cie electricity and magnetism sample pages

7/29/2019 CIE Electricity and Magnetism Sample Pages

1/31

www.gcephysics.com

CIE-Electricity-and-magnetism

Copyright 2011-2012 gcephysics.com. All Rights Reserved

http://www.gcephysics.com/

I
http://www.gcephysics.com/http://www.gcephysics.com/http://www.gcephysics.com/


2/31

www.gcephysics.com

Electricity-and-magnetism........................................................................................................................1

Chapter 1 Electric Fields and Capacitance................................................................................................1

3-1 Electric charge Coulomb law...................................................................................................1

1-2 Electric field Electric field strength .......................................................................................11-3 Electric field lines .......................................................................................................................1

1-4 Electric potential difference Electric potential.........................................................................1

1-5 Linking the electric field and the electric potential.....................................................................1

1-6 Capacitor Capacitance .............................................................................................................1

1-7 The motions of charged particles in the uniform electric field ...................................................1

1-8 Comparison of electric and gravitational fields ..........................................................................1

1-9 29 Worked examples........................................................................................................................1

Chapter 2 Magnetic fields .........................................................................................................................1

2-1 Magnetic flux density .................................................................................................................1

2-1-1 Magnets and fields...........................................................................................................1

2-1-2 Magnetic fields from currents..........................................................................................1

2-1-3 Magnetic force on moving charges and flux density (B).................................................1

2-1-4 Magnetic force on a current.............................................................................................1

2-2 Moving charges in a magnetic field............................................................................................2

2-2-1 Charged particles in circular orbits..................................................................................2

2-2-2 The cyclotron...................................................................................................................2

2-3 19 Worked examples........................................................................................................................2

Chapter 3 Electromagnetic induction........................................................................................................2

3-1 Electromagnetic induction phenomena.......................................................................................23-2 Faraday law of electromagnetic induction and Lenzs law.........................................................3

3-3 Motional emf ..............................................................................................................................4

3-4 AC Generators ............................................................................................................................5

3-5 The physics parameters of alternating current ............................................................................5

3-6 The transformer...........................................................................................................................6

3-7 18 Worked examples........................................................................................................................9

Chapter 4 Current Electricity ..................................................................................................................27

4-1 Charge, current and potential difference...................................................................................27

4-2 Resistance .................................................................................................................................27

4-3 Ohms law and measuring resistance........................................................................................27

4-3-1 Graph of V against I and I against V for the ideal resistor ............................................27

4-4 Resistivity .................................................................................................................................27

4.5 Circuit components symbols:....................................................................................................27

4-5-1 IV and VI graphs for different conductors.............................................................27

4-6 Circuits......................................................................................................................................27

4-7 Potential divider........................................................................................................................27

4-8 Electromotive force and internal resistance..............................................................................27

4-9 Alternating currents ..................................................................................................................27

4-9-1 AC Generators ...............................................................................................................274-9-2 The physics parameters of alternating current...............................................................27

I


3/31

www.gcephysics.com

II

4-9-3 The transformer .............................................................................................................27

4-10 39 Worked examples....................................................................................................................27

Chapter 5 Gravitation..............................................................................................................................27

5-1 Newtons law............................................................................................................................27

5-2 Gravitational field strength .......................................................................................................28

5-3 Gravitational potential ..............................................................................................................28

5-4 Orbits of planets and satellites ..................................................................................................28

5-4-1 An orbit equation ...........................................................................................................28

5-5 36 Worked examples......................................................................................................................28


4/31

www.gcephysics.com

Electricity-and-magnetism

Chapter 1 Electric Fields and Capacitance

3-1 Electric charge Coulomb law

1-2 Electric field Electric field strength

1-3 Electric field lines

1-4 Electric potential difference Electric potential

1-5 Linking the electric field and the electric potential

1-6 Capacitor Capacitance

1-7 The motions of charged particles in the uniform electric field

1-8 Comparison of electric and gravitational fields

1-9 29 Worked examples

Chapter 2 Magnetic fields

2-1 Magnetic flux density

2-1-1 Magnets and fields

2-1-2 Magnetic fields from currents

2-1-3 Magnetic force on moving charges and flux density (B)

2-1-4 Magnetic force on a current

1


5/31

www.gcephysics.com

2-2 Moving charges in a magnetic field

2-2-1 Charged particles in circular orbits

2-2-2 The cyclotron


Chapter 3 Electromagnetic induction

3-1 Electromagnetic induction phenomena

1.1 Magnetic flux and flux linkage

Consider a situation where the magnetic field is uniform in magnitude and

direction. Place a rectangular coil in the B-field. (Fig. 3.1)

Fig. 3.1 a rectangular coil in the B-field

Rectangular coil ofarea A

The magnetic flux, , is defined as the product of the field magnitude by the

area crossed by the field lines.

cosB A BA is the component of B perpendicular to the coil,Where B

is the angle

between B and the normal to the loop.

SI units: Weber (Wb) and 21 1Wb T m

0

Note:

if cos, then B A BA BA

if the coil has N turns in series, then there is a flux through a coil of N

turns. And we call the flux linkage.N

2


6/31

www.gcephysics.com

cosN N B A N BA

Flux linkage:

1.2 Electromagnetic induction phenomena

Provided that the magnetic flux varies in a circuit, there will be a current in

the circuit and no matter what method is used. Generating a current in amagnetic field is called electromagnetic induction, and the current is called

an induction current.

3-2 Faraday law of electromagnetic induction and Lenzs law

(i) Faradays law of magnetic induction (Fig. 3.2):

The emf induced in a loop of wire is proportional to the rate of change of

magnetic flux through the coil. This statement is known as Faradays law.

We often call the emf induced by a changing magnetic flux and induced emf,

and the current it produces is called an induced current or an induction

current.

Fig. 3.2 a current is set up in the circuit as

long as there is relative motion between the

magnet and the loop

(ii) Lenzs law

Lenzs law: the polarity of the induced emf is such that is produces a current

whose magnetic field opposes the change in magnetic flux through the loop.

That is, the induced current tends to maintain the original flux through thecircuit.

We can write Faradays law of induction mathematically with the equation:

t

Here E is the induced emf in one loop of wire,

t

is the change in

magnetic flux, and is the time interval over which the change takes place.

The minus sign indicates the direction of the emf and is in agreement with

Lenzs law. If we apply this equation to a coil containing N loops or turns of

3


7/31

www.gcephysics.com

wire, then

Nt

l

3-3 Motional emf

In our discussion of Faradays law, we examined the changing flux in a loop

of wire due to the relative motion of the field and the loop. Now we extend

our study to include the effects of a magnetic field on a straight piece of wire

or other conductor that moves relative to the field. We will see that an emf is

developed between the ends of the wire.

To investigate the induction of emf on a moving conductor, lets consider a

conducting bar of length sliding along a stationary U-shaped conductor

that is perpendicular to a uniform magnetic field B (Fig. 3.3). The bar moveswith a constant speed v in the x direction. As the bar moves, the area

enclosed by the loop consisting of the bar and the U-shaped conductor

increases. Consequently, the magnetic flux through the loop increases. The

emf developed around the loop is obtained from Faradays law:

( ) ( )BA A l x xB B Bl

t t t t t

Sincex

vt

E Blv

, we have

4


8/31

www.gcephysics.com

3-4 AC Generators

AC generator (Fig. 2.4)

Fig. 3.4 a simple AC generator

As the Fig. 2.4 shows, if the field B is uniform and the loop rotates with a

constant angular speed , the magnetic flux through the single loop of area

A may be expressed as

cos cosB A BA BA t

This flux is a function of time, from the Faradays law, we calculate the emf

to be

(cos )BA tE

t t

And

sincon t

t

t

sinE BA t

Thus,

sinE NBA t

Note: generalize the result to N loops:

So, the output of an alternator is a sinusoidal emf. When an alternator is

connected to a closed circuit, it produces a sinusoidally alternating current.

3-5 The physics parameters of alternating current

Alternating current: the current repeatedly reverses its direction.

5


9/31

www.gcephysics.com

Frequency ( ) of alternating current: number of cycle per second.f

Time period (T) of alternating current: the time for one full cycle.

And1 1

T or f f T

Peak value of alternating current (or potential difference): the maximum

current (or potential difference) that is the same value in either direction.

Peak-to-peak value: the difference of the peak value in the opposite direction.

Root mean square value of an alternating current: the value of direct current

that would give the same heating effect as the alternating current in the same

resistor.

The root mean square value of an alternating current:

0

1

2rms

I I

The root mean square value of an alternating potential difference:

0

1

2rms

V V

Where I0 is the peak current, V0 is the peak voltage.

Therefore:

The root mean square value of an alternating current or pd

=1

2the peak value

Then the mean powersupplied to the resistor:2

2 rmsrms rms rms

Vp I R I V

R

3-6 The transformer

One of the most important applications of mutual induction and

self-induction takes place in a transformer. A transformer is a device for

increasing or decreasing an AC voltage.

Fig. 3.5 shows a drawing of a transformer. The transformer consists of an

iron core on which two coils are wound: a primary coil with turns, and a

secondary coil with

pN

sN turns.

6


10/31

www.gcephysics.com

The primary coil is connected to an ac generator. And the alternating current

in the primary coil establishes a changing magnetic field in the iron core.

Because iron is easily magnetized, it greatly enhances the magnetic field

relative to that in an air core and guides the field lines to the secondary coil.

In a well-designed core, nearly all the magnetic flux that passes through

each turn of the primary coil also goes through each turn of the secondary

coil. Since the magnetic field is changing, the flux through the primary and

secondary coils is also changing, and consequently an emf is induced in both

coils. In the secondary coil the induced emf sE arises from mutual induction

and is given by Faradays law of electromagnetic induction as

s sE Nt

pE

In the primary coil the induced emf is due to self-induction and is

specified by Faradays law as

p pE Nt

The termt

is the same in both of these equations, since the same flux

penetrates each turn of both coils. So

s s

p p

E N

E N

In a high-quality transformer the resistances of the coils are negligible, so the

7


11/31

www.gcephysics.com

magnitudes of the emfs,sE pand E , are nearly equal to the terminal

voltages, sV pVand , across the coils. The relations s

p p

E N

E N is called the

transformer equation and is usually written in terms of the terminal voltages:Thus

Transformer equation:

sec s s

p p

V Nondary voltage

primary voltage V N

According to the transformer equation, if sN pNis greater than , the

secondary voltage is greater than the primary voltage. In this case we have a

step-up transformer. On the other hand, if sN pNis less than , the secondary

voltage is less than the primary voltage. In this case we have a step-down

transformer.

The transformer can change the voltage of the secondary coil, but the energy

conservation requires that the energy delivered to the secondary coil must be

the same as the energy delivered to the primary coil, provided no energy is

dissipated in heating these coils or is otherwise lost. In a well-designed

transformer, less than 1% of the input energy is lost in the form of heat.

Noting that power is energy per unit time, and assuming 100% energy

transfer, the average power pP delivered to the primary coil is equal to the

average power delivered to the secondary coil; sosP

p p s sI V I V

Or

ps s

p s p

IV N

V I N

But in fact, there is energy dissipated, so the efficiency of the transformer can

be calculated in this form:

100%s s

p p

I Vtransformer efficiency

I V

8


12/31

www.gcephysics.com


1. A magnetic field with 0.2B T pass through a circle with an area of

, as Fig. 1.1 shows, find the flux through the circle in each orientation

shown.

260cm

Fig. 1.1

Solution:

We know that B A

4 2 3(0.2 ) (60 10 ) 1.2 10

(a) B A BA T m Wb

(b) 0 0 3 33

( cos30 ) cos30 1.2 10 1.04 102

B A B A BA Wb Wb

2. A square coil 10 cm on a side has 20 turns of wire. Initially, it is at rest

between the poles of a magnet whose field is 0.25 T, with the plane of the coil

perpendicular to the field. What average voltage (emf) is produced on the coil

if it is withdrawn completely from the field in 0.10 s?

Solution:

To find the emf, we must first find the change in magnetic flux . Then

from knowledge of the time interval t , we can use the equation

Nt

to compute E. and we know that B A

l

2l

2

iB A Bl

0

, because the field is

perpendicular to the plane of the coil, and the initial flux is just the produce

of the field B with the area of the coil. For a square coil of side length , the

area is , thus the initial flux is

, thusAfter the coil is withdrawn from the magnet,

20 i i Bl

Therefore, the induced emf is

9


13/31

www.gcephysics.com

2 2 2(0.25 ) (10 10 )20 0.5

0.10

Bl T mN N V

t t s

E Blv

3. A 0.27-m aluminum rod is rapidly moved between the poles of alaboratory magnet that generates a magnetic field of 0.89 T. with what speed

must the rod be moved so that the emf developed between the ends is 1.5 V?

the rod and its motion are perpendicular to the direction of the magnetic field.

Solution:

From the equation , we can get the speed

1.56.2 /

0.89 0.27

E Vv m s

Bl T m

sinE NBA t

4. A coil made of 25 loops, each of area 0.01 m2, is rotated about an axis

perpendicular to the earths magnetic field (about ) at a frequency

of 60 Hz. what is the peak emf generated by the coil

55.0 10 T

Solution:

The peak emf in the coil is given by the equation , modified

by including the number of loops N and with sin t 1 :

E NBA2 2 60 377

The angular frequency f Hz Hz

5 35.0 10 0.01 377 4.71 10 4.7125 T V mV E NBA

, thus the emf becomes

5. (a) In an experiment to illustrate electromagnetic induction, a permanent

magnet is moved towards a coil, as shown in Fig. 5.1, causing an emf to be

induced across the coil.

Using Faradays law, explain why a larger emf would be induced in this

experiment if a stronger magnet were moved at the same speed.

10


14/31

www.gcephysics.com

You may be awarded additional marks to those shown in brackets for the

quality of written communication in your answer.

Solution:

A stronger magnet produces flux lines closer together, which causes more

flux lines cut per unit time. And the rate of change of flux is greater, thus the

emf is larger.

Review:

The emf induced in a loop of wire is proportional to the rate of change of

magnetic flux through the coil. This statement is known as Faradays law.

We often call the emf induced by a changing magnetic flux and induced emf,

and the current it produces is called an induced current or an induction

current.

(b) A conductor of length l is moved at a constant speed v so that it passes

perpendicularly through a uniform magnetic field of flux density B, as shown

in Fig. 5.2.

(i) Give an expression for the area of the magnetic field swept out by the

conductor in time t.

Solution:

The distance traveled in time t is given by s v t

A ls lv t

. Thus the area swept

out by the conductor is

(ii) Show that the induced emf, , across the ends of the conductor is given

by

Blv

Solution:

From Faradays law:

( )BA A lv tB B Blvt t t t

11


15/31

www.gcephysics.com

The minus sign indicates the direction of the emf and is in agreement with

Lenzs law.

(c) A simple electrical generator can be made from a copper disc, which is

rotated at right angles to a uniform magnetic field, directed into the plane of

the diagram (Fig. 5.3). An emf is developed across terminals P (connected to

the axle) and Q (connected to a contact on the edge of the disc).

Fig. 5.3

The radius of the disc is 64 mm and it is rotated at 16 revolutions per second

in a uniform magnetic field of flux density 28mT.

(i) Calculate the angular speed of the disc.

Solution:

The time taken for one revolution (time period T) is given by

1

16T

Therefore, the angular speed is given by

22 16 32 101 /rad s

T

(ii) Calculate the linear speed of the mid-point M of a radius of the disc.

Solution:

Distance between M and the axle is64

322 2

M

rr mm

332 10 101 3.23 /M M

v r m s

E Blv

Thus, the linear speed of M is given by

(iii) Hence, or otherwise, calculate the emf induced across terminals P and Q.

Solution:

The equation gives the induced emf, and

12


16/31

www.gcephysics.com

364 64 10PQl r r mm m

3.23 /M

v v m sAverage linear speed of the disc is

3 3 328 10 64 10 3.23 5.8 10E Blv V

cosB A BA

Therefore,

6. Which one of the following is not a unit of magnetic flux?

A NmA-1

B Wb

C Tm2

D Vs-1

Solution:The magnetic flux, , is defined as the product of the field magnitude by the

area crossed by the field lines.

Where is the component of B perpendicular to the coil,B

is the angle

between B and the normal to the loop.

SI units: Weber (Wb) and 21 1Wb T m

Andsec

1 ( ) ( )newton ond N stesla T

C mcoulomb meter

Thus

2 21 1 1 ( ) 1 1N s N s m N mWb T m mC m C A

As 1 1C

As

Choose (D).

7. The magnetic flux through a coil of 5 turns changes uniformly from

to in 0.50 s. What is the emf induced in the coil due to

this change in flux?

315 10 Wb

37.0 10 Wb

A 14mV

B 16mV

C 30mV

D 80mVSolution:

13


17/31

www.gcephysics.com

Using the Faradays law of induction, the magnitude of the induced emf is

given by3 315 10 7 10

5 0.08 800.5

N V mVt

Choose (D).

8. A wire of length 0.50 m, forming part of a complete circuit, is positioned at

right angles to a uniform magnetic field. The graph shows how the force

acting on the wire due to the magnetic field varies as the current through the

wire is increased.

What is the flux density of the magnetic field?

A 2 mT

B 4 mT

C 15 mT

D 25 mT

Solution:The magnetic force on the wire is given by:

( )F BIL BL I

And the gradient of the graph is10

2 /5

mNk mN A

A

2 / (0.5 ) 2 /

That is

BL mN A m B mN A

We can get

B = 4 mT.

14


18/31

www.gcephysics.com

Choose (B).

9. A uniform magnetic field is produced by mounting two flat magnets on a

U-shaped iron frame, so that the north and south poles are facing, as shown in

Fig. 9.1. The flux density of the magnetic field is 45 mT and may be assumed

to act only over the area of the pole faces, which measure 40 mm by 20 mm.

This magnet arrangement rests on the pan of a top pan balance.

Fig. 9.1

(a) A horizontal wire is placed in the centre of the magnetic field and aligned

to make it perpendicular to the flux lines. When a current is passed through

the wire, the balance reading increases by 31.4 10 kg .

Calculate the current in the wire.

Solution:

Magnetic force on a current:

sinF BIL 0sin sin 90 1

, the wire is perpendicular to the flux lines. Thus

Therefore, the current in the wire3

3 3

(1.4 10 )(9.81 / )

7.63(45 10 )(40 10 )

F kg N kg

I ABl T m

(b) The wire is disconnected from the current source and its ends are

connected to a sensitive voltmeter. When the wire is moved rapidly, vertically

upwards across the whole magnetic field, cutting all of the flux lines

perpendicularly, the voltmeter gives a reading.

Calculate

(b) (i) the magnetic flux change experienced by the wire during its movement

15


19/31

www.gcephysics.com

completely across the magnetic field,

(b) (ii) the time taken for the wire to pass completely across the magnetic

field, assuming it is moved at constant speed, if the voltmeter reads 0.15 mV.

Solution:

(i) The magnetic flux, cosB A BA

, where is the angle between B

and the normal to the area; and here cos cos0o 1

Thus

cosB A BA BA

3 3 4 240 10 20 10 8 10

The wire is moved across the whole magnetic field, so the area is given by

A m m m

3 4 2 545 10 8 10 3.6 10

BA T m Wb

(ii) From the equationt

, we can get the time taken

5

3

3.6 100.24

0.15 10

Wbt s

V

10. The magnetic flux through a coil ofNturns is increased uniformly from

zero to a maximum value in a time t. An emf, E, is induced across the coil.

What is the maximum value of the magnetic flux through the coil?A

Et

N

BN

Et

C EtN

DE

Nt

Solution:

Using the equation max max0

0N N N

t t t

Thus,

max

Et

N

Choose (A).

11. An aircraft, of wing span 60 m, flies horizontally at a speed of 150ms-1

, If

the vertical component of the Earths magnetic field in the region of the plane

16


20/31

www.gcephysics.com

is , what emf is induced across the wing tips of the plane?51.0 10 T

A 0.09V

B 0.90V

C 9.0V

D 90V

Solution:

Motional emf: 51.0 10 60 150 0.09E Blv V

Choose (A).

12. The graph shows how the magnetic flux passing through a loop of wire

changes with time.

What feature of the graph represents the magnitude of the emf induced in the

coil?

A the area enclosed between the graph line and the time axis

B the area enclosed between the graph line and the magnetic flux axis

C the inverse of the gradient of the graph

D the gradient of the graph

Solution:

The induced emf in one loop of wire is given by

t

Therefore, choose (D).

13. A 20-W high-intensity bulb in a desk lamp has a resistance of 7.2

when burning. The lamps power comes from the secondary of a small

transformer, the primary of which is connected to a 120-V circuit. (a) What is

the ratio of the number of turns on the primary winding to the number of

turns on the secondary winding? (b) What is the minimum current through

the primary when the lamp is on?

17


21/31

www.gcephysics.com

Solution:

To find the relative number of turns, we first need to find the secondary

voltage. We can calculate the secondary voltage from knowledge of the

lamps resistance and the power expended. Then we can use the primary and

secondary voltages to find the ratio of turns. We will assume that the power

input and the power output are equal and use the equationp p s s

I V I V to find

the primary current.

(a) The secondary voltages

V is found from

2s

s s ss

VP I V

R

(20 )(7.2 ) 12s s sV P R W V

According to the equationsec s s

p p

V Nondary voltage

primary voltage V N , the number of turns is

equal to the ratio of the voltages. Setting 120p

V V , we have

12010

12

p p

s s

N V V

N V V

20p p s s

That is, the primary coil has 10 turns of wire for each turn of wire in the

secondary coil.

(b) We now equate the input power to the output power to find the primary

current:

I V I V W

200.17

120p

WI A

V

pN

14. (a) (i) Outline the essential features of a step-down transformer when in

operation.

Solution:

A transformer is a device for increasing or decreasing an AC voltage. The

transformer consists of an iron core on which two coils are wound: a primary

coil with turns, and a secondary coil with sN turns. The primary coil is

connected to an ac generator. And if sN pNis less than , the secondary

voltage is less than the primary voltage. In this case we have a step-down

transformer.

18


22/31

www.gcephysics.com

(a) (ii) Describe two causes of the energy losses in a transformer and discuss

how these energy losses may be reduced by suitable design and choice of

materials.

The quality of your written communication will be assessed in this question.

Solution:

Any two from the following four:

(1) When a transformer is in operation, there are ac currents in the primary

and secondary coils. The coils have some resistance and the currents cause

heating of the coils, causing some energy to be lost. This loss may be reduced

by using low resistance wire for the coils. This is most important for the high

current winding (the secondary coil of a step-down transformer). Thick

copper wire is used for this winding, because thick wire of low resistivity has

a low resistance.

(2) The ac current in the primary coil magnetises, demagnetises and

remagnetises the core continuously in opposite directions. Energy is required

both to magnetise and to demagnetise the core and this energy is wasted

because it simply heats the core. The energy wasted may be reduced by

choosing a material for the core which is easily magnetised and demagnetised,ie a magnetically soft material such as iron, or a special alloy, rather than

steel.

(3) The magnetic flux passing through the core is changing continuously.

The metallic core is being cut by this flux and the continuous change of flux

induces emfs in the core. In a continuous core these induced emfs cause

currents known as eddy currents, which heat the core and cause energy to be

wasted. The eddy current effect may be reduced by laminating the core

instead of having a continuous solid core; the laminations are separated by

very thin layers of insulator. Currents cannot flow in a conductor which is

discontinuous (or which has a very high resistance).

(4) If a transformer is to be efficient, as much as possible of the magnetic flux

created by the primary current must pass through the secondary coil. This

will not happen if these coils are widely separated from each other on the

core. Magnetic losses may be reduced by adopting a design which has the

two coils close together, eg by better core design, such as winding them on

19


23/31

www.gcephysics.com

top of each other around the same part of a common core which also

surrounds them.

(b) Electronic equipment, such as a TV set, may usually be left in standby

mode so that it is available for instant use when needed. Equipment left in

standby mode continues to consume a small amount of power. The internal

circuits operate at low voltage, supplied from a transformer. The transformer

is disconnected from the mains supply only when the power switch on the

equipment is turned off. This arrangement is outlined in Fig. 14.1.

Fig. 14.1

When in standby mode, the transformer supplies an output current of

300 mA at 9.0V to the internal circuits of the TV set.

(b) (i) Calculate the power wasted in the internal circuits when the TV set is

left in standby mode.

Solution:

The power wasted is given by39 300 10 2.7P VI W

(b) (ii) If the efficiency of the transformer is 0.90, show that the current

supplied by the 230 V mains supply under these conditions is 13mA.

(b) (iii) The TV set is left in standby mode for 80% of the time. Calculate the

amount of energy, in J, that is wasted in one year through the use of the

standby mode.71 3.15 10year s

Solution:

(ii)

The power in the mains supply is given by

2.73

0.9 0.9s

PP W

Thus the current supplied by the mains supply is

30.013 13

230

P WI A mA

V V

(iii)

20


24/31

www.gcephysics.com

The total time 7 73.15 10 80% 2.52 10t s

7 73 2.52 10 7.6 10

Thus, the energy wasted is given by

E Pt J

(b) (iv) Show that the cost of this wasted energy will be about 4, if electrical

energy is charged at 20 p per kWh.

Solution:7

37.6 10Energy wasted 21 10 ( ) 21( )3600

JWh kWh

s

Cost of wasted energy 21 20 420 4.20p

(c) The power consumption of an inactive desktop computer is typically

double that of a TV set in standby mode. This waste of energy may be

avoided by switching off the computer every time it is not in use. Discuss one

advantage and one disadvantage of doing this.

Solution:

Advantage: Cost saving, saving essential fuel resources, reduced global

warming.

Disadvantage: energy required to reboot may exceed energy saved by

switching off.

15. A coil rotating in a magnetic field produces the following voltage

waveform when connected to an oscilloscope.

With the same oscilloscope settings, which one of the following voltage

waveforms would be produced if the coil were rotated at twice the original

speed?

21


25/31

www.gcephysics.com

Solution:

The induced emf is given by

sinE BA t

Where is angular speed

And now the speed is doubled, thus the induced emf now is given by

1(2 )sin(2 ) 2 sin(2 )E BA t BA t

Thus, the amplitude is doubled, but the time period is halved.

Choose (C).

16. A 230V, 60 W lamp is connected to the output terminals of a transformer

which has a 200 turn primary coil and a 2000 turn secondary coil. The

primary coil is connected to an ac source with a variable output pd. The lamp

lights at its normal brightness when the primary coil is supplied with an

alternating current of 2.7 A.What is the percentage efficiency of the transformer?

A 3%

B 10%

C 97%

D 100%

Solution:

The current passing through the lamp is given by

22


26/31

www.gcephysics.com

600.26

230

PI A

V

And

ps

s p

I N

I N

The current passing through the secondary coil is given by

2.7 2000.27

2000

p p

s

s

I NI A

N

Therefore, the efficiency is given by

0.26100% 100% 97%

0.27s

I

I

Choose (C).

17. Fig. 17.1 shows an end view of a simple electrical generator. A

rectangular coil is rotated in a uniform magnetic field with the axle at right

angles to the field direction. When in the position shown in Fig. 17.1 the

angle between the direction of the magnetic field and the normal to the plane

of the coil is.

Fig. 17.1

(a) The coil has 50 turns and an area of 1.9103

m2. The flux density of the

magnetic field is 2.8102

T. Calculate the flux linkage for the coil when

is 35, expressing your answer to an appropriate number of significant

figures.

Solution:

cosN N B A N BA

Flux linkage:

Where is the angle between B and the normal to the loop.

Therefore

2 3 0 3cos 50 2.8 10 1.9 10 cos 35 2.2 10N NBA Wb

23


27/31

www.gcephysics.com

(b) The coil is rotated at constant speed, causing an emf to be induced.

(b) (i) Sketch a graph on the outline axes to show how the induced emf varies

with angle during one complete rotation of the coil, starting when 0 .

Values are not required on the emf axis of the graph.

Solution:

The induced emf of the generator is given by

sinE NBA t

sin

Thus, it is a reasonable sine curve drawn on axes, showing just one cycle,

starting at emf = 0.

(b) (ii) Give the value of the flux linkage for the coil at the positions where

the emf has its greatest values.

Solution:


E NBA t sin 1t, when the emf at its greatest values, , thus

cos 0t

cos 0N NBA t

, and

Flux linkage:

Therefore, the flux linkage in these positions is zero.

(b) (iii) Explain why the magnitude of the emf is greatest at the values of

shown in your answer to part (b)(i).Solution:


sinE NBA t sin 1t, when 090t 0270t

, the emf has greatest values, thus in one cycle,

or .

18. Domestic users in the United Kingdom are supplied with mains electricity

at a root mean square voltage of 230V.

(a) State what is meant by root mean square voltage.

24


28/31

www.gcephysics.com

(b) (i) Calculate the peak value of the supply voltage.

(b) (ii) Calculate the average power dissipated in a lamp connected to the

mains supply when the rms current is 0.26 A.

(c) The frequency of the voltage supply is 50 Hz. On the axes below draw the

waveform of the supplied voltage labelling the axes with appropriate values.

Review:

Root mean square value of an alternating current: the value of direct current

that would give the same heating effect as the alternating current in the same

resistor.

The root mean square value of an alternating current:

0

1

2rmsI I

The root mean square value of an alternating potential difference:

0

1

2rms

V V

Where I0 is the peak current, V0 is the peak voltage.

Therefore:

The root mean square value of an alternating current or pd

=1

2the peak value

Then the mean powersupplied to the resistor:2

2 rmsrms rms rms

Vp I R I V

R

25


29/31

www.gcephysics.com

Solution:

(a) Root mean square value of an alternating current: the value of direct

current that would give the same heating effect as the alternating current in

the same resistor.

(b)(i) From 01

2rms

V V , 0 2 2 230 325rmsV V V

(b)(ii)2

2 0.26 230 60rmsrms rms rms

Vp I R I V

R W

(c) Time period is given by1 1

0.0250

T sf

26


30/31

www.gcephysics.com

Chapter 4 Current Electricity

4-1 Charge, current and potential difference

4-2 Resistance

4-3 Ohms law and measuring resistance

4-3-1 Graph of V against I and I against V for the ideal resistor

4-4 Resistivity

4.5 Circuit components symbols:

4-5-1 IV and VI graphs for different conductors

4-6 Circuits

4-7 Potential divider

4-8 Electromotive force and internal resistance

4-9 Alternating currents

4-9-1 AC Generators

4-9-2 The physics parameters of alternating current

4-9-3 The transformer


Chapter 5 Gravitation

5-1 Newtons law

27


31/31

www.gcephysics.com

5-2 Gravitational field strength

5-3 Gravitational potential

5-4 Orbits of planets and satellites

5-4-1 An orbit equation


cie electricity and magnetism sample pages

Documents