Exam Revision Workshop: Physics ¾ Unit 3 AOS 1 Fields

Tutor: Angad Singh Website: contoureducation.com.au | Phone: +61 491 128 767 | Email: [email protected]

Exam Revision Workshop: Physics ¾

Unit 3 AOS 1 Fields

✓ Fields Concepts and Comparing Fields

✓ Gravitational Fields and Orbital Mechanics

✓ Electric Fields and Particle Accelerators

✓ Magnetic Fields and DC Motors

Lesson Outline:

Gravitational Fields:

Characteristics:

➢ Radially Inwards

➢ Static

➢ Points Towards C.O.M

➢ Influenced by Mass & Distance

➢ Inverse Square Law, 𝒚 ∝ 𝟏

𝒙𝟐

Types:

➢ Radial (Non-Uniform) – Orbital Mechanics

➢ Uniform – Surface of Earth

Energy:

➢ Area Under Force Distance Graph

➢ Potential Energy Increases Further Away

from Mass

Electric Fields:

Characteristics:

➢ Radially Inwards/Outwards

➢ Static (unless moving charge)

➢ Points Towards (+) Test Charge Motion

➢ Influenced by Charge & Distance

➢ Inverse Square Law, 𝑦 ∝ 1

𝑥2

Types:

➢ Radial (Non-Uniform) – Point Charge

➢ Uniform – Capacitors/Parallel Plates

Energy:

➢ Area Under Force Distance Graph

➢ Potential Energy Increases Further Away from

Charge

Magnetic Fields:

➢ Only exist as dipoles

➢ Static (in VCE at least)

➢ From North to South Pole

➢ Influenced by Distance & Material

➢ Inverse Something Law 𝒚 ∝ 𝟏

𝒙??

➢ Types:

o Bar Magnet – Acts like a Dipole

o Uniform – Opposite Bar Magnets

o Current-Induced – Solenoids/Loops

Applications of Magnetic Fields:

➢ DC/AC Motors

➢ Cathode-Ray Tubes

➢ Magnets

➢ Solenoids

➢ Wire-loops

➢ Moving Charged Particles in regions of Uniform

B-Fields.

➢ Currenty Carrying Wires

➢ Mass Spectrometers

DC Motors:

➢ Torque and Rotation

➢ Split Ring Commutators/Slip-Rings

➢ Electrical Energy → Mechanical Energy

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Notes By: Amitav Madan

𝒆 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪

𝒈𝑬𝒂𝒓𝒕𝒉′𝒔 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 = 𝟗. 𝟖 𝒎 𝒔−𝟐 = 𝟗. 𝟖 𝑵 𝒌𝒈−𝟏

𝑮 = 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 𝑵 𝒎𝟐 𝒌𝒈−𝟐

𝒌 = 𝟖. 𝟗𝟗 × 𝟏𝟎𝟗 𝑵 𝒎𝟐 𝑪−𝟐

𝟏 𝒆𝑽 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱

𝒈 =𝑮𝑴

𝑹𝟐

𝑭𝒈 =𝑮𝑴𝒎

𝑹𝟐 𝑼𝒈 = 𝒎𝒈∆𝒉

𝑲𝑬 = 𝟏

𝟐𝒎𝒗𝟐 𝑾 = ∫ �⃗� ∙ 𝑑𝑥⃗⃗⃗⃗⃗ = 𝑭 𝒙 𝒄𝒐𝒔(𝜽) 𝑾 = ∆𝑲𝑬

𝑈𝑔 = −𝐺𝑀𝑚

𝑅 𝒗 =

𝟐𝝅𝑹

𝑻 𝒂 =

𝒗𝟐

𝑹

𝒂 =𝟒𝝅𝟐𝑹

𝑻𝟐 𝑭𝒏𝒆𝒕 =

𝟒𝝅𝟐𝑹𝒎

𝑻𝟐 𝑭𝒏𝒆𝒕 =

𝒎𝒗𝟐

𝑹

𝑹𝟑

𝑻𝟐=

𝑮𝑴

𝟒𝝅𝟐 𝒗𝒐𝒓𝒃𝒊𝒕 = √

𝑮𝑴

𝑹 𝑭𝒆 =

𝒌𝑸𝒒

𝑹𝟐

𝑈𝑒 = −𝑘𝑄𝑞

𝑅 𝑈𝑒 = −

𝑘𝑄𝑞

𝑅 𝑬 =

𝑭

𝒒=

𝒌𝑸

𝑹𝟐

𝑬 =𝑽

𝒅 𝑾 = 𝒒𝑬𝒅 𝑾 = 𝒒𝑽

𝑭𝒆 = 𝒒𝑬 =𝒒𝑽

𝒅 𝑭𝑳 = 𝒒𝒗𝑩 𝒔𝒊𝒏(𝜶) 𝐵𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑 = 𝜇

𝑁

𝐿𝑖 = 𝜇𝑛𝑖

�⃗� = 𝑞�⃗⃗� + 𝑞�⃗� × �⃗⃗� 𝑹 =𝒎𝒗

𝒒𝑩=

𝒑

𝒒𝑩 𝟏𝒆𝑽 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗𝑱

𝒗𝒇 = √𝟐𝑬𝒌

𝒎= √

𝟐𝒒𝑽

𝒎 𝑭𝒄𝒖𝒓𝒓𝒆𝒏𝒕−𝒘𝒊𝒓𝒆 = 𝒏𝑰𝑳𝑩 𝒔𝒊𝒏(𝜶) 𝜏 = 𝑟 × �⃗�

Lesson Formulae:

Useful Constants:


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Study Design Key Knowledge:



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Section A: Brief Introduction to Fields

What are Fields?

➢ Fields are simply models we use in physics to describe forces and the interactions between forces and

particles.

➢ They can be described as a set of mathematical constructs that match our experimental observations –

and we can use fields to explain the world around us.

➢ VCE Physics covers three types of fields:

o Gravitational

o Electric

o Magnetic

Representing Fields – Field Lines

➢ Gravitational, Electric, and Magnetic Fields permeate through all space.

➢ At any particular point in space, the local field describes the magnitude and the direction of the

gravitational/electric/magnetic force that a particle would experience at that point.

➢ Rather than having a numerical value for a field at the infinitely many points in space, we represent and

visualise fields with field lines.

➢ At any point, the magnitude of the field (and hence of the corresponding force) is determined by the

_________________________ of the field lines aka “how close are they packed together”.

➢ The direction of the field (and hence of the corresponding force) is determined by the ________________

______________________________ of the field lines.

NOTE: Field lines, like the forces they describe, are

vectors – they have both magnitude and direction.

➢ This means the field lines from multiple sources

can simply be added together like vectors to get a

“resultant” or “net” field.

Question 1. Conceptual.

(a) Where the field lines are denser, the force

experienced will be __________________.

(b) Where the field lines are less dense, the force

experienced will be ___________________.



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Section B: Classifying and Comparing Fields

Uniform and Non-Uniform Fields

Uniform fields have field lines that are:

(i) _______________________, AND

(ii) _______________________

This means that ANYWHERE inside a uniform field, the force experienced has exactly the same

_____________________ AND ___________________.

➢ Any field that is NOT uniform is classified as a non-uniform field.

➢ In VCE, the most common type of non-uniform field is a radial field – one that emanates out in all

directions in 3D space from a single point-sized centre.

➢ Both gravitational fields from a point mass, and electric fields from a point charge are radial fields.

NOTE: While the field lines appear two-dimensional on a diagram, remember that they are actually 3D!

ALSO NOTE: In fact, we can use the 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 = 4𝜋𝑅2 to determine that the density of field

lines (i.e. the magnitude of a radial field) at a distance R away is inversely proportional to R2.

𝑭𝒊𝒆𝒍𝒅 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 ∝𝟏

𝑹𝟐



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Static vs Non-Static Fields

➢ A static field is fixed and does not change either magnitude or direction.

➢ Non-static fields are changing fields. However, we only cover static fields within VCE physics.

Field Line Conventions

➢ There are some rules and conventions as to how field lines exist and are represented:

➢ Electric field lines travel _______________

positive charge _________________

negative charge.

o The direction of the electric field

line represents the direction of

the electric force that would be

experienced by a positive test

charge.

➢ Gravitational field lines always travel _________________

masses.

o The direction of the gravitational field line

represents the direction of the gravitational force

that would be experienced by a point test mass.

o The gravitational force is always attractive – hence

the field lines are always inwards towards masses.

2

3

1 _______________________________________________________

_

_______________________________________________________

_______________________________________________________

4 _______________________________________________________



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➢ Magnetic field lines travel ______________________ the

North Pole and ______________ the South Pole.

o The direction of the magnetic field line represents

the direction of the magnetic force that would be

experienced by a positive North monopole.

NOTE: That a magnetic monopole is only theoretical – and does not actually exist! This means magnetic

fields always exist in dipoles.

Special Field Diagrams: Dipoles

When two monopoles (either gravitational or electric) are placed close to each other, their field lines affect

each other and create special patterns.

➢ Recall that field lines cannot intersect!

Unlike Poles Dipole:

NOTE: This type of dipole can only exist with electric and magnetic fields – not gravitational fields (since

gravitational fields only have one type of pole – attractive).

Like Poles Dipole:

NOTE: This type of dipole can only exist with electric fields (since magnetic fields cannot have two like

poles together, and gravitational field lines can only point towards the mass, never away.

Like Poles Dipole v2:

NOTE: This type of dipole can only exist with gravitational and electric fields – not magnetic.



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Section C: Sketching Fields

Question 2. The following diagrams show field lines around masses or groups of masses. For each

diagram, state which of (i) gravitational, (ii) electric, and (iii) magnetic fields it cannot possibly represent?

(Multiple answers are possible for each!)

Question 3. Sketch the magnetic field around each of the following magnet configurations, drawing at

least 5 field lines in each case.



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Question 4. Complete the following table with 𝑌 and 𝑁 for Yes and No, corresponding to when each of the

types of fields can or cannot have the given properties.

The shading is for the next question; ignore for this question.

Property Gravitational Electric Magnetic

Uniform field

Non-uniform field

Radial field

Attractive monopoles

Repulsive monopoles

Attractive dipoles (i.e. opposite poles at each end)

Question 5. [Optional HW] Sketch example field lines for the highlighted boxes in the table above.



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Section D: All About Gravitational Fields

Gravitational Field Strength (𝒈)

➢ The gravitational field is an attractive field that is

created around every object with mass, which attracts

other masses towards it (by exerting a gravitational

force).

➢ The gravitational field strength is denoted by 𝒈 and

has SI units of 𝑵 𝒌𝒈−𝟏 𝒐𝒓 𝒎𝒔−𝟐.

➢ The strength of the gravitational field created by a

mass depends on two factors:

o __________________________________

o __________________________________

➢ Thus, the gravitational field strength has the following proportionalities:

𝒈 ∝

𝒈 ∝𝟏

➢ The constant of proportionality is the Universal gravitational constant (aka Newton’s gravitational

constant) which was empirically derived from Newton’s work on gravity. It is denoted by 𝑮 and is:

𝑮 = 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 𝑵 𝒎𝟐 𝒌𝒈−𝟐

➢ The gravitational field, 𝒈, created by a mass 𝑴 at a distance 𝑹 away from it is given by:

𝒈 =𝑮𝑴

𝑹𝟐

➢ Since the gravitational field strength has the same units as acceleration, it is also sometimes referred to

as the gravitational acceleration.

➢ The field has an inverse square proportionality with distance because the emanating field lines spread

over a spherical region with surface area 4𝜋𝑅2.

It’s Just a Model!

➢ Remember that gravitational fields – at the end of the day – are simply a model. In fact, we know today

that gravity is more accurately described by Einstein’s General Relativity!



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Question 6. Remembering field line conventions, draw a sketch of the gravitational field lines between

the Earth-Moon system. Assume 𝑀𝑒𝑎𝑟𝑡ℎ ≈ 5𝑀𝑚𝑜𝑜𝑛 , and ignore all other gravitational effects.

Where Does g = 9.8 come from?

Exploration 1. We know that the Earth has a radius of 6370 𝑘𝑚, and a mass 𝑚𝐸 = 5.98 × 1024 𝑘𝑔. Find the

gravitational field experienced by a person:

(a) At sea level.

(b) Atop Mount Everest, which is 8848 𝑚 above sea level.

(c) In the International Space Station, which orbits at an altitude of 408 𝑘𝑚 above the surface of the

Earth.

IMPORTANT NOTE: Beware of the difference between ALTITUDE (i.e. height above the surface) and

RADIUS (i.e. distance from the centre of the Earth). Remember that the formulae for gravitational fields

(and forces) will always use the RADIUS.



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➢ So 𝑔 = 9.8 𝑚𝑠−2 is an __________________________ which only works ______________________________

_________________________.

➢ Once we get sufficiently far away, the gravitational field is no longer 9.8 𝑚𝑠−2, and we must use

Newton’s formula to find it.

➢ This can also be seen visually when we represent gravitational fields close to the surface of the Earth

vs. sufficiently far away.



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Force in a Gravitational Field

➢ A gravitational field exerts a force on every mass within it.

➢ The force experienced by a mass inside a gravitational field is the gravitational force, commonly

denoted by 𝑭𝒈.

➢ The magnitude of the gravitational force (aka force due to gravity) experienced by an object depends on

two factors:

o ___________________________________

o ___________________________________

➢ We can use Newton’s second law to derive the relationship for the gravitational force on an object of

mass 𝑚 in a gravitational field 𝑔:

𝑭𝒈 = 𝒎𝒈

But 𝑔 is no longer always 9.8 𝑚𝑠−2! We can substitute the complete relationship for 𝑔:

𝑭𝒈 =𝑮𝑴𝒎

𝑹𝟐

➢ Like any other force, the gravitational force is a vector quantity, and is measured in Newtons.

Question 7. Find the magnitude of the gravitational force exerted by Earth on an electron at its surface,

given 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 9.1 × 10−31 𝑘𝑔, 𝑚𝐸𝑎𝑟𝑡ℎ = 5.98 × 1024 𝑘𝑔, 𝑅𝐸𝑎𝑟𝑡ℎ = 6370 𝑘𝑚.

NOTE: Always remember to convert to SI Units!



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Question 8. At exactly what altitude above Earth would an astronaut experience a gravitational force 2/3

that of the force experienced at sea level, to the nearest km? Use 𝑚𝐸𝑎𝑟𝑡ℎ = 5.98 × 1024 𝑘𝑔, 𝑎𝑛𝑑 𝑅𝐸𝑎𝑟𝑡ℎ =6370 𝑘𝑚.

Energy in Gravitational Fields

➢ The system between a central mass (such as a planet) 𝑀 and another mass (such as a satellite or comet)

𝑚 is an isolated system, which means the mechanical energy of the system is ____________________.

∑𝑬𝒎𝒆𝒄𝒉 = ∑𝑬𝒌 + ∑𝑼𝒈 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

VERY IMPORTANT NOTE: Since the gravitational field 𝑔 is no longer constant once we are high above the

Earth’s surface, the approximation 𝑼𝒈 = 𝒎𝒈𝚫𝐡 can NO LONGER BE APPLIED.

➢ Instead, the equation for gravitational potential energy becomes:

𝑼𝒈 = −𝑮𝑴𝒎

𝑹

NOTE: The formula is not assessable within VCE. The reason it is negative is because it is defined as the

work done against gravity to bring a mass from infinitely far away to a given distance 𝑅 away from 𝑀.



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Energy Conservation for Satellites

➢ For an object in space, such as a satellite, its total mechanical energy is ____________________.

➢ This means any increase in kinetic energy must come from a corresponding ____________________ in the

object’s gravitational potential energy, and vice versa.

𝚫𝑬𝒌 = −𝚫𝑼𝒈

Question 9. Consider a satellite in an elliptical orbit around the Earth. At which point(s) along the orbit will

the:

(a) Gravitational potential energy be maximal?

(b) Gravitational potential energy be minimal?

(c) Orbital speed be maximal?

(d) Orbital speed be minimal?

(e) If the satellite stops orbiting and instead just falls directly towards Earth, its 𝑈𝑔 will be

_____________________________ and its 𝐸𝑘 will be _______________________________ by the same

corresponding amount.

NOTE: A satellite is isolated because it does not have any other external forces acting on it. A rocket on

the other hand is NO LONGER AN ISOLATED SYSTEM!

(f) When a rocket is launched from the surface into orbit, its 𝑈𝑔 has to _____________________ and its 𝐸𝑘

has to _________________________. That is, its total mechanical energy is _________________________.



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Work Recap

Recall that the work done on an object by a constant force moving through a displacement d is given by:

𝑾 = 𝑭𝒅 𝐜𝐨𝐬 (𝜽)

However, when the force is varying, we use the area under a force-distance graph to find the work done.

𝑾 = ∫ �⃗⃗⃗� 𝒅�⃗⃗⃗�

Also recall that a work done is simply a change in energy (using the work-energy theorem):

𝑾𝑭𝒏𝒆𝒕= 𝚫𝑬𝒌

Gravitational Force vs. Distance Graphs

Consider a graph of the gravitational force vs. distance on an object of mass 𝑚 = 100 𝑘𝑔 at various

distances away from Earth’s surface.



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➢ At the surface (𝑅 = 6.37 × 106 𝑚), the gravitational force is just 𝐹𝑔 = 𝑚𝑔 = 100 × 9.8 = 980 𝑁.

➢ The graph stops at the surface, as a satellite cannot realistically go below sea level.

➢ If the satellite goes from B to A

o It is [rising] / [falling]

o Its gravitational potential energy [increases] / [decreases]

o Its kinetic energy [increases] / [decreases]

o The shaded area under the 𝐹𝑔 𝑣𝑠 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 graph represents the [gained] / [lost] kinetic energy,

which is equivalent to the [gained] / [lost] gravitational potential energy.

Question 10. Consider the 100 𝑘𝑔 satellite from the graph

above. Alex calculates the area under the force-distance

graph between 𝐴 and 𝐵 to be 2.03 × 109 𝐽.

(a) If the satellite falls from 𝐵 to 𝐴, what is its change in

gravitational potential energy? Is this an increase or a

decrease?

(b) If the satellite was travelling at 8.44 𝑘𝑚/𝑠 at point 𝐵, and no other forces act on it, what must be its

speed at point 𝐴?



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Gravitational FIELD vs. Distance Graphs

➢ A field-distance graph appears more commonly in VCE. It looks very similar to a force-distance graph so

pay attention to the axes!

➢ At the surface (𝑅 = 6.37 × 106 𝑚), the gravitational field is 𝑔 = 9.8 𝑁 𝑘𝑔−1.

➢ The graph stops at the surface, as a satellite cannot realistically go below sea level. However, we can

continue the theoretical gravitational field graph below the surface linearly if we assume that Earth’s

density is constant.

➢ On a field-distance graph, the area under the graph represents the change in energy PER UNIT MASS.

o This means we have to multiply by the mass of the satellite to obtain the actual change in

energy.

𝑨𝒓𝒆𝒂 =𝚫𝑼𝒈

𝒎

𝚫𝑼𝒈 = 𝑨𝒓𝒆𝒂 × 𝒎𝒔𝒂𝒕



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Section E: Orbital Mechanics

Newton’s Cannon

➢ If a projectile is launched horizontally with a high

enough velocity, it can be placed into orbit.

Uniform Circular Motion

1. Uniform = constant speed.

2. Circular = describes the path of the motion; this

means the object changes direction continuously, i.e.

changes velocity continuously.

3. Motion = motion.

NOTE: While real orbits are always at least slightly elliptical, we can approximate them as perfectly circular.

Orbital Motion

➢ When a satellite is in orbit, it is travelling in uniform circular motion.

o Constant radius/altitude of orbit.

o Constant speed of orbit.

o The only force acting in orbital motion is __________________________________________.

➢ We know that uniform circular motion needs a centre-seeking centripetal force. In an orbit, the

centripetal force is provided by the gravitational force.

𝑭𝒄𝒑 = 𝑭𝒈 Thus, we can substitute:

𝒎𝒈 =𝑮𝑴𝒎

𝑹𝟐 =

𝒎𝒗𝟐

𝑹=

𝟒𝝅𝟐𝑹𝒎

𝑻𝟐

➢ Remember that 𝑔 is the gravitational field AT THAT PARTICULAR RADIUS, and NOT 𝟗. 𝟖.

NOTE: The mass of the satellite, 𝑚, is actually irrelevant to the equations of motion that describe orbits:

𝒈 =𝑮𝑴

𝑹𝟐 =

𝒗𝟐

𝑹=

𝟒𝝅𝟐𝑹

𝑻𝟐



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➢ So, the mass of a satellite does not affect its motion in any way!

Question 11. The International Space Station orbits the Earth at an average altitude of 408 𝑘𝑚. Find:

(a) The centripetal force that keeps the ISS in orbit. What force provides this centripetal force?

(b) The orbital speed of the ISS?

(c) How long it takes for the ISS to complete one full revolution around the Earth?

Kepler’s Law

We can rearrange and combine 𝑮𝑴

𝑹𝟐 =𝟒𝝅𝟐𝑹

𝑻𝟐 to get Kepler’s Law:

𝑹𝟑

𝑻𝟐=

𝑮𝑴

𝟒𝝅𝟐

Which tells us that the value of the ratio 𝑹𝟑/𝑻𝟐 is a CONSTANT for all objects orbiting around the same

central body of mass M.

➢ This means we can equate any ratio (𝑹𝟑

𝑻𝟐)𝒑𝒍𝒂𝒏𝒆𝒕 𝟏

= (𝑹𝟑

𝑻𝟐)𝒑𝒍𝒂𝒏𝒆𝒕 𝟐

if they orbit the same central body.



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Question 12. Given that the Earth is 147.6 million kilometres away from the Sun, and Uranus is 5.8 times

that distance away from the Sun, what is the length of an (orbital) year on Uranus, in terms of Earth-years

correct to the nearest integer?

Orbital Speed

We can rearrange and combine 𝑮𝑴

𝑹𝟐 =𝒗𝟐

𝑹 to obtain the orbital speed for a satellite:

𝒗𝒔𝒂𝒕 = √𝑮𝑴

𝑹

Energy Required to Place a Satellite into Orbit

➢ To place a satellite which is initially stationary on the surface of the Earth into stable orbit, we need:

o To increase its gravitational potential energy to the orbital radius, AND

o To increase its kinetic energy to provide the orbital speed.

NOTE: A very common question asks us to find the energy required to place a satellite into orbit. The

biggest mistake among students is to forget to include the increase in kinetic energy which is also required

to keep the satellite in orbit!

➢ BEWARE of altitude vs orbital radius!

➢ BEWARE of field-distance vs force-distance graphs!



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For the gravitational potential energy Ug, we need the area under the graph (times the mass of the

satellite). The general, best-practice steps for solving questions involving areas under graphs for VCE

physics are:

2

1 ________________________________________________

________________________________________________

___

3 ________________________________________________

4 ________________________________________________

___

5 ________________________________________________



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Section F: Weight and Weightlessness

Weight

➢ This is the “true weight” of an object, given by the weight force.

𝑭𝒈 = 𝒎𝒈

➢ Any object with non-zero mass in a non-zero gravitational field has a weight.

Weightlessness

➢ An object is “truly” weightless if its weight force is zero.

➢ For an object with mass, 𝐹𝑔 = 0 can only be true if 𝑔 = 0.

➢ Thus, true weightlessness only occurs in deep space, where the gravitational field is negligible, OR at

special points where the net gravitational field due to multiple bodies is zero.

The FEELING of Weightlessness

➢ This is fundamentally different to TRUE weightlessness!

➢ We experience our weight through a contact force – the normal reaction force (and not the actual

gravitational force that acts on us!)

➢ Our perception of how “heavy” we are – i.e. our perception of our weight comes from the magnitude of

the normal reaction force we experience.

➢ Thus, we can FEEL weightless if 𝑭𝑵 = 𝟎. This is NOT the same as bring truly weightless because W does

not have to be zero.

➢ This is how astronauts aboard the ISS feel weightless – they are in freefall and they only have one force

acting on them (𝐹𝑔), which means the normal reaction force on them is zero and hence they feel

weightless.

o But they are not actually weightless since they are within Earth’s gravitational field!

𝑾𝒆𝒊𝒈𝒉𝒕𝒍𝒆𝒔𝒔𝒏𝒆𝒔𝒔 𝒐𝒓 𝑻𝒓𝒖𝒆 𝑾𝒆𝒊𝒈𝒉𝒕𝒍𝒆𝒔𝒔𝒏𝒆𝒔𝒔: 𝑭𝒈 = 𝟎

𝑻𝒉𝒆 𝑭𝒆𝒆𝒍𝒊𝒏𝒈 𝒐𝒇 𝑾𝒆𝒊𝒈𝒉𝒕𝒍𝒆𝒔𝒔𝒏𝒆𝒔𝒔: 𝑭𝑵 = 𝟎



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Section G: All About Electric Fields

Electric Field Strength (𝑬)

➢ The electric field is an attractive/repulsive field that is

created around every object with electrical charge (𝑸), which

attracts/repels other electric charge (by exerting an electric

force).

➢ The electric field strength is denoted by 𝑬 and has SI units of

𝑵 𝑪−𝟏 𝒐𝒓 𝑽 𝒎−𝟏.

➢ The strength of the electric field created by a charge depends

on two factors:

o __________________________________

o __________________________________

➢ Thus, the electric field strength has the following

proportionalities:

𝑬 ∝

𝑬 ∝𝟏

➢ The constant of proportionality is the Coulomb’s constant (aka the electrostatic constant) which was

empirically derived from Coulomb’s work on electricity. It is denoted by 𝒌 and is:

𝒌 = 𝟖. 𝟗𝟗 × 𝟏𝟎𝟗 𝑵 𝒎𝟐 𝑪−𝟐

➢ The electric field, 𝑬, created by a charge 𝑸 at a distance 𝑹 away from it is given by:

𝑬 =𝒌𝑸

𝑹𝟐

➢ Similar to gravitational fields, the electric field has an inverse square proportionality with distance

because the emanating field lines spread over a spherical region with surface area 4𝜋𝑅2.



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Question 13. Coulomb’s constant and the Gravitational constant vary vastly (by 20 orders of magnitude!).

Find the magnitude of both the Electric and Gravitational fields, in SI units, felt 60.5cm away from a box

of mass 2.70kg, that has +2.70C of charge deposited on it.

Smallest Unit of Electric Charge

➢ Recall that in physics, electric charge is measured in Coulombs.

➢ It is discrete, with the smallest unit being the ____________________ _________________, 𝒆, which has

the value

𝒆 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪

➢ All charges on all objects are integer multiples of the elementary charge.

➢ The charge of a proton is +𝑒, and the charge of an electron is −𝑒.

Force in an Electric Field

➢ An electric field exerts a force on every charged object within it.

➢ The force experienced by a charge inside an electric field is the electric/electrostatic force, commonly

denoted by 𝑭𝒆.

➢ The magnitude of the electrostatic force (aka force due to the electric field) experienced by an object

with a certain amount of charge depends on two factors:

o ___________________________________

o ___________________________________



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➢ Thus, the electrostatic force on an object with charge 𝑞 in an electric field 𝐸 is given by:

𝑭𝒆 = 𝒒𝑬

We can substitute the complete relationship for 𝐸:

𝑭𝒆 =𝒌𝑸𝒒

𝑹𝟐

➢ Like any other force, the electrostatic force is a vector quantity, and is measured in Newtons.



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UNIFORM Electric Fields

Aka Charged Plates

Aka Capacitors

➢ There exists a uniform electric field between two oppositely charged plates

separated by a potential difference 𝑽 and a distance 𝒅.

➢ The magnitude of this uniform electric field between two charged plates

(aka a capacitor) is given by:

𝑬 =𝑽

𝒅

➢ So electric field can be measured in both 𝑁 𝐶−1 and 𝑉 𝑚−1.

➢ If we know the voltage and the distance, we can find the electric field, and

hence the electric force, and hence the acceleration of a charged particle!

Work and Energy in an Electric Field

➢ In a constant electric field, the electrostatic force is constant.

➢ The work done by a constant force is given by 𝐹 𝑑 cos (𝜃), so the work done on a charge 𝑞 in a uniform

electric field 𝐸 is:

𝑾 = 𝑭𝒆𝒅 = 𝒒𝑬𝒅

➢ We can substitute 𝐸 =𝑉

𝑑 to get:

𝑾 = 𝒒𝑽

➢ Using the work-energy theorem, this means that the change in kinetic energy for a charge 𝑞 moved

through a potential difference of 𝑉 is always given by:

𝚫𝑬𝒌 = 𝒒𝑽

Non-Uniform Electric Fields

➢ Similar to gravitational fields, the change in energy in an electric field is given by the area under a force-

distance graph.

➢ The area under an electric-FIELD vs. distance graph would give the change in energy per unit charge.



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Millikan’s Oil Drop Experiment

Question 14. An oil droplet with 2.7 × 106 excess electrons that weighs 0.015 𝑚𝑖𝑐𝑟𝑜𝑔𝑟𝑎𝑚𝑠 is suspended

somewhere between the two electric plates shown below, which are 0.025 𝑚𝑚 apart.

(a) What is the charge of the oil droplet?

(b) Label the sign of the charge that must exist on either plate if the droplet is suspended.

(c) Hence, sketch the electric field that must exist between the plates.

(d) What must be the potential difference between the two plates?

(e) In what area(s) between the plates is it possible to suspend the oil droplet?



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Work in an Electric Field

Question 15. A proton, 𝑚𝑝 = 1.67 × 10−27 𝑘𝑔, is accelerated through a potential difference of 5.05 𝑘𝑉

separated by 4.0 𝑐𝑚.

(a) What force acts on the proton?

(b) What is its acceleration through the electric field?

(c) What is the total work done by the electric field on the proton?

(d) Hence, what is the proton’s final speed if it started from rest, ignoring relativistic effects?



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MISCONCEPTION #1: The field lines represent the path a test mass (or test charge) would take.

➢ Consider the gravitational field near the surface of the Earth. It is a uniform field, with all field lines

parallel and pointing directly downwards towards the ground.

➢ While yes, if we release a ball from rest, it actually takes a path which follows a gravitational field line.

➢ That is no longer true, for instance, for a projectile launched at an angle – the projectile does not

follow the path along any single gravitational field line.

𝑻𝑹𝑼𝑻𝑯: 𝑭𝒊𝒆𝒍𝒅 𝒍𝒊𝒏𝒆𝒔 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕 𝒕𝒉𝒆 𝑭𝑶𝑹𝑪𝑬𝑺 𝒂 𝒕𝒆𝒔𝒕 𝒎𝒂𝒔𝒔 𝒐𝒓 𝒄𝒉𝒂𝒓𝒈𝒆 𝒘𝒊𝒍𝒍 𝒆𝒙𝒑𝒆𝒓𝒊𝒆𝒏𝒄𝒆 𝒂𝒕 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕

𝒑𝒐𝒊𝒏𝒕𝒔 𝒊𝒏 𝒔𝒑𝒂𝒄𝒆 𝒘𝒊𝒕𝒉𝒊𝒏 𝒕𝒉𝒆 𝒇𝒊𝒆𝒍𝒅. 𝑻𝒉𝒆𝒚 𝒅𝒐 𝑵𝑶𝑻 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕 𝒕𝒉𝒆 𝑷𝑨𝑻𝑯 𝒐𝒇 𝒂 𝒕𝒆𝒔𝒕 𝒎𝒂𝒔𝒔 𝒐𝒓 𝒄𝒉𝒂𝒓𝒈𝒆.

Fields add as Vectors

Electron-Volt

➢ We know that the work done by the electric field on a charge 𝑞 moving through a potential difference 𝑉

is given by 𝑊 = 𝑞𝑉.

➢ If the relevant charge is an elementary charge, this gives the work done to move a single elementary

charge through 𝑉, which is a unit of ENERGY.

➢ It represents the energy gained or lost by a single electron moving through 1 Volt.

𝟏 𝒆𝑽 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱



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Particle Accelerators

➢ These are simply devices which apply electric fields to accelerate (or decelerate) charged particles –

usually electrons, protons, etc.

➢ If a positive charge moves IN the direction of electric field, the work done on it is _________________ and

hence it __________________ kinetic energy.

➢ If a negative charge moves IN the direction of electric field, the work done on it is _________________

and hence it ___________________ kinetic energy.

Question 16. An electron gun is used to accelerate an electron through a potential difference of 84 kV.

(a) If the electron started from rest, what is its final kinetic energy, in eV?

(b) If the electron was already travelling at 8.0 x 107 m/s before it started accelerating, what is its final

speed (ignoring relativistic effects)?

(c) What is the acceleration experienced by the electron, if the electron gun is 90 cm long (i.e. the

charged plates are 90 cm apart)?



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Section H: All About Magnetic Fields and Electromagnetism

Magnetic Fields in VCE Physics

➢ Magnetic fields only exist as

___________________ and magnetic monopoles

do not exist.

➢ Magnetic field lines travel from the ___________

pole to the ____________ pole outside a magnet.

➢ Not all materials can be magnets. Only

ferromagnetic materials – iron, nickel, cobalt, and

their alloys – can be permanent magnets.

➢ The magnetic field strength at a point is denoted by 𝑩, and measured in SI Units of Tesla (𝑻).

➢ We do not quantify the strength of the magnetic field (like we do for gravitational and electric fields) in

VCE Physics. But the strength still depends on distance – as well as the material of the magnet.

o The magnetic field strength falls off faster than 1/𝑅2.

Uniform Magnetic Fields

➢ Just like a uniform gravitational field can be created near the surface of the Earth, and a uniform electric

field between two charged plates – a uniform magnetic field can be created between two very large

(and opposite) magnetic poles.

Electromagnetism Basics

Electric Current Induces Magnetic Field

➢ Danish physicist Hans Christian Oersted

noticed that a compass needle (which is a

tiny magnet) gets deflected near a

current-carrying wire.

➢ He found that a current carrying wire

actually produces a static, circular

magnetic field around it.

➢ The direction of the magnetic field around

a current-carrying wire is found using the

right hand rule. Thumb in the direction of

current. Fingers wrap around in the

direction of magnetic field.

NOTE: Remember, the direction of

conventional current is OPPOSITE to the

direction of motion of electrons!



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Current-Carrying Loop of Wire

➢ The magnetic field created around a current-carrying

LOOP of wire can be found by first finding the

direction of the local magnetic field for each section

of the wire.

➢ The overall magnetic field is similar to that of a bar

magnet.

➢ This is the first step to creating an electromagnet –

which (as opposed to a permanent magnet) can be

controlled with electric current.

Solenoids

➢ A solenoid is simply wire which is tightly packed in a

helical structure – it acts as multiple current-carrying

loops of wire next to each other.

➢ When current flows through a solenoid, the “mini bar magnets” of each coil add up to create a much

stronger total magnetic field.

➢ The combined effect of multiple coils is such that the magnetic field of a solenoid is equivalent to that

of a bar magnet.

Magnitude of Solenoid Magnetic Field

➢ The magnetic field strength of a solenoid has the following proportionalities:

𝑩𝒔𝒐𝒍𝒆𝒏𝒐𝒊𝒅 ∝






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Direction of Solenoid Magnetic Field

➢ One way of finding the direction of the magnetic field for a loop/solenoid is to apply the original right

hand rule to any section of wire.

➢ An alternative method is to use the right hand solenoid rule. Fingers wrap around in the direction of

current in the loop – then thumb points towards the direction of the North Pole of the solenoid.

Question 17. Max has a solenoid with 500 coils of copper wire looped around a 14 𝑐𝑚 long nickel core. He

passes a current of 11.0 𝑚𝐴 through the solenoid.

Max now wishes to decrease the strength of the magnetic field of the solenoid. Which of the following

steps (all judged independently of each other) will help him achieve this?

1. Increasing the gaps between successive coils by using a longer nickel core of length 25 𝑐𝑚.

2. Increasing the number of loops of copper wire to 650.

3. Replacing the nickel core with a soft iron core.

4. Increasing the current to 15.0 𝑚𝐴.

5. Removing the nickel core.

Magnetic Force (aka Lorentz Force)

➢ Dutch physicist Hendrik Lorentz found that when electric charge, 𝑞, moves within a magnetic field, the

magnetic field applies a force on the moving charge (known as the Lorentz force).

➢ The Lorentz force on a charge 𝑞 moving through a magnetic field 𝐵 at a velocity 𝑣 is given by:

𝑭𝒍 = 𝒒𝒗𝑩 𝐬𝐢𝐧 (𝜶)

Where 𝛼 is the angle between the velocity and the magnetic field.



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Magnitude of the Lorentz Force

Question 18.

(a) What is the magnitude of the Lorentz force when the charged particle is moving along the magnetic

field lines (i.e., the velocity vector 𝑣 is parallel to the magnetic field vector 𝐵)

(b) What is the magnitude of the Lorentz force when the charged particle is moving perpendicular to the

magnetic field lines?

Direction of Lorentz Force – Right Hand Slap Rule

➢ To find the direction of the magnetic force on a charged particle, we use the right hand slap rule. Lay

your hand flat with thumb outstretched, like a slap.

o Thumb points in the direction of the velocity 𝒗 of POSITIVE CHARGE, fingers point in the

direction of the magnetic field lines.

o Then, the direction of the Lorentz force comes out of the palm.

➢ Importantly, remember that 𝒗 is the direction of motion of POSITIVE charge. If the moving charge is

negative, you can either reverse the direction of the force (i.e. it comes out the back of your hand); OR

reverse the direction of the velocity (i.e. thumb points OPPOSITE to the motion of negative charge).



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Question 19. A proton travelling towards the right at 4.7 × 106 𝑚/𝑠 enters a magnetic field of magnitude

4.0 𝑚𝑇 into the page as shown.

(a) Find the magnitude and label the direction of the Lorentz force that initially acts on the proton.

(b) Where does the force point if the field is out of the page?

(c) Where does the force point if an electron enters the field instead of a proton?

(d) What about both of those at the same time?

× × × × ×

× × × × ×

× × × × ×

What would happen if the proton kept travelling through the region of the magnetic field? What path would it take?

Circular Motion in Magnetic Fields

➢ The direction of the Lorentz force will always be perpendicular to both the velocity and the magnetic

field.

➢ When a force on an object acts perpendicular to a velocity, it does _____________ work on the object.

➢ If the force has just the right magnitude, it can create circular motion – where the centripetal force is

provided by the Lorentz force.

➢ We can find the radius of the circular path taken by a charge 𝑞 moving at velocity 𝑣 through a magnetic

field 𝐵 by equating the Lorentz force with the centripetal force:

𝒒𝒗𝑩 =𝒎𝒗𝟐

𝑹



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𝑹 =𝒎𝒗

𝒒𝑩

➢ Note that circular motion involves constant speed, and the centripetal force does not do work (i.e.

change the kinetic energy) of the object.

o Thus, the Lorentz force never does work on the charged particle – it can never change the speed

or kinetic energy of the charged particle.

Question 20. An electron (𝑚𝑒 = 9.1 × 10−31 𝑘𝑔) inside a magnetic field moves in uniform circular motion

at a speed of 𝑣 = 5.5 × 105 𝑚/𝑠 as shown in the diagram below.

(a) Find the magnitude, and label the direction, of the magnetic field that the electron must be in.

(b) Using proportionalities, what would happen to the radius of the path if the strength of the magnetic

field was tripled?

(c) What would happen to the radius of the path if two electrons (stuck together) moved through the B-

field instead of one?



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Force on Current-Carrying Wire

➢ Lorentz’s discovery was that a magnetic force acts on a charge moving through a magnetic field. Recall

that electric current is simply the rate of flow of charge.

o So, a current-carrying wire in a magnetic field is just a bunch of charge moving in a B-field!

➢ We can derive the formula for the Lorentz force on a current-carrying wire of length 𝐿 carrying current 𝐼

in a magnetic field 𝐵. If there are 𝑁 wires together, we can get the general formula:

𝑭𝒄𝒖𝒓𝒓𝒆𝒏𝒕−𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒘𝒊𝒓𝒆 = 𝑵𝑰𝑳𝑩 𝐬𝐢𝐧 (𝜶)

Where 𝛼 is the angle between the current and the magnetic

field.

➢ For the direction, we can just replace “direction of motion

of positive charge” with “direction of current” – since

current is conventionally defined as the direction of

moving positive charges.

➢ Always remember that electrons move in the opposite

direction to conventional current.

Question 21. A power line runs 250 𝐴 from West to East along the equator, where the strength of Earth’s

magnetic field is 3.05 × 10−3 𝑇.

(a) What is the magnitude AND direction of the Lorentz force on a 4.5 𝑚 long section of wire?

NOTE: There is an effective “bar magnet” inside the Earth – with the magnetic North near the geographic

South Pole.



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(b) [Optional HW] Imagine the 4.5 𝑚 section of wire is isolated in the air. It is possible for it to float in the

air if the net force on it is zero. What is the maximum allowable mass of the 4.5 𝑚 section of wire

which would allow this to occur when the power line is active?

Earth’s Magnetic Field

Two Current-Carrying Wires Close Together

➢ Each wire creates a magnetic field around it. This magnetic field reaches the other wire – which has

moving charged particles (a current). Thus, the magnetic field of the first wire creates a Lorentz force on

the second wire. And vice versa.

➢ When the current in both wires points in the same direction, the wires are attracted to each other.

➢ When the current is in opposite directions, the wires are repelled from each other.



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Optional HW Questions:



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Section I: DC Motors

➢ A motor is simply a device that converts electrical energy to mechanical energy. Usually, this mechanical

energy is in the form of (rotational) kinetic energy.

➢ A DC motor uses Direct Current (e.g. from a battery) to convert electrical energy into mechanical energy.

Motor Schematic

➢ Consider a current-carrying loop of wire – which is free to rotate about a central axis – placed inside a

magnetic field.

Exploration 2. DC Motor.

➢ When the circuit is closed, current flows through the wires. Label the direction of current in the wires.

➢ Label the direction of the magnetic field due to the magnets.

➢ Hence, label the Lorentz forces that act on each section of the main square loop of wire.

➢ What effect do these Lorentz forces have on the loop of wire – which is free to rotate about the

central axis?

➢ This loop will start to rotate [clockwise] / [anticlockwise].

N S



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Does the Motor Spin?

Exploration 3.

➢ Say the coil starts spinning anticlockwise. What happens at exactly a quarter turn (when the loop

becomes vertical) to:

o The direction of current flow through the loop?

o The Lorentz forces on the sections of the loop?

o The direction of the torque on the loop? Hence, the rotational acceleration of the loop?

o The actual rotational motion of the coil?

➢ What happens as the loop moves past a quarter turn and approaches half a turn to:

o The direction of current flow through the loop?

o The Lorentz forces on the sections of the loop?

o The direction of the torque on the loop? Hence, the rotational acceleration of the loop?

o The actual rotational motion of the coil?

➢ So, does the motor work as intended?

No! Even though the motor starts spinning as intended – once it crosses over the vertical orientation, the

torque due to the Lorentz forces actually starts going in the opposite direction to the motor’s spin.

This means the motor starts slowing down instead of speeding up – and it is eventually completely

stopped at the halfway point. Then, it starts up spinning in this new opposite direction – but only until it

crosses the vertical orientation and begins slowing down to a stop again.

This cycle repeats, and the motor actually oscillates. If there is friction, it will eventually come to a

complete stop in the vertical orientation.

The Split Ring Commutator (SRC)

➢ The SRC is a mechanical device consisting of a conducting ring

which is split in the middle.

➢ When the coil spins, it disconnects and then immediately

reconnects the source DC current to the opposing wire.

➢ Thus, when the SRC is included in the circuit, the direction of the

input current into the coil gets reversed every half turn (or half

period) of the coil.



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The DC Motor With SRC

➢ We know that the DC motor schematic from earlier starts slowing down after it crosses the vertical

orientation (due to the torque being in the opposite direction).

➢ However, if the SRC reverses the direction of current, that will reverse the direction of Lorentz forces –

which would result in the torque staying in the same direction!

➢ The SRC continues to flip the direction of input current every half period, which maintains a constant

torque on the motor. Instead of oscillating, the motor can now spin as intended.

➢ Additionally, the split-ring is connected with metallic conducting brushes, which allow the main coil to

spin without tangling the wires.

➢ The workings of the SRC in a DC motor can be seen at

https://www.scienceflip.com.au/subjects/physics/electromagnetism/learn9/ and at

https://www.animations.physics.unsw.edu.au/jw/electricmotors.html#DCmotors

NOTE: The split ring commutator reverses the direction of input current every half period. This reversal

occurs when the coil is in the vertical position – that is when the torque would otherwise have reversed!

➢ If we know the lengths/dimensions of the coil, we can find the individual Lorentz forces on each side of

the coil.

NOTE: Real-world DC motors usually have MANY coils (𝑁 coils) wrapped together, which increases the

rotational kinetic energy available – as more coils are spinning. Remember to multiply by 𝑁 when finding the

Lorentz force on a side of the coil in this case!

N S



https://www.scienceflip.com.au/subjects/physics/electromagnetism/learn9/

https://www.animations.physics.unsw.edu.au/jw/electricmotors.html#DCmotors

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1. The split-ring commutator reverses the direction of input current every half period.

2. This also reverses the direction of the Lorentz forces on each arm every half period.

3. Which, in turn, keeps the torque in the same direction, allowing the motor to spin.

4. (Additionally, the commutator allows the wires to spin without tangling).



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DC Motor Practice Question:

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Exam Revision Workshop: Physics ¾ Unit 3 AOS 1 Fields

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