Download Chapter 1 Magnetic and Electric basics

Electricity and Magnetism
Chapter 1
Magnetic and Electric basics
Question 1
As they are all in parallel, then the voltage across each circuit element will be 120 volt.
V2
Power:
P = VI = I2R =
R
120 2
1202
1202
+
+
120
60
30
= 120 + 240 + 480
= 840 Watt
=
A quicker way to come to this answer is to replace the three circuit elements with one of the same effective
resistance (17)
1202
Power:
P=
17
= 840 Watt
Question 2
When they are connected in series then the total resistance becomes 120 + 60 + 30 = 210
Circuit elements in series all have the same current, because every electron passes through each element.
V=I×R
120 = I × 210
 I = 0.57 Amp.
Question 3
No. For each appliance to operate correctly they need to have a P.D of 120 Volt across each one. This is not
the case.
PDheater = I × R = 120 × 0.57 = 69 Volt. (make sure you use the real figures in your calculations, not rounded off
figures)
PDfood processor = I × R = 60 × 0.57 = 34 Volt.
PDmicrowave = I × R = 30 × 0.57 = 17 Volt.
69 + 34 + 17 = 120 Volt.
Question 4
B
The direction of the field due to the coil is given by the right hand grip rule. If the thumb of your right hand is
pointing in the direction of the arrow on the left hand end of the coil, then your fingers will tend to coil and point to
the left. This is when your fingers are where the maximum field is, and this is inside the coil.
So the field produced by the coil is to the left. The force on the wire is given by the right hand rule.
So the current is the direction of the thumb, the field the direction of the fingers. This means that your palm will
be pointing out of the page. Your thumb is up, fingers to the left.  The force is given by the direction your palm
is facing.
Question 5
A
The electron will experience a force in the same direction as the current experiences a force, because it is the
same force acting.
With the right hand thumb pointing to the right (in the direction of the current) the fingers point into the page (the
direction of the field), then the force is projecting upwards from the palm of your hand.
Question 6
D
The flow of electrons is down the conductor, so the conventional current is 'up'. The direction of the field is to the
North. If your right hand thumb points up, your fingers are in the direction of the field, (North), then the force
must be in a "Westerly" direction, coming straight out of your palm. Refer back to the notes on Magnetic Fields,
if you need further clarification.
Question 7
This question is unusual, it relies on you having an appreciation of the strength of the Earth's magnetic field.
From your notes it is 5 × 10-5T. It might be useful to put this fact on to your cheat sheet.
The lightning conductor is going to be reasonably long, because it needs to be high to attract the lightning,
because it wants to be the easiest path for the charge to reach ground. This also means that the old saying that
'Lightning never strikes the same place twice" is quite incorrect.
F = nBIL
= 1 × 5 × 10-5 × 104 × 10
=5N
The answer in the book uses B = 2 × 10-5T. Just make sure that you show your working out and thinking very
clearly.
Question 8
C
Because the conductor is now bent over at an angle of 300, the effective length of the conductor perpendicular to
the field is less.

 this is now the effective length.


So the new force will be less, but not zero.
Question 9
The heat energy produced each second by the heater is given by P = VI = I2R = V2/R. Using either supply the
resistance will remain constant, so the energy is proportional to V2.
For a 180 V (peak) AC supply VRMS = 180/ 2 = 127V.
 the 150 V DC supply will provide more energy, therefore more infra-red radiation.
Question 10
F = nBIL,
where F is the force (in Newton), n is the number of turns or number of wires, B is the strength of
the magnetic field (in Tesla), I is the current (in amp) and L is the length of the wire (in metres).
Substituting
F = 3N, n= 1,
3 = 1 × 0.15 × I × 2
 I = 10 amp.
B = 0.15T, and L = 2m,
Question 11
The current flowing in coil 2 is creating the force acting on coil 1.
In coil 2 the current is flowing from the right hand side of the cell (battery). This is produces a field directed to
the left.
The current in coil 1 is in the opposite direction, this produces a field to the right. These fields are in the opposite
directions, like two North poles, so they will repel each other.
So the force on coil 1 (from coil 2) is a repulsive force, acting to the left.
Question 12
AC
The expression one or more means exactly that, there may be 1, 2, 3 or 4 correct answers. If it does not say
one or more then there is only one 'best' answer.
If the wire is being pushed then F = nBIL, implies that n, B, I, L are all non-zero,  they exist.
Answer C is correct because it is actually the definition, the force acts on the moving charges, which in this case
is the electrons that are constrained to move inside the wire.
Question 13
D
Magnetic fields are vectors, so the two fields due to the individual bar magnets will cancel each other out. There
is still the magnetic field due to the Earth at this point, so the best answer is D. C is also correct but D is much
better.
Question 14
If the wire is parallel to the field, then the force on the current carrying wire = 0.
Question 15
Power:
P = VI = I2R =
V2
R
P = I2R
 16 000 = I2 × 11
 I2 = 16 000/ 11
 I = 38 Amp
This answer is already RMS, because we always assume values are RMS unless otherwise stated.
Question 16
Power:
P = VI = I2R =
V2
R
P = VI
 16 000 = V × 38.1385 (don't use rounded off values, use your calculator)
 V = 16 000 / 38.1385
 V = 420 V
If you didn't know the value for I, you could have just as easily used the formula
V2
Power:
P=
R
 16 000 × 11 = V2
 V = 420 V
(make sure that you always round off properly).
Question 17
The peak value of the current is the RMS × 2
 I = 38. 1385 × 1.4.14
 I = 54 Amp.
Question 18
The peak value of the voltage is the RMS × 2
 V = 420 × 2
 V = 593 Volt.
Question 19
Power is VI. You are given RMS values, so
 P = 240 × 150 × 10-3
 P = 36 Watt.
(always be careful with the units)
Question 20
Power is the rate at which energy is being used, i.e. in joules per second.
Power = VI, assuming both V and I are measured in RMS
Ipeak
Vpeak
P=

2
2
10
400
P=

2
2
4000
P=
2
 P = 2000 Watt
 2.0 × 103 Joules of energy are being used every second.
Question 21
Power = VI, assuming both V and I are measured in RMS
 40 = 80 × I
 I = 0.5 amp.
Question 22
If the EMF of the supply is 240 Volt, then the total potential differences around the circuit must also be 240 V.
This means that the PD across the resistor must be 240 - 80 = 160 V.
As the two elements are in series the current must be the same in both. Iresistor = 0.5 amp.
Using
V = IR
160 = 0.5 R
 R = 320
Question 23
P = VI
= 240 × 0.5
= 120 Watt
(because the current is the same in both elements and the total PD = 240 V)
Question 24
The power of the resistor is given by P = i2R
Where I = 0.5 A, and R = 320
 P = 0.52  320
= 80W
Question 25
The distance between A and B represents half a period.
 Period = 7.2ms
1
1
f=
=
= 139 Hz
T
7.2  10 3
Question 26
The RMS voltage =
Vpeak
2
=
1.2
2
= 0.85V
Question 27
Each vertical division = 0.4V.  The DC voltage = 1.2V.
Question 28
This signal shows an AC voltage with a DC voltage of 0.8V (1 division). The magnitude of the AC voltage is 2
divisions,  1.6Vpeak.
C
Question 29
The total resistance is
V2
Rtotal 
P
2402
960
 60 
Rtotal 
Rtotal
The two heating elements are joined in series so each of them must have a resistance of 30 ohm.
Question 30
When the resistors are connected in parallel their total resistance becomes
1
1
1
1
1
1
1
1
Rtotal  15





Rtotal
R1 R2
Rtotal 30 30
Rtotal 15
Use P 
V2
R
R
PSeries
V2

 Parallel
PParallel RSeries
V2
PSeries
PParallel

RParallel
RSeries
PSeries
15
1


PParallel
60 4
Question 31
This is a voltage divider question. To have 120v across the light bulb you need to halve the supply voltage and
you do this by placing the same resistance in series with it; hence the resistor must have the same resistance as
the bulb.
From
P  VI 
V2
1202
we have R 
 240
R
60
Question 32
The voltage drop across group P will be 16 x 10 = 160V
The voltage drop across both group Q and R (the groups are in parallel) will thus be 240 – 160 = 80V
If each bulb in group Q has a voltage drop across it of 10V there will thus be 80/10 = 8 bulbs in the group.
Group R will also have 8 bulbs
Question 33
The string of lights is the total combination of P + Q + R and this is supplied with 0.50A of current
Question 34
The current through P is 0.50A and as Q and R are identical they will each have 0.25A flowing through them.
Each bulb has a voltage drop of 10V across it and hence the power dissipated by each bulb in groups Q and R
will be P = VI = 10 x 0.25 = 2.5W
Question 35
Since Q and R are in parallel, if group Q is ‘removed’ from the circuit this will result in doubling the resistance of
this part of the circuit which will result in a reduction of the current flowing through part P. Part R however will
have an increase in the current that flows through it. Hence
Group
ON/OFF
Brightness
P
ON
Dimmer
Q
OFF
R
ON
Brighter
Question 36
Since the elements in the circuit are in series, the current through both will be the same.
Power in globe = I2R,  9 = I2  8
 I = 1.06.
The voltage drop across the globe will be
V = IR  V = 1.06  8 = 8.49V.
This means that the voltage drop across the variable resistor must now be 12 – 8.49 = 3.51.
 Vresistor = IR  3.52 = 1.06  Rresistor
 R = 3.31
 R = 3.3 (to 2 sig figs.)
Question 37
The fuse burns out when the current is 2A. This means that the effective resistance of the two elements in
parallel must be 6
1
1
1

 1 – 1 = 1
 2 = 1
 R = 24
=
+
48
R
6
8
R
6
8
R
Question 38
When justifying your answer it is preferable if you can prove your case and disprove the others.
In circuit A, when altering the variable resistor, the voltage drop (and current) of the globe will change, this will
cause the brightness to change. In Circuit B, altering the variable resistor will not change the voltage (and hence
current) of the globe.
A
Question 39
P = VI  48 = 240  I,  I = 0.2A
Question 40
The three sets of globes are in parallel, therefore the current of 0.2A will be divided between the three sets.
 0.067A
this should be 6.7  10-2A
Question 41
There are 12 globes in each set. The potential drop across each set is 240, therefore the voltage drop across
each globe must be 240/12 = 20V
Question 42
If the power rating of the 36 globes is 48W, then each globe has a power rating of 48/36 = 1.33W
Question 43
If the globe (circled) is removed, then the middle set will not operate at all.
The voltage across the two other sets remains at 240V and their resistance remains the same, so they will be at
the same brightness.
B
Question 44
The force on the wire is given by F = nBiL
F = 200  0.4  0.5  π  0.4 = 50.3N.
On the exam, the diameter was 0.04m. This meant that a much more realistic answer of 5.03N was correct.
Question 45
Perpendicular; magnetic; increases.
Question 46
B
(Use of right hand (palm) rule.)
Question 47
1.0 T
From B = F/IL = 0.01/(1x0.01)
Be careful with the measurements of the wire (10mm = 0.01m)
Question 48
With an AC power supply the magnetic force alternates up/down at 100 Hz.
The wire may vibrate up and down rapidly at 100 Hz or it may do nothing if it is reasonably heavy or stiff.
Question 49
D
Magnetic field
The field is into the page at point P. With the current “up” the right hand slap gives the force In direction “D”
Question 50
A to F
B to D
C to E
The secret to doing multiple choice questions is to not look at the answers, but to work your own out and then
find the one that agrees with you. This is by far the safest way to do multiple choice questions.
Diagram A is of a coil, we use coils to become 'bar' magnets, so the field around a coil must look like the field
around a bar magnet. F is the best option.
The next easiest one to deal with is 'C'. (Don't feel constrained to do the questions in the order that they are
written, sometimes it is easier to do them in another order.)
So 'C' is a straight current carrying conductor, so the filed is given by the direction that the fingers curl if the
thumb is in the direction of the current. This means that they are concentric circles, the direction of the field is
controlled by the direction of the current (which is not shown)
That means that 'B' must behave like two versions of 'C'. The field lines will be in the opposite direction if the
direction of the current was given, then you could show this. The non-concentric circles are a result of the
interaction between the two fields.
Question 51
You needed to draw the arrows to indicate the direction both inside and outside the coil. The field lines were not
allowed to join each other, or to cross each other.
Question 52
None.
The section ‘cd’ of the wire is parallel to the field.
Question 53
F= BiL
= 2.0  10-2  5.0  2.0  10-2
= 2.0  10-3 N
The direction is given by the right hand slap rule. The field inside the coil is from the left to the right. The current
is from ‘b’ to ‘c’. so the force is upwards.
Q
(The examiner also accepted the direction ‘T’, because some found the diagram confusing).