This document discusses key concepts related to electric current. It defines current as the flow of electric charge, measured in coulombs per second. It explains Ohm's Law, which states that voltage is directly proportional to current. Resistance controls the flow of current and is defined as voltage divided by current. Circuits can have components connected in series or parallel, and the document explains how voltage and current are distributed in these configurations. Key electronic components like diodes and LEDs are also summarized.
1. CURRENT ELECTRICITY
1. Understand the nature of electric current in terms of a flow of moving
charge - applied to both conventional current & electron flow
2. Define electric current as Coulombs per second, I = q/t
3. Understand how an electric current carries energy in terms of potential
difference and emf
4. Understand the nature of resistance in terms of the control of electric
current
5. Understand Ohm’s Law, V = IR and its limitations
6. Understand how voltage and current divide up in series and parallel
7. Calculate total resistance for resistors in series using RT = R1 + R2 + ...
and in parallel using RT = 1/R1 + 1/R2 + ...
8. Calculate electric power as P = VI
9. Construct a simple circuit from a simple sketch of the components in the
circuit
10. Draw a circuit diagram from a sketch of the components in the circuit
11. Describe the operation of a diode, a thermistor, an LDR and an LED
Wednesday, 25 August 2010
2. ELECTRIC CURRENT
Power Supply
An electric current is a
flow of charge + -
Electric charge flows around a simple circuit
and under the influence of a power supply
(the energy source).
A simple circuit is a conducting path around
which charge can flow.
A conducting path
Two models of electric current:
• Electron flow - electrons in wires flow from the positive to the negative terminal
of a power supply.
• Conventional current - The direction in which positive charge would move if it
could - this is the model that we use in physics
http://regentsprep.org/Regents/
physics/phys03/bsimplcir/default.htm
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3. ELECTRIC FIELDS
When there is a power supply and a wire conducting path an electric field is set up
in the wire.
We consider that positive charge would flow from the positive to the negative end of
the electric field:
Power Supply
+ - +++
Potential difference
+ + + A
(i.e. voltage)
is how much more potential
energy a unit of charge has
B at A than it has at B
- - -
A conducting path
Note
(i) The electric field in a wire is uniform.
(ii) As charge flows in the field it loses potential energy.
(iii) The minimum potential energy that charge has is at the negative terminal of the power
supply.
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4. CIRCUITS
Current - is the rate of flow of electrical charge
- it is the number of coulombs of electrical charge that passes a point in
one second.
where I = electric current (Cs-1 or A)
I=Q
Q = electric charge (C)
t
t = time (s)
Note
Cs-1 reads Coulomb per Second V
+ -
Consider the simple circuit
I
I
When voltage, V is increased the energy difference between a coulomb of charge on
either side of the power supply will increase. This energy difference drives electrons
around the circuit faster.
In other words, as supply voltage increases then current will also increase (provided
that the resistance remains constant)
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5. Potential difference (i.e. Voltage) is the energy difference that a coulomb of
charge has on either side of a component (Formal definition)
where V = potential difference or voltage (Volts, V)
V = ∆Ep
∆Ep = change in potential energy that a charge
q
experiences when it moves from one side to the
other side of a component (Joule, J)
q = the unit of charge (Coulomb, C)
Unit of Voltage: Joule per Coulomb or Volt
(JC-1) or (V)
Example
V
Consider the voltage across a lamp: -
+
A B
1A
If V = 6V then a coulomb of charge has 6J more electrical potential energy at
point A than it does at point B
Assignment questions 1 & 2
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6. RESISTANCE
Definitions
1. Resistance, R is a measure of the “electrical friction” in a conductor. (the
opposition to the flow of current)
2. It is the ratio of the voltage across a conductor to the current through it.
Resistance = Voltage
Current R=V
Unit of resistance is the ohm, Ω
I
Resistance is given by the slope or gradient of a voltage - current graph
Example
In an experiment, the voltage across a lamp is measured and recorded as the current
is increased 1 A at a time. Calculate the resistance of the lamp.
V (V)
24
20
16
12
8
4
0 1 2 3 4 5 6
I (A)
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7. OHM’S LAW
Ohm’s Law states that the voltage across a resistor is proportional to the current
through it. i.e. V α I
A resistor that obeys Ohm’s Law has a voltage - current graph that is a straight line
so the resistance is always a constant value:
V Vα I
The gradient of the graph is R
So V = RI
or
V = IR
I
Ohm’s law is usually written this way
Note
Ohm’s law allows us to calculate the correct voltage when the current in the circuit
changes. This requires knowledge of the resistance and requires the value of the
resistance to stay the same regardless of the current.
A conductor which obeys Ohm’s Law is called an Ohmic conductor.
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8. LIMITATIONS OF OHM’S LAW
24
V (V) 20
When a temperature of a lamp increases its
16 resistance increases
12
8
4
I (A)
0 1 2 3 4 5 6
For most conductors, as the temperature increases the increased vibration of particles
impedes the flow of electrons. Resistance in the conductor will therefore increase. The
graph slopes upwards.
V (V) 24
The resistance of a thermistor decreases as its
20
16
temperature decreases
12
8
4
I (A)
0 1 2 3 4 5 6
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9. RESISTANCE CALCULATIONS
Resistors which are connected end to end are in series with one
another
R1 R2
The total resistance of the series combination, Rs is the sum of the
resistances R1 and R2.
For two or more resistors in series: Rs = R1 + R2 + ...........
Resistors which are connected side by side are in parallel with each other.
R1
R2
The total resistance of the parallel combination, Rp is less than any individual
resistor in the combination.
For two or more resistors in parallel 1 1 1 + ....
the total resistance,Rp is given by: RP = R1 + R2
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11. VOLTAGE AND CURRENT IN SERIES CIRCUITS
+ -
VT
A1 A3
I1 I3
I2
A1
V1 V2
Current in series is constant I1 = I2 = I3
Voltage in series is shared VT = V1 + V2
Note
Voltage is shared in proportion to the size of the resistance
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12. PARALLEL CIRCUITS
+ -
Current in parallel is shared
IT VT IT
IT = I1 + I2
I1 R1
in other words “charge splits
up as it enters a junction in
a circuit”
V1
I2 R2
Voltage in parallel is constant
V2 VT = V1 = V2
Note
Current is shared in an inverse proportion to the size of the resistance.
For example:
If R1 = 5 and R2 = 10
and IT = 3
“Double the resistance then halve the current”
then I1 = 2 and R2 = 1
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13. http://phet.colorado.edu/simulations/ CIRCUIT CONSTRUCTION
sims.php?
sim=Circuit_Construction_Kit_DC_Only
1. Enter the URL (above) into the address bar of your internet browser.
2. Use the simulation tools to construct each of the following 3 circuits (ensure that you use
identical lamps and an the same power supply for each circuit).
3. Record the current in each circuit and explain your observation.
4. Repeat this exercise for the second set of 3 circuits.
1 2 3
+ - + - + -
A A A
3
+ -
2
A
1 + -
+ - A
A
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14. Examples CIRCUIT CALCULATIONS
1 + 9V -
A1 V1 A3
5Ω 10Ω
A2
V2 V3
For the circuit represented by the circuit diagram above, what is the reading on:
(a) V1
(b) A2
(c) V2
(d) V3
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15. 2 + 15V -
V1
5Ω A3
V2 A1
V3
R 10Ω
A2
For the circuit represented by the circuit diagram above, what is the reading on:
(a) V3 if V2 = 10 V
(b) A1
(c) A2
(d) A3
(e) What is the value of resistor R?
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16. 3 + 12V -
V1
2Ω 4.8Ω V3
A1
V2
3Ω
A2
For the circuit represented by the circuit diagram above, what is the reading on:
(a) V1
(b) V3
(c) V2
(d) A1
(e) A2
Assignment
questions 4, 5 & 6
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17. Examples APPLIANCES IN THE HOME p167 ABA Q5 & 6
1. The diagrams opposite show two different heating
circuits for a hot plate. Both circuits use two similar
heating elements, A and B, of equal resistance.
(b) Draw a circuit diagram for each circuit.
240 V
240 V
(b) Consider circuit 1. The current flowing through
element A is measured to be 1.2 A. What is the
current through element B?
_____________________________________
(c) How much current is drawn by circuit 1 from the mains?
_________________________________________________________________
(d) Explain why the voltage across element A is 120 V.
_________________________________________________________________
_________________________________________________________________
(e) Calculate the resistance of each heating element.
_________________________________________________________________
_________________________________________________________________
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18. Consider circuit 2
(f) Explain why the current through element A is 2.4 A.
_________________________________________________________________
_________________________________________________________________
(g) How much current is drawn from the mains?
_________________________________________________________________
(h) Calculate how much electric power is turned to heat by the circuit.
_________________________________________________________________
_________________________________________________________________
(i) How many times more heat is generated in circuit 2 than in circuit 1.
_________________________________________________________________
2. A stereo uses 240 V and the combined resistance of all its internal components is
60Ω.
(c) Calculate the power rating of the stereo.
__________________________________________________________________
(d) Calculate the amount of energy used to operate the stereo for half an hour.
__________________________________________________________________
__________________________________________________________________
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19. 3. A set of 10 Christmas tree lights operate from a 20 V supply. They are all similar
1.0 W bulbs, connected in parallel.
(a) Calculate the voltage across each bulb.
(b) Calculate the current through each bulb, in mA.
(c) Calculate the resistance of each bulb.
(d) Calculate the total resistance in the circuit.
4. A 1000 W iron is connected to a 120 V supply. Should the iron need to be used on
a 240V supply calculate the size of the resistance that will need to be added in
series to the iron so that the iron continues to draw the same current.
Assignment question 7
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20. SPECIALIZED COMPONENTS
Thermistor
A thermistor is a resistor which is sensitive to heat. Unlike most resistors though its
resistance decreases as its temperature increases. This makes it useful in the circuit in
your car that contains the temperature gauge. The temperature gauge is an ammeter
calibrated to read temperature instead of Amps and the thermistor is in contact with
the engine and connected in series with the gauge. As the engine temperature
increases the resistance of the thermistor decreases thereby allowing the current in
the circuit to increase. This increase in current is reflected in the reading on the
gauge.
Light dependent resistor
An LDR is a resistor which is very sensitive to light. Its resistance decreases with light
intensity. They are useful in light meters where the meter is essentially an ammeter
re-calibrated to read lux instead of Amps.
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21. Diode
Because a diode allows current to flow in one direction only, it is called a
semiconductor. Diodes require only a low voltage (about 0.6 V) and will only allow a
small current to flow through them. They are useful in circuits that convert AC current
to DC.
Light emitting diode
These give off light as they allow current to flow one way through a circuit. They
require about 2 V to function. Because of their low power input they are useful for
lights (eg. they are finding their way into the tail light clusters of motor vehicles)
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22. 12 PHYSICS ELECTRICITY ASSIGNMENT Name ______________________
1. 1. Draw using arrows, the direction of charge flow in each circuit
2. For each circuit, highlight the lamps that will
+ - + -
(a) (b)
+ - + -
(c) (d)
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23. + - + -
(e) (f)
2. Write the correct term (from the word list) into the column (left)
Term Definition Word list
the unit of charge impede
the unit of work or energy parallel
the difference in the potential energy per coulomb of charge between
electric field
two points in an electrical circuit
the unit for potential difference constant
a region within which a charge experiences a force Ohm’s law
electrical friction or a measure of the ability of a conductor to conduct potential
electricity difference
relationship between resistance, voltage & current where the resistance
series
remains constant.
to slow down or resist volt
connected “one after the other” in a circuit coulomb
connected “side by side” in a circuit joule
used to describe quantities that remain the same resistance
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24. 3. Calculate the effective resistance, R of the following combination of resistors:
(a)
R
<=>
R=
(b)
R R=
<=>
(c)
R
<=> R=
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25. (d) A 12V battery supplies 1.5 A to a circuit as shown in the diagram.
12 V
A voltmeter will read 10 V when it is
2A connected between two points in the
1Ω 2Ω 3Ω above circuit. Which two points? (Show
all working below)
A B C D
(e)
(i) Describe the change in the total
resistance in the circuit as R increases
from 1Ω to 2Ω to 3Ω.
2Ω R V (ii) Explain how the total current, I changes
as R changes.
(iii) How is the voltage, V affected by these
changes?
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26. 4. For the following circuit answer the questions which follow:
(a) Calculate the total current, IT which flows through the circuit.
(b) Calculate the current through the 1Ω resistor.
(c) Calculate the size of the current which would flow through the circuit if another 1Ω
resistor was added in parallel to the 1Ω and 2Ω already in the circuit.
Wednesday, 25 August 2010
27. 5. A 30V battery supplies a current of 3A to the circuit shown:
30 V (a) Calculate the voltage across AB.
+ -
3A R 8Ω
R
A B (b) Calculate the current through R
3Ω
(c) Determine the value of R
(d) Explain why adding a 100000 Ω resistor in parallel to the lamp will not cause the
brightness of the lamp to change
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28. 6. In the circuit shown the ammeter has negligible resistance and reads 2.5 A.
(a) Calculate the current, I
(b) Calculate the value of R.
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29. 7. Questions (a) to (e)
refer to the circuit
(right):
(a) Calculate the total
resistance in the circuit
(b) Show that the unknown resistance R is
3.0 x 104 Ω
(c) Calculate the power dissipated by the 1.0 x 104 Ω resistor
(d) Calculate the power delivered by the battery
(e) The battery is left on for 24 hours
(f) Calculate the energy delivered in this period of time
(g) Where did this energy go?
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32. ELECTROMAGNETISM
1. Understand that an electric current creates a magnetic field around itself
2. Describe the magnetic field created by a straight current-carrying wire
3. Describe the magnetic field created by a solenoid
4. Use the right-hand slap rule to predict the direction of the magnetic
force on a current-carrying wire in a magnetic field
5. Use F = Bqv to calculate the size and direction of the magnetic force on
a moving charge inside a magnetic field
6. Describe the circular motion of a charged particle inside a magnetic field
7. Understand electromagnetic induction in terms of the relative motion
between a coil and a magnetic field.
8. Use V = Bvl to calculate the voltage induced across a wire moving
through a magnetic field
9. Describe the operation of the electric generator
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33. ELECTRIC CURRENTS AND MAGNETIC FIELDS
Charm compass
N
Magnetic field lines The compass needle is
itself a tiny magnet (the
S North pole of this magnet
points towards the South
end of the magnet)
Area of strong magnetic field strength
• Magnetic field strength depends on the density of the magnetic field lines.
• Uniform field: field lines are parallel
A magnetic field is produced when electrons in an object
display a common pattern of motion. The smallest
magnetic field is the magnetic field of a single electron.
• When there is random electron motion, the magnetic fields of the electrons cancel
each other. This results in no net field being produced.
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34. A circular magnetic field is formed around a straight current - carrying conductor:
For each of the views, draw the magnetic field lines around the wire
3D View
View from above
•
The direction of the magnetic
field lines is given by the
Magnetic Field Strength Symbol: B
~ Right-hand Thumb Rule
Magnetic field strength is a vector quantity since it has both magnitude and direction
Unit of B: Newton per Amp per metre (N A-1 m-1)
or Tesla (T)
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35. THE SOLENOID
The magnetic field of a solenoid is similar in shape to that of a bar magnet:
Draw the magnetic
field lines around the
solenoid
If the current is known, the poles of the solenoid
can be determined using the right hand thumb
rule for solenoids:
Thumb points to
North
pole of the solenoid
from inside the coil
Curled fingers indicate
the direction of the
current
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36. Field lines are parallel in the core of the solenoid which --> the magnetic field in the
core is uniform.
The density of magnetic field lines is greatest in the core --> the magnetic field
strength is greatest in the core.
Magnetic Field Strength, B is:
- proportional to current, I B α I
- proportional to the number of turns of wire per unit length of the coil, n
i.e. B α n
http://www.phy.ntnu.edu.tw/ntnujava/main.php?t=276 Ex 17A: Q.1 to 5
Wednesday, 25 August 2010
37. MOTOR EFFECT
• A current carrying conductor in a magnetic field will experience a force.
• This is seen as movement when the conductor hangs in the magnetic field of a bar
magnet
The force is maximum when the angle between the conductor and the magnetic field
lines is 90o
The size of the force can be increased by:
(i) increasing the current
(ii) increasing the magnetic field strength
(iii) increasing the length of the wire in the field
The direction of the force can be predicted using the right hand slap rule.
The DC motor: http://www.walter-fendt.de/ph14e/electricmotor.htm
Wednesday, 25 August 2010
38. THE RIGHT HAND SLAP RULE FOR CONVENTIONAL CURRENT
Direction of
Direction that a Magnetic field
positive charge
would flow (direction
of conventional
current)
The direction of the force is given
by the direction of the slap -
into the board in this case
A LEFT HAND SLAP RULE CAN BE USED FOR ELECTRON FLOW
Wednesday, 25 August 2010
39. Side view of the hanging wire:
N Magnetic field lines on LHS of wire
oppose each other and repel ----> weak
field
F
Magnetic field lines on the RHS of the
wire bunch up ----> strong field
Draw the magnetic field lines
S
which explain the direction of the
The wire experiences a force which is to the left force
Examples
1. The following images all show conductors in a magnetic field. In which of the
following situations is the force on the conductor directed out of the page.
magnetic field electric current
out of page out of page
I I
A B B C D E
~ • • • •
B • • • •
I ~ • •
• • •
• • • •
• • • •
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40. 2. A copper wire is carrying an electric current. It is placed between the poles of a
magnet. The magnetic force on the wire pushes it into the page, as shown.
(a) Label the poles of the magnet as North and South.
(b) Where is the stronger field, in front or behind the wire? __________________
(c) What would happen if the wire was made of aluminium?
_______________________________________________________________
(d) What would happen if the wire was rotated (to the right) so it is horizontal.
_______________________________________________________________
Wednesday, 25 August 2010
41. 3. The current balance is a device that is like a see-saw. One end of the see-saw is
inside a solenoid. The other end has a mass (m) on it that can be adjusted to
achieve a balance. It is able to measure the magnetic force on a conductor in a
magnetic field. The uniform field inside the solenoid of the current balance
(shown) has a strength of 10-2 NA-1m-1. CD (a straight section of the conductor at
right angles to the magnetic field) is 2 cm long and carries a current of 1 A.
Side view
m
x
Board
Pivots Pivot Current into
the page
(a) Show the direction of the force on CD using an arrow drawn on the side view .
(b) What is the magnitude of the force on CD?
(c) What mass, m is needed to be placed on the other end to maintain balance?
Wednesday, 25 August 2010
42. 4. For each of the devices illustrated (the mechanism that drives an ammeter needle
on the left and a loudspeaker on the right), explain how they work. Assume the
ammeter needle rotates clockwise and the loudspeaker is pushing outwards to the
right. (Draw on the diagram to assist you in your explanation)
ammeter loud speaker
This coil is attached to the cone
cone
Cylindrical S
magnet
N S
N
S
View of cylindrical magnet from
the front
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43. F α B
F = B Il applies to a conductor which is perpendicular
F α I to the magnetic field
F α l where: F = force on the conductor (N)
B = magnetic field strength (T)
I = Electric current (A)
l = length of the conductor in the
magnetic field (m)
I
N
B
~
S
l
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45. Maximum torque exists F
when the force is at
right angles to the coil
F
F
As the coil rotates, the
angle that the force
makes with the coil
changes and the torque
decreases.
F
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46. F
The torque is zero
now because there
is no component of
the force that is
perpendicular to the
coil.
No rotation
F
Decrease in torque
as the coil continues
F
to rotate
F
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47. Maximum torque again
F
F
Note that the size of the force does not change as the coil rotates but the angle that
the force makes with the coil does.
The reason that the size of the force doesn’t change is because B, I and l do not
change (Remember that F = BIl)
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48. THE MOTOR EFFECT AND THE SPEED OF MOVING CHARGES
Consider a charge, q. It is one of many charges in a current-carrying wire moving
with a velocity, v at right-angles to a magnetic field of strength, B:
Since a moving charge
constitutes an electric current
in a magnetic field, the charge must
experience a force which is at right angles
to the velocity and the magnetic field.
Apply the Left hand slap rule
to the situation shown and confirm
that the direction of the force is as it
is represented in the diagram
Deriving an expression for force: F = B Il (Equation 1)
but I = q (Equation 2)
t
Substituting Eqn. 1 into Eqn. 2 gives: F = Bql => F = Bqv since v = l
t t
If the charge is an electron then: F = Bev
where e is the charge on an electron and v is its speed.
Wednesday, 25 August 2010
49. Example THE TV TUBE
The diagram to the right shows the path of
electrons along a TV tube. They travel at 20% of
the speed of light. The TV tube is designed to x x x x xxxx
provide a magnetic field in the direction shown, e
x x x x xBx x
x
through which the electrons must travel. The x x x x x~x x
field is 0.06 T into the page. x x x x xxxx
(a) How fast are the electrons traveling in ms-1?
____________________________________
e = 1.602 x 10-19C
(b) Calculate the force exerted on the electrons
c = 3 x 108 ms-1
as a result of the magnetic field.
____________________________________
____________________________________
____________________________________
(c) What is the direction of this force? _______________________
(d) Calculate the weight force on each electron. ______________________________
(e) Is this weight force significant? ________________________________________
(f) Explain the shape of the electron’s path.
__________________________________________________________________
__________________________________________________________________
(a) 6 x 107 ms-1 (b) 6 x 10-13 N (c) downwards initially and always at right angles to the path of the electron (d) 9 x 10-30 N (e) This
weight is not significant. Much smaller that the magnetic force (f) The shape of the path is circular because there is a force at right angles
to the electron’s velocity.
Wednesday, 25 August 2010
50. Example CHARGE DEFLECTION IN A MAGNETIC FIELD
One of the problems associated with the commercial harnessing of energy from a
nuclear fusion reaction is that the enormously high temperatures make it very
difficult to contain the hydrogen needed for the reaction. Magnetic fields provide a
solution to the problem. At the temperatures involved the hydrogen atoms are
ionized to positively charged protons.
http://
surendranath.tripod.com/
Applets/Electricity/
MovChgMag/
MovChgMagApplet.html
9b.mass of electron
A stream of protons is directed towards an area of magnetic field directed into
the page as shown in the diagram. A proton is shown entering the field.
(a) In which initial direction will it experience a force? ______________________
(b) On the diagram sketch the path of the proton while it is in the field.
(c) Briefly explain how this phenomenon can help in solving the nuclear fusion
problem.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
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51. INDUCTION
A current will be induced in a hanging wire when the wire is moved in a magnetic
field
This is the direction of a
magnetic force on a
positive charge
A very sensitive meter could measure this current.
If the hanging wire remained stationery and the magnet moved there would also be a
current induced in the wire.
An electric current will be induced in a conductor as long as there is
relative movement between the _________ and the _________ field.
There are 3 obvious ways of increasing the induced current:
(i) ________________________________________________
(ii) ________________________________________________
(iii) ________________________________________________
Wednesday, 25 August 2010
52. INDUCTION and the right hand slap rule
A current will be induced in a hanging wire when the wire is moved in a magnetic
field. A very sensitive meter could measure this current.
If the hanging wire remained stationery and the magnet moved there would also be
a current induced in the wire.
An electric current will be induced in a conductor as long as there is
relative movement between the conductor and the magnetic field.
The thumb points in Direction of the slap
the direction in which Palm of shows the direction of the
right hand magnetic force on a
the wire is moved
positive charge
(conventional current)
N
B
~
S
The fingers point in the
l direction of Magnetic field
Wednesday, 25 August 2010
53. Demo: Induction in the solenoid
Observations
North pole moved into the solenoid
_______________________________________
The magnet is held stationery whilst the coil is moved towards it
_______________
North pole stops inside the solenoid
______________________________________
North pole pulled out from inside the solenoid
______________________________
The rate of movement of the magnet is doubled
____________________________
Wednesday, 25 August 2010
54. INDUCED VOLTAGE
Extra for experts
DERIVATION OF V = Bvl
Consider a wire moving with constant velocity, v in an external magnetic
field. The wire is not part of a circuit.
Electrons jump briefly to the right and accumulate at the right hand end of the wire
and stop. These electrons jump because they are under the influence of a magnetic
force, Fmag.
“Thinking it through”
x x x x v x x x x
x x x x x x x x
The left hand slap rule for e
Fmag
induction predicts the x x x x x x x x
direction that the electron
x x x x x x x x
will jump. In this example
they jump to the right.
Wednesday, 25 August 2010
55. WHY DOES THE ELECTRON STOP?
x x x x v x x x x
Electrons quickly accumulate at
the right end of the wire. This x x x x x x x x
accumulation of electrons leads + -
+ e -
to the formation of an electric -
+
force (as a result of the electric x x x Felec
x x Fmag
x x x
field that forms) x x x x x x x x
+ x x x x v x x x x
The electric force Felec quickly -
+ x x x x x x x x -
increases in size as the charge
+ -
accumulation increases until it is
+ e -
equal in size to the magnetic -
+
force, Fmag. x x x Felec
x x Fmag
x x x -
+ -
+ x x x x x x x x -
But because this fixed length of wire is not part of a circuit, no current will flow. The
electrons will have stopped. When it is connected to a circuit, Fmag drives the current
separation which causes it to act as a power supply. We say that there is a potential
difference across the wire.
Wednesday, 25 August 2010
56. WHY DOES THE ELECTRON JUMP IN THE FIRST PLACE??
Remember that the electron (with balanced forces acting on it) is not moving
along the wire
BUT
it is moving across the magnetic field with speed v and so is experiencing a force
given by F = Bev
x x x x v x x x x
x x x x x x x x
e
Fmag
x x x x x x x x
x x x x x x x x
The direction of this magnetic force can also be predicted using the left hand rule for
the motor effect.
Wednesday, 25 August 2010
57. Felec = Fmag
Ee = Bev (where B = the strength of the external magnetic
field and
e = the charge on an electron)
Ve = Bev
l
Where V = the induced voltage (V)
=> V = Bv l
B = the strength of the magnetic field (T)
v = the speed with which the wire is
moved
l = the length of the wire
In summary, the voltage induced in a wire in a magnetic field depends on the
following factors:
(i) The strength of the magnetic field
(ii) The speed with which the wire is moved and
(iii) The length of the wire in the magnetic field
http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter32/chapter32.html
Wednesday, 25 August 2010
58. A MOVING CONDUCTIVE LOOP IN A MAGNETIC FIELD DOES NOT HAVE A NET
VOLTAGE INDUCED ACROSS IT
v
x x x x x x x x
x x x x x x x x
+ e -
+ -
x
+ x x x x x
Fmag x x
-
x x x x x x x x
LAMP WILL
x x x x x x x x
NOT GLOW
x x x x x x x x
x x x x x x x x
x x x x x x x x
+
x x x x xe x x -
x
+ -
+
x x x x x Fmag
x x -
x
x x x x x x x x
Current can not flow in the circuit
Wednesday, 25 August 2010
59. Example
An aircraft is flying directly over the North Magnetic Pole with a velocity of
300 kmh-1. The aircraft is pictured below as it would be viewed from above.
(a) Which of the following diagrams best represents the direction of the magnetic
field directly over the North Magnetic Pole as “viewed” from someone standing at
this North magnetic pole
(b) The free electrons in the wings of the aircraft experience a force.
Is this force from P to Q or Q to P? _____________
Wednesday, 25 August 2010
60. (c) Would this change if the plane was going the opposite way? ______
(d) If the magnetic field is 2.0 NA-1m-1, what is the force on the single electron?
_________________________________________________________________
_________________________________________________________________
(e) If the wing span is 8 m, what is the induced voltage across the wings when the
electrons stop moving relative to the wing?
_________________________________________________________________
_________________________________________________________________
(f) What are the two equal and opposite forces acting on the electrons when they
stop moving relative to the wing?
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
Answers: (a) up (b) Q --> P (c) Yes P --> Q (d) 2.7 x 10-17 N (e) 1333.3 V (f) Magnetic force = Electric force
Wednesday, 25 August 2010
61. THE ELECTRIC GENERATOR
A generator is based on a coil of wire which spins in a magnetic field. Because there is
always relative motion between the wire and the field there will always be an induced
current.
Apply the Right hand slap rule for induction to
1 draw the direction of current flow through the coil
in each of the following diagrams.
Explain why the induced current through the coil
during this phase of the generators rotation is zero
2
The alternator that
charges the battery of a
3 car is an AC generator
see web page below for
the animation
4
http://www.walter-fendt.de/ph14e/generator_e.htm
Wednesday, 25 August 2010
62. Example
Diagram 1 (below) shows a loop of wire spinning between the two poles of a
magnetic. Diagram 2 shows an electron in the right side of the loop.
0.4
0.25
0.35
(a) In what direction does the right vertical side of the loop move? (into or out of the
page?) ___________________________________________
(b) Explain why the electron experiences a magnetic force.
__________________________________________________________________
__________________________________________________________________
(c) Use the right hand slap rule for induction to determine the direction relative to
the wire that the electron experiences this force.
__________________________________________________________________
(d) Explain why there is no induced current in the loop for a brief instant when the
loop has rotated 90o from the pictured position.
__________________________________________________________________
__________________________________________________________________
Wednesday, 25 August 2010
63. (e) Explain why there is no induced current in the horizontal parts of the loop at any
time during the rotation of the loop.
__________________________________________________________________
__________________________________________________________________
The vertical sections of the loop are in uniform circular motion. The loop has a
rotational speed of 50 rpm.
(f) Calculate the period of the circular motion of one vertical section of the loop.
Wednesday, 25 August 2010