Magnetic Fields
| Key Words Magnetic, Flux, Flux Density, Flux Linkage |
|
In
the exam you
are expected to know about:
|
Force
On A Current Carrying Wire In A Magnetic Field
You will be familiar with the basic notion of a magnetic field, in which magnetic materials experience a magnetic force. However it is worth revising some of the basic ideas that you will have come across in early secondary school.
Revision
Magnetic
fields can be shown by field lines, which go from North to South.
The field lines in a strong magnetic field are more closely packed than in a weak field.

Unmagnetised
materials are attracted to either pole.
Like
poles repel; unlike poles attract.
In the Earth’s magnetic field, the North pole will align itself to point to the North, if the magnet is allowed to swing freely.
The Earth has a magnetic field like a bar magnet. Notice that the S-pole is under the North geographic pole. Be careful not to be confused by this.

We
never get single magnetic poles; if there is an N-pole, there must also be
an S-pole to go with it.
Magnets
are thought to result from the action of tiny atomic magnets called domains.
This can be explained by the movement of electrons that represent a tiny
electric current that results in magnetism. In most materials, the
currents cancel out.
Only
iron, cobalt, and nickel and their alloys are magnetic.
When
a magnetic material is unmagnetised, the
domains are all jumbled up. If some
of the domains are lined up, then the
material is partially magnetised. If
the domains are fully lined up, the magnet
is saturated, and cannot be magnetised
further.
Some
materials like soft iron lose their
magnetism quickly. These are used for temporary
magnets. Permanent or hard
magnetic materials do not lose their magnetism.
Electric currents always produce a magnetic field, even if the wire itself is not made of a magnetic material. The magnetic field of a single current carrying wire is like this:

The
direction of the current is determined by the Screwdriver
Rule.
The magnetic field of a solenoid is like a bar magnet.
You will be familiar with the motor effect. If we put a current carrying wire in a magnetic field, we see that there is a force.


We
can work out the force that is exerted on the wire quite simply.
Experiment shows us that the force is proportional to:
The current;
The
strength of the magnetic field
The
length of wire within the magnetic field.
This
is summed up in a simple formula:
F = BIl
[B – magnetic field strength;
I
– current in A;
l – length in m]
The
term
`
is called the
magnetic field strength,
or the flux density, and is measured
in Tesla, T. The magnetic flux
density can be thought of as the concentration of field lines.
We can increase the force by increasing any of the terms within the
equation. If we coil up the wire,
we increase its length within the magnetic field.
Worked
Example
| A current of 8.5 A flowing through a magnetic field is found to exert a force of 0.275 N. The length of wire in the magnetic field is 5 cm. What is the value of the magnetic field? |
| Formula first: F = BIl. |
|
Þ
B = F
= ___0.275 N ___ = 0.647
T Il 8.5 A ´ 0.05 m |
| Question 1 |
In a demonstration of the above equation, the length of wire in a magnetic field is 0.05 m. When a current of 2.5 A flows, a force of 0.01 N is shown. What is the magnetic field strength? |
ANSWER |
This
relationship holds true as long as the current is at 90o to the
magnetic field. If the wire is at
an angle to the field, the relationship takes this into account by changing to:
F
= BIl
sin
q
We
know that a magnetic field and an electric current interact to produce a force.
Since a current is a flow of charge, it is reasonable to suppose that a
magnetic field exerts a force on individual charge carriers.
We
find that in a magnetic field, the force acts on a stream of electrons always at
90o to the direction of
the movement. Therefore the path is
circular.
Consider
a charge q
moving through a magnetic field B at a
constant velocity v.
The charge forms a current that moves a certain distance, l, in a time t.
We
know:
Velocity
= distance ¸
time
Current
= charge ¸
time
F
= BIl
So
we can substitute into this relationship to give us:
F = B ´ (q/t) ´ vt = Bqv
So
the formula now becomes:
F = Bqv
[
F- force in N;
B
– field strength in
T;
q –
charge in C;
v –
speed in m/s]
The charge is usually the electronic charge, 1.6 ´ 10-19 C.
| Question 2 |
An electron accelerated to 6.0 ´ 106 m/s is deflected by a magnetic field of strength 0.82 T. What is the force acting on the electron? Would it be any different for a proton? |
ANSWER |
Path
of charged particles in a Magnetic Field
We
have seen that the force always acts on the wire at 90o, and that
gives us the condition for circular
motion. We can combine the
relationship a = v2/r with
Newton II to give us:

Therefore:

This
rearranges to give us:

Worked
Example
| An electron passes through a cathode ray tube with a velocity of 3.7 × 107 m/s. It enters a magnetic field of flux density 0.47 mT at a right angle. What is the radius of curvature of the path in the magnetic field? |
|
F
= Bqv and F = mv2/r Þ
r = mv Bq |
|
r
= 9.11 × 10-31 kg × 3.7 × 107 m/s = 0.39
m = 39 cm 0.47 × 10-3 T × 1.6 × 10-19 C |
| Question 3 |
In a particle physics experiment, a detector is place in a magnetic field of 0.92 T. A particle is found to produce a track of radius 0.5 m. Other experiments have shown that the particle carries a charge of +1.6 ´ 10-19 C and that its speed was 3.0 ´ 107 m/s. What is the mass of the particle? How does it compare to the mass of an electron (9.11 ´ 10-31 kg)? |
ANSWER |
The cyclotron is a particle accelerator that relies on this idea. The machine’s main components are two D-shaped electrodes ("Dees") in an evacuated chamber, placed between the poles of a large electromagnet

From the top it looks like this:

Notice
that the beam of particles is not circular, but a spiral.
This is because the particles are being accelerated by the electric field
between each D-shaped electrode (called a dee).
As their speed increases, so does the radius of the curved path.
If
a particle of charge Q
enters one of the dees with a speed v,
it will move in a semi-circular path of radius r.
Þ rearranging gives us

We
can work out from
t = s/v
what time it takes for the charge to travel:

The
voltage will change twice in each alternating current cycle, therefore in the
period of one cycle, the time is
2t.
Since
f =
1/T, we can write the expression
as:

Magnetic
Flux Density, Flux, and Flux Linkage
We
have seen how
B is called the magnetic
field strength, or the flux density,
and is measured in Tesla, T. The
magnetic flux density can be thought of as the concentration of field lines.
We can increase the force by increasing any of the terms within the
equation. If we coil up the wire, we increase its length within the
magnetic field.
If we look at the magnetic field of a solenoid, we know that it is like a bar magnet:

We
can see that the magnetic field strength is uniform within the solenoid.
However the flux density becomes less at the ends, as the field lines get
spread out.
We
need a term that tells us the number of field lines, and it is called the magnetic flux. It is
given the physics code
F
(‘Phi’, a Greek capital letter ‘Ph’), and has the units Weber (Wb).
The formal definition is:
The product
between the magnetic flux density and the area when the field is at right angles
to the area.
In
code we write:
F
= BA
Remember
that flux density is the number of field line per unit area, not unit volume!
The
flux linkage is the flux multiplied by the number of turns of wire. If each turn cuts (or links) flux
F,
the total flux linkage for
N
turns must be
NF.
We can also write this as
NBA. In
other words:
Flux
linkage = number of turns of wire ´
magnetic field strength ´
area
| How much flux links a 200 turn coil of area 0.1 m2 when it is placed at 90o to a magnetic field of strength 2.5 ´ 10-3 T? |
The
flux linkage can be changed in two ways:
We
can alter the strength of the magnetic field;
We
can alter the area at 90o to the magnetic field by moving the
coil. If we are turning the
coil, the new flux linkage is given by
NBA
sin
q
where
q
is the angle the area makes to the magnetic field.
When we move a coil across a magnetic field, the area swept is the
change in area (just like the change in distance is the distance moved).
We
give the change in flux linkage the physics code
DF.
| The coil in question 4 is now turned so that it makes an angle of 60 o with the magnetic field lines. What is the change in flux linkage? |
If we pass a current in a wire in a magnetic field, we know that the wire will move. It is therefore reasonable to suppose that if we move the wire in a magnetic field, and the wire is connected to an outside circuit, a voltage and current are induced. If the wire is not connected, a voltage only is induced. Consider this demonstration:

If
we move the magnet parallel to the wire, the galvanometer hardly responds.
However, if we move the magnet across the wire, then we see a definite
reading on the galvanometer. The
current (and voltage) induced on a single wire is rather small, but is increased
by having more turns of wire. For
any voltage to be induced, we must move the magnet.
We call this voltage the induced
electromotive force. It is
often given the code Î,
a fancy letter ‘E’.
Faraday’s
Law
and Lenz’s Law are two important
rules that govern this effect.
Faraday’s
Law is a formal definition of the effect:
The
induced e.m.f. across a conductor is equal to the rate at which flux is cut.
Lenz’s
Law says:
The
direction of any induced current is such as to oppose the flux change that
caused it.
The
induced e.m.f. sets up a current that would oppose the force that is pulling the
wire. If the force were to assist
the motion, we would get acceleration, and an increase in kinetic energy.
This would break the Law of Conservation of Energy.
In other words, we cannot get something for nothing.
Lenz’s
Law is important in motors and generators.
As a motor speeds up, it acts like a generator to produce a back e.m.f. to oppose the current flowing in the motor.
Therefore the current through a fast running motor is quite small.
When it is running slowly, a big current flows.
Electric motors are therefore very suited to railway use, where big
currents are needed to get trains moving, and there is no need for a gearbox
that is needed for a diesel engine.
The
effect that we have seen is summed up in the relationship:

[N
- number of turns;
E– e.m.f (V);
DF/Dt
– rate of change in flux (Wb/s)]
Worked Example
| A single turn of wire of cross-sectional area 5.0 cm2 is at 90o to a magnetic field of 0.02 T, which is reduced to 0 in 10 s at a steady rate. What is the e.m.f. induced? |
|
Two
formulae to use: F
= BA
and
Î =
- NDF
Dt |
|
We
need to work out the flux: F = BA = 0.02 T × 5 × 10-4 m2 = 1 ×10-5 Wb |
|
Now
we can work out the e.m.f:
Î =
- NDF
= 1×
1 ×10-5 Wb
= 1 ×10-5 V Dt 10 s |
| A search coil has 2500 turns and an area of 1.5 ´ 10-4 m2. It is placed between the poles of a large horseshoe magnet. It is rapidly pulled out of the field in a time of 0.30 s. A datalogger records an average value for the emf of 0.75 V. What is the flux density between the poles of the magnet? |
We
can use a magnetic field to induce a voltage in two ways:
1.
Relative movement. The size
of the voltage depends on:
Speed
the magnet passes through a coil or vice
versa.
Number
of turns in the coil.
Strength
of the magnet.
2.
Changing a magnetic field. We
don’t have to make the magnetic field move.
If we turn the current on or off, there is a change in the magnetic
field, and that induces a voltage in a second unconnected
coil. This is called the transformer
effect or mutual induction.
The transformer is a machine that is simplicity itself. It consists of:
A primary coil connected to the alternating power source. This provides the changing magnetic field.
A secondary coil connected to the load (in this case, the rectifier and control equipment in the locomotive).
A laminated soft iron core.
The two coils are electrically completely different circuits. Either of the coils can act as a primary. The laminated core is made up of layers of soft iron separated by an insulating layer of varnish or glue. This reduces losses from eddy currents. Soft iron is NOT soft like putty; it is heavy and hard. However "soft" means that it loses its magnetism immediately the current is turned off. Therefore the magnetic field can change forwards to backwards as the current changes.
The ratio of the input voltage to the output voltage is the same as the ratio of the number of turns on the primary to the number of turns on the secondary. We can write this as:
Number of turns on the primary = Primary Voltage
Number of turns on the secondary Secondary voltage
In Physics code:
N1 = V1
N2 V2
If N1 is greater than N2, we have a step-down transformer, because the voltage is reduced. A step-up transformer increases the voltage.
If a transformer is 100 % efficient (and it nearly is) we can say that:
power in = power out
V1I1 = V2I2
Therefore we can say that when the voltage is lower, the current is bigger. We can rewrite the transformer equation in terms of current to give us:
N1 = I2
N2 I1
In practice, the transformer is about 97 % efficient. When a large transformer is transferring a lot of energy, even 3 % losses produce a fair amount of heat. Therefore the transformer is cooled with oil which is pumped to heat exchangers.
The picture above shows the huge transformers used at a power station. You can see the massive cables coming out from the generator (they are like pipes). In this power station, the generator voltage is 15 000 V, and this transformer steps the voltage up to 275 000 V.
|
A power station generator generates 500
MW at a voltage of 15 000 V. (a) What is the current? (b) The voltage is stepped up to 275 000 volts. Assuming that the transformer is 100 % efficient, what is the current in the secondary? (c) What is the turns ratio in the transformer? |
Transmission of Electricity
To carry the kinds of currents you worked out in Question 7, you would need very thick wires. Heavy currents make even thick wires get hot. The power lost is worked out using:
P = I2R
| A low voltage transmission line is carrying a current of 30 000 A. Over its whole length, the resistance of the transmission line is 1.5 W. How much power is lost? |
Your answer would be about the whole output of a big power station. That energy would be lost as heat, simply to warm up the countryside. So electrical energy is distributed at very high voltages, and relatively low currents.
| A high voltage transmission line is carrying a current of 1000 A. Over its whole length, the resistance of the transmission line is 1.5 W. How much power is lost? |
You can see that the lost power is much less.
In power stations, alternating current is generated typically at 25 000 V (with a current of 30 000 A) from each alternator (ac generator). The alternators are connected by short and massive cables to a step up transformer immediately outside the building. The voltage is stepped up to 275 kV. Much thinner cables can carry the electricity to where it's needed, using a network of high voltage transmission lines called the National Grid. The transmission lines are carried by transmission towers (pylons) to substations where the voltage is reduced by step-down transformers:
33000 V for local distribution;
25000 V for railways;
11000 V for heavy industry;
415 V for light industry;
230 V for our homes.
While such high voltages are potentially extremely dangerous, the distribution from large power stations is much more efficient that lots of small local power stations, so less fuel needs to be burned overall.
|
Summary Force on a current carrying conductor: F = BIl Force
on a charged particle: F = Bqv Trajectory
of a Charged Particle in a magnetic field: v = Bqr m
Flux
Density Magnetic
field strength,
B. “Concentration
of field lines” Flux Product
of field strength and area.
F
= BA. Total number of field
lines. Flux
Linkage Product
of flux and the number of turns Faraday’s
and Lenz’s Laws
E
=
- NDF
Dt
The minus sign tells us that the emf opposes the change.
Transformer Equation Nprim = Vprim = I sec
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Now try the Topic Quiz |
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