long, thin, straight wire At a distance a from such a wire (as on the top)


Contoh 3

Arus sebanyak 10A mengalir melalui seutas dawai mencancang lurus yang panjang. Kirakan komponen mengufuk medan magnet Bumi. Jika terdapat satu titik neutral 8.0cm dari dawai itu. [m0= 4p ´ 10-7 Hm-1]


B at the centre of a thin coil Or the axis of such a coil, of N turns and radius r (as on the top),


Contoh 4

Arus yang sama magnitud mengalir secara bergilir-gilir melalui seutas dawai tegak lurus yang panjang dan satu gegelung membulat berjejari 10cm dimana satahnya tegak dan selari dengan komponen mengufuk medan magnet pada titik berjarak 10cm di utara atau di selatan dari dawai itu terhadap paduan medan magnet di pusat gegelung itu.

Contoh 5

Satu gegelung membulat diletakkan, dimana satahnya adalah mencancang dan paksinya membuat sudut kecil q dengan meridian magnet itu. Arus dilalukan mengelilingi gegelung supaya medan yang dihasilkan berkecenderungan untuk menentang medan Bumi. Satu magnet mengufuk yang kecil di letakkan di pusat gegelung itu.

Perihalkan dan terangkan kesan ke atas magnet itu apabila kekuatan arus bertambah. Gegelung itu mempunyai jejari min 10cm,50 lilitan dan arus sebanyak 0.5A mengalir didalamnya. Jika q=12° dan magnet mengarah ke Timur-Barat.Kirakan komponen mengufuk medan magnet Bumi.


B inside an infinitely long solenoid (coil) The field inside such a solenoid (as on above) is uniform. If n is the number of turn per unit length (per m), then


This equation is a reasonable approximation for any solenoid which is at least ten times longer than it is wide.

Contoh 6

Berapa banyakkah lilitan mesti dipunyai oleh satu solenoid jika panjang solenoid itu 2.0m dan arus 5.0A mengalir melaluinya supaya arus menghasilkan ketumpatan fluks magnet bermagnitud 2.0 ´ 10-2 T di pusat solenoid ?

B inside a solenoid with a core The value of B is changed by a core. For example, with a pure iron core, B is increased by a factor of about 1000 (depending on the temperature). The previous equation then becomes:

B = µrµ0 nI

µr, is called the relative permeability of the material. So, for pure iron, µr, is about 1000.

An electromagnet is a solenoid with a core of high µr.

21.5 Force between two current-carrying conductors

Force between two current-carrying wires


X and Y above are two infinitely long, straight wires in a vacuum. The current in X produces a magnetic field, whose flux density is B at Y. As a result, there is a force on Y. F is the force acting on length l.

From equation (3) clip_image036

From equation (1) FB = BI2 l



• The above equation gives the force of X on Y. Working out the force of Y on X gives exactly the same result.

• If the two currents are in the same direction (as above), then the wires attract each other. If the two currents are in opposite directions, then the wires repel.

Contoh 7


Tiga konduktor yang selari, P, Q dan R berada pada satah yang sama.Jarak antara P dan Q adalah 4cm dan antara Q dan R ialah 2 cm. Arus dalam P dan Q adalah 10A dan 5A masing-masing dan mengalir pada arah yang berlawanan.Daya paduan ke atas Q adalah 5 ´ 10-4N per meter dan bertindak ke arah P. Kirakan magnitud dan arah arus dalam R.

21.6 Definition of ampere

Defining the ampere

The SI unit of current is defined as follows:

One ampere is the current which, flowing through two infinitely long, thin, straight wires placed one metre apart in a vacuum, produces a force of 2 x 10-7 newtons on each metre length of wire.

Using the various factors in equation

clip_image038[1] ,

the above definition can be expressed in the following form (for simplicity, units have been omitted):

If I1=I2=1A,a=1m and l=1m,then F= 2×10-7N. Substituting these in equation above gives µ0 = 4p x 10-7. The above definition is not a practical way of fixing a standard ampere. This is done by measuring the force between two current-carrying coils.

Current Balance

Measure current by using the principle of the force between two paralell wire using mechanical




m = jisim pemberat W

b = jarak F ke tuas

c = jarak berat ke tuas

L = jarak XY

apabila sistem tuas seimbang kita dapati


Known that



Maka FB = µ0n I . IL

Then clip_image048


21.7 Torque on a coil


Lets consider the force on PQ and RS that involve in producing the torque.

Force that produce on PQ and RS have the same magnitude but different direction.

so force,F



L = height of coil(length of the coil in the magnetic field) and N = no. of turns carrying current,I.





q = angle (below the horizontal from the plan view) between magnetic flux density and the coil.

[coil area,A = Lb]

So the equation for torque will be



q = 0°

t = BIAN

Contoh 8

21.8 Determination of ratio e/m and q/m

Using an electron gun

Lets consider the electron gun.


Lets consider the movement of electron in an electro gun

1. The kinetic energy gain by electron = Electric potential energy of electron

2. Electron velocity :



3. Electric force, FE = Magnetic Force, FB , when it’s in equilibrium



4. From equation (2) and (3) we can conclude:



Todetermine any positive charge,q/m using spectrometer.

1. any positivly charge velocity,


2. The charges follow the circular path, the centripetal force of the charge.


3.Magnetic force of charges



Also clip_image084



clip_image090 for any specific charges.

21.9 Hall effect

1. apabila arus mantap I melalui konduktor/semikonduktor


2. kemudian diletakkan kedalam medan magnet yang kuat dimana I berserenjang dengan B.

3. Medan magnet melencongkan cas positif ke bahagian atas dan cas negatif ke bahagian bawah


4. Pemisahan ini menghasilkan medan elektrik E yang berserenjang dengan arah aliran arus I

– Bila cas bertambah

– Medan elektrik juga bertambah

– Daya gerak elektrik juga meningkat


5. Pemisahan akan berhenti bila daya E, FE=FB

Magnetic force = Electric force


clip_image100 where [clip_image102 n ialah halaju hanyut elektron,n bilangan elektron bebas]


Hall voltage clip_image106


[1] Oxford Dictionary

[2] Fajar Bakti STPM V2 m.s 113


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