Lesson 28: Magnetic fields

20. Magnetic Fields

20.1 Concept of a magnetic field
Learning outcomes
understand that a magnetic field is an example of a field of force
1 produced either by moving charges or by permanent magnets

2 represent a magnetic field by field lines

20.4 Magnetic field due to currents

sketch magnetic field patterns due to the currents in a long straight wire, a flat
1 circular coil and a long solenoid

understand that the magnetic field due to the current in a solenoid is

2 increased by a ferrous core
MAGNETIC FIELDS

• Magnetic fields are very similar to electric fields, as they deal with the
poles and magnetism from an object to create a field which exhibits a
force on another magnetised object. You can get uniform magnetic
fields, however radial magnetic fields are not possible due the fact that
a N and S pole cannot be separated.
• When two fields interact, they can form interesting patterns, where the
lines (always N to S) will overlap and resultant field can from all sorts of
funky patterns, as well as neutral points, where there is no magnetism.
Pop a compass in here and it won’t know where to point.
Which way does a compass point?
Experiment: Investigating Magnetic field lines
1) A single magnet
2) Two magnets with like poles facing each other
3) Two magnets with opposite poles facing each other
Electromagnets
Another important aspect is that an electrical current will produce a magnetic
field, where a wire with a flowing current will produce a field in the form of a
closed loop.
Field around a wire
Prediction Result
Field around a Coil
Prediction Result
Field around a Solenoid
Prediction Result
Determining Field Directions
• The direction of a magnetic field
produced by a current carrying wire
can be determined with the right hand
rule – if you put your right hand in the
‘thumbs up’ position, then your thumb
is the direction of current, and your
fingers show the direction of the field
produced.
 Determining Field Directions

• The direction of a magnetic field produced by a solenoid can be

determined with the right hand rule – if you put your right hand in the
‘thumbs up’ position, then your thumb is the direction of north pole, and
your fingers show the direction of the current.
Strengthening the electromagnetic field

• A solenoid wrapped around an iron core can increase the strength of

the magnetic field
Lesson 29: Left-Hand rule
 20. Magnetic fields
20.2 Force on a current-carrying conductor

Learning outcomes
understand that a force might act on a current-carrying conductor
1 placed in a magnetic field

recall and use the equation F = BIL sin θ, with directions as interpreted
2 by Fleming’s left-hand rule

define magnetic flux density as the force acting per unit current per
3 unit length on a wire placed at right-angles to the magnetic field

20.4 Magnetic field due to currents

explain the origin of the forces between current-carrying conductors and
3 determine the direction of the forces
FORCE ON A CURRENT CARRYING
CONDUCTOR
• If a wire is present in a magnetic field, the wire will experience a force
due to electromagnetism. This relies on all three components being a
right angles to each other. We can use Flemming’s left hand rule to
determine the direction of the force.
 MAGNETIC FLUX DENSITY (B)
• Magnetic flux density tells us the total number of magnetic field lines (ie.
magnetic flux) passing through a given area, and this determines how
strong the field is. (Therefore magnetic flux density is also known as
Magnetic field strength)

Units of B is the TESLA

Or 1N A-1 m-1

Magnetic Flux Density can be defined as the magnetic flux per

unit area.
 MAGNETIC FLUX DENSITY (B)

In electromagnetism, Magnetic flux density can similarly be defined as:

the force experienced per unit length by a current carrying

conductor placed at right angles to a magnetic field

𝐹
𝐵=
𝐼𝐿
How the induced E.M.F varies with the
angle of the current carrying wire
Forces between currents

So far, we have looked at how 2 magnets interact with each

other and how both a current carrying conductor behaves in a
permanent magnetic field.

Let us now look and see how 2 current carrying wires interact
with each other
Anti-parallel currents using Right hand grip rule

Parallel currents using Right hand grip rule

Parallel currents using LHR

In this example we focus only

on the radial magnetic field
produced by I1. We then look
at the direction of the magnetic
field of I1 at I2. Since the
current is going into the page,
we can use L.H.R to determine
the direction of the force
exerted on I2
Lesson 30 : Force on a moving charge
 20. Magnetic fields
20.3 Force on a moving charge

Learning outcomes
1 determine the direction of the force on a charge moving in a magnetic field

2 recall and use F = BQv sin θ

understand the origin of the Hall voltage and derive and use the expression
3 VH = BI / (ntq), where t = thickness

4 understand the use of a Hall probe to measure magnetic flux density

describe the motion of a charged particle moving in a uniform magnetic field

5 perpendicular to the direction of motion of the particle

6 explain how electric and magnetic fields can be used in velocity selection
 FORCE ON A MOVING CHARGE
• An electron beam emits an electron into a magnetic field.
Here the electron is travelling at right angles to the
direction of the field. This will cause the electron to start
undergoing circular motion due a force at right angles to
its motion.

USE flemmings LHR

and check to see
whether you agree
with the direction
of the force on the
electron!

Magnetic field into paper

FORCE ON A MOVING CHARGE
At IGCSE we learnt the Left-hand rule to determine the direction of the
force on a wire. However, note that this was using a CONVENTIONAL
CURRENT or a moving POSITIVE CHARGE.
Therefore, to determine the force on a moving ELECTRON we use
Flemming’s right-hand rule

USE flemmings RHR and

Magnetic field into paper
check to see whether you
agree with the direction
of the force on the electron!

Electron flow
Q1.
The force experienced on a charge
• Recall force on a wire:

Force = Magnetic Flux Density (T) X Current X Length of Wire X Sin θ

Q2. P424
The Hall effect

Consider a current flowing through a metal

conductor with a magnetic field perpendicular
to the plate and current. The electrons
experience a magnetic force, which make them
drift to one side of the conductor where they
will gather. The other side is deficient of electrons.

Hall effect: the production of a potential

difference across an electrical conductor
when an external magnetic field is applied
in a direction perpendicular to the direction
of the current
The Hall Voltage (VH)
Hall voltage: the production of a potential difference
across an electrical conductor when an external
magnetic field is applied in a direction perpendicular
to the direction of the current. The Hall Voltage is
directly proportional to the mag flux density B.

Imagine a single electron as it travels with drift

velocity (v) through the slice (ie. Conventional down,
electron flow Up)

As a result of this P.D, an electric field is also created

from + to – plate. As the + plate is on the left, the
electron will experience an electric force to the left
(opposites attract)

Due to mag. Field Flemmings RHR, the electron

experience a magnetic force to the right.
 The Hall Voltage (VH)
Electric field strength (E)=
d is width

As the electric force builds up, eventually:

Electric force = magnetic force

or
eE = Bev

Substituting E we get

Drift velocity recall v= I/nAe

Substituting v we get

or

Charge of electron More general equation, where q is charge

of individual charge carrier
Measuring magnetic flux density (B):

• With a hall probe

Velocity Selection
We will now consider what happens when charged particles (all with
varying speeds) passes through an electric field
Can you derive an equation for the speed of a
charge particle in a field that goes through
undeflected?
What happens if speed too slow or too
fast?

Since Electric force = Mag Force (for particles going straight)

QE = BQv

If velocity increases, then the Magnetic force will be larger than electric force
and cause particle to go up (from previous example) and vice versa

Ie if v> then particle goes up

if v< then particle goes down

Lesson 31 : Magnetic flux
 20. Magnetic fields
20.5 Electromagnetic induction

Learning outcomes
define magnetic flux as the product of the magnetic flux density and the
1 cross-sectional area perpendicular to the direction of the magnetic flux
density

2 recall and use Φ = BA

3 understand and use the concept of magnetic flux linkage

 MAGNETIC FLUX () & MAGNETIC
FLUX DENSITY
• Magnetic flux density tells us how close together the magnetic field lines are
together in a field, which is how strong the field is. (ie magnetic field strength = Flux
per unit area)

Green shaded area=

ALL field lines mag.Flux density
• Magnetic flux tells us the hitting south side=
total flux in a given area. Magnetic flux
The units Weber (Wb)

  = BA
A ) LOOP A
B) LOOP B
C) The fluxes are the same
D) Not enough information to
tell
Magnetic flux linkage ()
• We now know that the amount of flux through one loop of wire is:
 = BA
If we have a coil of wire made up of N loops of wire the total flux is given
by:
N  = BAN
The total amount of flux, N  , is called the Magnetic Flux Linkage ();
this is because we consider each loop of wire to be linked with a certain
amount of magnetic flux. Sometimes flux linkage is represented by  ,

so
 = BAN
Textbook Questions 4-9 (p532)
Lesson 32 : Lenz’s law
 20. Magnetic fields
20.5 Electromagnetic induction

Learning outcomes

understand and explain experiments that demonstrate:

• that a changing magnetic flux can induce an e.m.f. in a circuit
4 • that the induced e.m.f. is in such a direction as to oppose the change
producing it
• the factors affecting the magnitude of the induced e.m.f.

5 recall and use Faraday’s and Lenz’s laws of electromagnetic induction

Electromagnetic Induction
 Fleming's Right-hand rule

This is used to predict the

direction of the current
caused by induced
e.m.f. in a conductor
moved at right angles to a
magnetic field:

Induced (conventional)
current
Origin of E-M Induction
Since we push the wire down, The
free electrons in the wire moving
down will be the electric current. (ie
conventional current is upwards).

We can use the LHR to determine

the the direction of the force on
the free electrons.

We therefore get a force on the

electrons from X to Y and therefore
induced (conventional) current.
Determining the direction of the induced
current in the coil and thus direction of its
field
 Faraday’s Law
The magnitude of the induced e.m.f is directly proportional to the
rate of change of magnetic flux linkage

∆ ( 𝑁 𝛷)
𝐸=
𝑡
 Lenz’s law
Lenz’s law: An induced e.m.f is in a direction so as to produce
effects that oppose The change producing it
 Lenz’s Law

∆ ( 𝑁 𝛷)
𝐸 =−
𝑡
The minus sign indicates that this induced emf
causes effects to oppose the change producing it

Lenz’s law: An induced e.m.f is in a direction so as to produce

effects that oppose The change producing it
Direction of induced E.M.F.
Diagram (a) is incorrect because if the
induced emf does NOT oppose the
motion, (ie south pole
inducedattracts the magnet), then a
little change in mag. Flux would
produce an induced current which
would help the change of flux further
thereby producing more current. The
Increased emf would then cause
further change of flux and it would
increase the current and so on.

This would create energy out of

nothing which would violate the law
of conservation of energy.
Thus diagram (b) is correct and a north
pole is induced at the left end of coil.
Question 15 (textbook)

A bar magnet is dropped vertically

downwards through a long solenoid which is
connected to an oscilloscope. The graph
shows how the emf induced in the coil varies
with time as the magnet accelerates
downwards.

• Explain why an emf is induced in the coil as the

magnet enters. (AB)
• Explain why no emf is induced is induced while
the magnet is entirely inside
the coil (BC)
• Explain why section CD shows a negative trace.
Why it shows a larger peak. Why the time
interval of CD is such a shorter time interval.
Question 15 (textbook)
Question 15 (Answer)
• 1) As the magnet enters it generates a
current in the loop that sets up a magnetic
field to oppose the entry of the magnet.

2) When the magnet is in the middle of

the coil the it is at the point where the
magnetic poles will switch. At this point
there is no current flowing in the
coil since the p.d is zero.

3) When exiting the coil the magnet is

moving faster (because of its acceleration
due to gravity) and thus the width is
narrow and larger. It induces a current in
the loop that sets up a magnetic field to
oppose the magnet moving away i.e. the
magnetic poles of the coil change. Thus, a
negative emf.
Alternating current
Generators
Lesson 33 :
Alternating
currents
 21. Alternating currents
21.1 Characteristics of alternating currents

Learning outcomes
understand and use the terms period, frequency and peak value as applied
1 to an alternating current or voltage

use equations of the form x = x0 sin ωt representing a sinusoidally alternating

2 current or voltage

recall and use the fact that the mean power in a resistive load is half the
3 maximum power for a sinusoidal alternating current

distinguish between root-mean-square (r.m.s.) and peak values and

4 recall and use I r.m.s. = I0 / 2 and Vr.m.s. = V0 / 2 for a sinusoidal
alternating current
Recall Time period & Frequency
 Alternating Current

+I/A

-I/A

Since an alternating current follows a sine wave pattern, we can use the following equation
to determine the current:
Alternating Voltage

𝑽 =𝑽 𝒐 𝒔𝒊𝒏 𝝎 𝒕
Root-mean-square (r.m.s) values

Current
• The r.m.s value of an alternating current is that steady
current which delivers the same average power as the A.C
to a resistive load.

𝐼𝑜
𝐼 𝑟 . 𝑚 . 𝑠=
2
Root-mean-square (r.m.s) values

Voltage
• The r.m.s value of a voltage is that steady voltage which
delivers the same average power as the alternating voltage
to a resistive load

𝑉𝑜
𝑉 𝑟 . 𝑚 . 𝑠=
2
Explaining R.M.S
Power
Recall equations for power:
2
𝑃= 𝐼 𝑅

𝑃=𝑉𝐼
2
𝑉
𝑃=
𝑅

When calculating power it is important to use the r.m.s

Values for both VOLTAGE & CURRENT.
 The variation with time of an alternating
current in a resistor of 120Ω is shown.
(a) What is the value of its r.m.s. current?

(b) What is the mean power dissipated by the

resistor?
Measurements using an
Oscilloscope
Lesson 34 : Rectification and smoothing
 21. Alternating currents
21.2 Rectification and smoothing

Learning outcomes

1 distinguish graphically between half-wave and full-wave rectification

explain the use of a single diode for the half-wave rectification of an alternating
2 current

explain the use of four diodes (bridge rectifier) for the full-wave rectification of an
3 alternating current

analyse the effect of a single capacitor in smoothing, including the effect of the
4 values of capacitance and the load resistance
Half-wave rectification

For some electrical appliances

a direct current is required. This
can be achieved by using a diode
in an A.C. circuit. As current can
only flow in one direction, Vout
is 0V for half the time.
Full-wave Rectification:The bridge-rectifier
Smoothing The Capacitor charges up and
maintains the voltage
at a high voltage level. It
discharges gradually when then
the rectified voltage drops, but
soon rises again and the
capacitor charges up again.

You can control the size of

this ripple by controlling the
capacitor and resistance. A
larger capacitor discharges more
slowly. A larger resistor will
similarly cause a slower
discharge of the capacitor

Recall from capacitors

Thus, increasing the time

constant () reduces the size of
ripple.