# BMAT Physics Notes: Magnetism

Advice & Insight From BMAT Specialists

## â€‹Magnetic Fields

A bar magnetic produces aÂ magnetic fieldÂ Â as shown.
â€‹
If the magnet is suspended by string then it will line up pointingÂ north-south.Â The end that points towards the Earthâ€™s north pole is called the â€˜ north (seeking) poleâ€™ and is labelledÂ N.Â The other end is called the â€˜south (seeking) poleâ€™ and is labelledÂ S.

The magnetic field has aÂ directionÂ Â which is fromÂ north to southÂ as shown.

If two N poles or two S poles are placed near each other they willÂ repel.

If aÂ NÂ pole is placed near aÂ SÂ pole they willÂ attract.

A lump ofÂ ironÂ isÂ easyÂ to magnetise and demagnetise. For this reason it is called aÂ â€˜softâ€™Â magnetic material.

If an unmagnetized iron bar is placed near a strong permanent magnet it may become magnetised. If the permanent magnet is removed then the iron bar will lose its magnetism. This is calledÂ inducedÂ Â magnetism.

A lump ofÂ steelÂ isÂ difficultÂ to magnetise and demagnetise. For this reason it is called aÂ â€˜hardâ€™Â magnetic material.
PermanentÂ magnets are usually made of steel.

## â€‹Magnetic Fields due to current-carrying conductors

(a)Â Â Long straight conductor

The magnetic field consists of:
(i) Concentric circles withÂ increasingÂ separation (because the field strength is decreasing).Â
(ii)Â  TheÂ planeÂ of the field is 90o Â to the wire.
(iii) The direction of the magnetic field is given by theÂ corkscrewÂ rule.

â€‹(b) Long solenoid
(i)Â The fieldÂ insideÂ the coil isÂ uniformÂ (shown byÂ equally spaced parallel lines).Â Â
(ii)Â The field getsÂ weakerÂ near the ends. Â
(iii)Â The strength of the magnetic field can be increased by wrapping the coils around an iron core, and Â
â€‹(iv)Â having more turns of coil.

(c) Flat circular coil

The field at theÂ centreÂ of the coil isÂ uniform.

TheÂ strengthÂ of the magnetic fields produced depends on
(i)Â The magnitude of theÂ currentÂ in the coil,Â
(ii)Â TheÂ distanceÂ from the conductor.

##### Intensive BMAT Course

Comprehensive 5 in 1 Package with a Full Day Intensive BMAT Course, Online BMAT Course Access & Much More…

##### Online BMAT Course

Online BMAT Tutorials, Expert Techniques & BMAT Mock Examinations With Our Popular BMAT Portal

##### Private BMAT Tuition

1-1 BMAT Support To Optimise Your BMAT Score & Convert Your Weakest BMAT Section To Your Strongest.

## Force on a current-carrying conductor

When a current-carrying conductor is placed in aÂ magneticÂ field it will experience aÂ force.
This force exists because the current in the conductor creates its own magnetic field and this interacts with the permanent magnetic field.

TheÂ directionÂ of the force is given byÂ Flemingâ€™s LEFT-HANDÂ rule.

FirstÂ finger ————-Â Field
seCondÂ finger ———–Â Current
thuMbÂ ——————–Â Motion

Examples 1:

â€‹Draw an arrow to show the direction of the force on the wire in the diagrams below.

A

B

Learn time-efficient BMAT strategies and practice with reflective BMAT questions & worked solutions.

## Equation for Force

The magnitude of the force on a conductor carrying a constant current can be show to beÂ directly proportionalÂ to:

(i)Â TheÂ current (I), Â Â Â (ii) TheÂ length of the conductor in the field (L),Â Â Â Â (iii) TheÂ strength of the magnetic field (B).Â

From this we can produce the equation:Â Â Â Â Â Â ForceÂ  F = B I L

This is equation is only valid when the conductor isÂ at 90Â°Â to the field,Â  as shown.

In the above equation:

FÂ is theÂ forceÂ measured inÂ newtonsÂ (N),
BÂ is theÂ magnetic flux densityÂ measured inÂ tesla (T),
IÂ is theÂ currentÂ measured inÂ amps (A),
LÂ is theÂ lengthÂ of the conductor measured inÂ metres (m).

â€‹Worked Example

A conductor of length 50 cm is placed at 90o to a uniform magnetic field of flux density 0.2 T.
If the conductor carries a current of 300 mA calculate theÂ forceÂ acting on the wire.

F = B I L = 0.2 x (0.3) x (0.5) =Â Â 0.03 N

## â€‹d.c. MOTOR

The diagram shows a simplified diagram of a d.c. motor. AÂ d.c.Â current flows from a battery (not shown) into the coil. AÂ forceÂ will then act on sides AB and CD.Â NOÂ force acts on sides BC and AD because they areÂ parallelÂ to the magnetic field.
By applying the Left-hand rule to side AB it will experience aÂ downwardÂ vertical force and side CD anÂ upwardÂ force. Hence the coil will spin in anÂ anticlockwiseÂ direction.

Â The magnitude of the force acting on each side of the coil depends on:
(i) TheÂ current,Â Â Â Â
â€‹
(ii) The strength of theÂ magnetic field (B),
If the coil is wrapped around aÂ soft-iron coreÂ this will also increase the force.

## Electromagnet

A coil is wrapped around an iron core as shown. A d.c. current is passed through the coil. The core becomes magnetized.
When the current is switched off the core will lose its magnetism.
Electromagnets are used in scrapyards to moveÂ  heavy objects such as cars.

## Electromagnetic Induction

Experiment 1
Â

Suppose a conductor is moved up and down in a magnetic field as shown. Experiment shows that aÂ voltageÂ (or emf) is created across the wire. We say a voltage isÂ inducedÂ across the wire.
If the ends of the conductor are joined together to form a closed loop then aÂ currentÂ Â will flow (as shown by the ammeter).

This effect is calledÂ electromagnetic induction.

The magnitude of the induced voltage depends on:

• TheÂ speedÂ of the conductor,
• TheÂ strengthÂ of the magnetic field,
• TheÂ lengthÂ of the conductor (within the field).

In general an induced voltage occurs whenever a conductorÂ cutsÂ through a magnetic field.

TheÂ directionÂ of the inducedÂ currentÂ (if there is a current) is given by Flemingâ€™sÂ right-hand rule.

Experiment 2

If a magnet is moved towards or away from a coil as shown then again aÂ voltageÂ (or emf) isÂ inducedÂ across the coil.
When the magnet movesÂ towardsÂ the coil the voltmeter will deflect in one direction.
When the magnet movesÂ awayÂ from the coil the voltmeter will deflect in theÂ oppositeÂ direction.

The magnitude of the induce voltage depends on:

• TheÂ speedÂ of the magnet,
• TheÂ strengthÂ of the magnet,
• TheÂ number of turnsÂ of the coil.

As the magnet movesÂ towards/awayÂ Â note that the magnetic field passing through the coil isÂ changing.
In general an induced voltage occurs whenever a conductor Â is placed in aÂ changingÂ magnetic field.

## The a.c. Generator (Dynamo)

â€‹The diagram shows a schematic diagram of a generator/dynamo.
It is basically a rectangularÂ coilÂ which isÂ made to rotateÂ by some external means (not shown).
As the coil spins the sides of the coilÂ cutÂ through the magnetic field and so anÂ inducedÂ voltageÂ is produced across the coil. If the ends of the coil are connected toÂ  anÂ externalÂ circuit then aÂ currentÂ will flow.

In one complete revolution of the coil, each side will moveÂ UPÂ andÂ DOWN.
This means the current in each sideÂ changes directionÂ and so anÂ alternating current (a.c.)Â is produced.

MAXIMUM currentÂ  ( and voltage)Â occurs when the coil isÂ horizontalÂ (Diagram A).

ZERO current (and voltage)Â occurs when the coil isÂ verticalÂ (Diagram C).

If the output from the generator is connected to a CRO (oscilloscope) a trace is produced similar to that shown below.

â€‹The maximum voltage can be increased by:

• Increasing theÂ frequencyÂ of rotation,
• Having moreÂ turnsÂ on the coil,
• Having aÂ strongerÂ magnetic field.

## Transformers

A transformer consists of aÂ soft-ironÂ core around which two coils ofÂ insulatedÂ wire are wrapped. One coil, theÂ primary, is connected to a source ofÂ a.c.Â The other coil, theÂ secondaryÂ is connected to a load (e.g. a lamp). When an alternating voltage is applied to the primary coil an alternating current passes through it. This sets up aÂ changing magnetic fluxÂ through the core.Â

This changing magnetic field passes through the secondary coil and hence there is anÂ induced voltageÂ across the coil.Â â€‹If theÂ numberÂ of turns in the secondary isÂ greaterÂ than the number of turns in the primary then theÂ voltageÂ across the secondary coilÂ  isÂ greaterÂ than the voltage across the primary. This is called aÂ step-upÂ transformer.

If theÂ numberÂ of turns in the secondary isÂ lessÂ than the number of turns in the primary then theÂ voltageÂ across the secondary coilÂ  isÂ lessÂ than the voltage across the primary. This is called aÂ step-downÂ transformer.

AnÂ idealÂ transformer is one where there isÂ noÂ energy loss. All the energy supplied by the primary coil is transferred to the secondary. TheÂ powerÂ in the primary coil equals theÂ powerÂ in the secondary coil. It is 100% efficient.
In practice theÂ  core becomes hot and this causes energy to be lost.

##### Transformer Equations
For an ideal transformer it can be shown that:
â€‹Also, for an ideal transformerÂ Â  Power of primary = power of secondary
â€‹[You need to be able to use these equations in calculations].

Example 2

â€‹An ideal transformer has 300 turns on the primary coil and 8100 turns on the secondary. TheÂ  input voltage to the primary is 9.0V. Calculate the voltage across the secondary coil.

Example 3

â€‹A transformer is used to change the mains power supply in order to run a lamp which is rated atÂ  50W, 12V.

â€‹â€‹(i)Â Calculate the number of turns in the primary coil of the transformer. Â

â€‹(ii) Calculate the current in the primary coil.

Example 4

â€‹AnÂ idealÂ transformer has 1500 turns on its primary coil and 750 turns on its secondary coil. The alternating voltage across the primary is 12V and the primary current is 0.18A.Â

â€‹Choose which pair of data correctly shows the values for the secondary coil:

(A)Â Â 24VÂ Â  0.09AÂ  Â  Â  Â  Â Â Â  (B)Â  Â 24VÂ Â  0.36AÂ  Â  Â  Â  Â Â (C)Â  Â 6VÂ Â Â  0.36AÂ  Â  Â  Â  Â (D)Â  Â 6VÂ Â Â  0.09A

Example 5

In the above diagram the primary coil has 400 turns and the secondary coil has 20 turns.
The p.d. across the lamp is 12V and it dissipates 24W.
The transformer is 100% efficient.

Calculate the current in the primary coil.

## â€‹Transmission of electrical energy

When electrical energy is transferred from power stations it has to travel through many kilometres of cable. Although the cable has aÂ low electrical resistanceÂ energy is stillÂ lostÂ in the cables by the heating effect of the current. This would decrease theÂ powerÂ available to the consumer. This would then require extra power stations to be built which would then increase the cost.
Â
Since power lossÂ P = I2 RÂ if the current in the cables is made as small as possible the energy loss (per second) will be reduced (since the resistance of the cable, R, is constant).
Â
Now,Â  Secondary power =Â  VsIs Â so if we make VsÂ largeÂ then Is Â will beÂ smallÂ Â for a given input of power.
Now, if the power station were to produceÂ d.c.Â it would be very difficult to change the current. However ifÂ a.c.Â is used then it is relatively easy to change the current by using aÂ transformer.
Â
For this reason,Â step-upÂ transformers are used at power stations toÂ reduceÂ the secondary current Is (andÂ increaseÂ the secondary voltage Vs) which is then transmitted through the overhead cables.
Â
At various places along the systemÂ step-downÂ transformers are used to reduce the voltage to safer levels for use in factories, houses, etc.

Examples 1â€‹
(a) Arrow vertically upwards.Â Â Â Â Â Â  (b)Â  Arrow INTO plane of paper (90oÂ  to plane).

Examples 2:Â Vs Â = 243V â€‹

Examples 3:Â NpÂ  =Â  5600Â Â Â Â Â Â Â Â Â Â Â Â  Ip = 0.21A Â
â€‹
Example 4:Â  Â (C)Â  Â 6VÂ  Â 0.36A

Example 5:Â Â Â 0.1A

#### BMAT Physics Notes: Magnetism

Intensive BMAT (5 in 1) Course

Private BMAT Tuition

Online BMAT Course

Shopping Cart
Scroll to Top