Advice & Insight From BMAT Specialists
A bar magnetic produces a magnetic field  as shown.
​
If the magnet is suspended by string then it will line up pointing north-south. The end that points towards the Earth’s north pole is called the ‘ north (seeking) pole’ and is labelled N. The other end is called the ‘south (seeking) pole’ and is labelled S.
The magnetic field has a direction  which is from north to south as shown.
If two N poles or two S poles are placed near each other they will repel.
If a N pole is placed near a S pole they will attract.
A lump of iron is easy to magnetise and demagnetise. For this reason it is called a ‘soft’ 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 called induced  magnetism.
A lump of steel is difficult to magnetise and demagnetise. For this reason it is called a ‘hard’ magnetic material.
Permanent magnets are usually made of steel.
(a)Â Â Long straight conductor
The magnetic field consists of:
(i) Concentric circles with increasing separation (because the field strength is decreasing).Â
(ii) The plane of the field is 90o  to the wire.
(iii) The direction of the magnetic field is given by the corkscrew rule.
​(b) Long solenoid
(i) The field inside the coil is uniform (shown by equally spaced parallel lines). Â
(ii) The field gets weaker 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
​(iv) having more turns of coil.
(c) Flat circular coil
The field at the centre of the coil is uniform.
The strength of the magnetic fields produced depends on
(i) The magnitude of the current in the coil,Â
(ii) The distance from the conductor.
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When a current-carrying conductor is placed in a magnetic field it will experience a force.
This force exists because the current in the conductor creates its own magnetic field and this interacts with the permanent magnetic field.
The direction of the force is given by Fleming’s LEFT-HAND rule.
First finger ————- Field
seCond finger ———– Current
thuMb ——————– Motion
Examples 1:
​Draw an arrow to show the direction of the force on the wire in the diagrams below.
A
B
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The magnitude of the force on a conductor carrying a constant current can be show to be directly proportional to:
(i) The current (I),    (ii) The length of the conductor in the field (L),    (iii) The strength of the magnetic field (B).Â
From this we can produce the equation:      Force F = B I L
This is equation is only valid when the conductor is at 90° to the field, as shown.
In the above equation:
F is the force measured in newtons (N),
B is the magnetic flux density measured in tesla (T),
I is the current measured in amps (A),
L is the length of the conductor measured in metres (m).
​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 the force acting on the wire.
F = B I L = 0.2 x (0.3) x (0.5) =Â Â 0.03 N
The diagram shows a simplified diagram of a d.c. motor. A d.c. current flows from a battery (not shown) into the coil. A force will then act on sides AB and CD. NO force acts on sides BC and AD because they are parallel to the magnetic field.
By applying the Left-hand rule to side AB it will experience a downward vertical force and side CD an upward force. Hence the coil will spin in an anticlockwise direction.
 The magnitude of the force acting on each side of the coil depends on:
(i) The current,   Â
​(ii) The strength of the magnetic field (B),
If the coil is wrapped around a soft-iron core this will also increase the force.
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 move heavy objects such as cars.
Experiment 1
Â
Suppose a conductor is moved up and down in a magnetic field as shown. Experiment shows that a voltage (or emf) is created across the wire. We say a voltage is induced across the wire.
If the ends of the conductor are joined together to form a closed loop then a current  will flow (as shown by the ammeter).
This effect is called electromagnetic induction.
The magnitude of the induced voltage depends on:
In general an induced voltage occurs whenever a conductor cuts through a magnetic field.
The direction of the induced current (if there is a current) is given by Fleming’s right-hand rule.
Experiment 2
If a magnet is moved towards or away from a coil as shown then again a voltage (or emf) is induced across the coil.
When the magnet moves towards the coil the voltmeter will deflect in one direction.
When the magnet moves away from the coil the voltmeter will deflect in the opposite direction.
The magnitude of the induce voltage depends on:
As the magnet moves towards/away  note that the magnetic field passing through the coil is changing.
In general an induced voltage occurs whenever a conductor  is placed in a changing magnetic field.
​The diagram shows a schematic diagram of a generator/dynamo.
It is basically a rectangular coil which is made to rotate by some external means (not shown).
As the coil spins the sides of the coil cut through the magnetic field and so an induced voltage is produced across the coil. If the ends of the coil are connected to an external circuit then a current will flow.
In one complete revolution of the coil, each side will move UP and DOWN.
This means the current in each side changes direction and so an alternating current (a.c.) is produced.
MAXIMUM current ( and voltage) occurs when the coil is horizontal (Diagram A).
ZERO current (and voltage) occurs when the coil is vertical (Diagram C).
If the output from the generator is connected to a CRO (oscilloscope) a trace is produced similar to that shown below.
​The maximum voltage can be increased by:
A transformer consists of a soft-iron core around which two coils of insulated wire are wrapped. One coil, the primary, is connected to a source of a.c. The other coil, the secondary 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 a changing magnetic flux through the core.Â
This changing magnetic field passes through the secondary coil and hence there is an induced voltage across the coil. ​If the number of turns in the secondary is greater than the number of turns in the primary then the voltage across the secondary coil is greater than the voltage across the primary. This is called a step-up transformer.
If the number of turns in the secondary is less than the number of turns in the primary then the voltage across the secondary coil is less than the voltage across the primary. This is called a step-down transformer.
An ideal transformer is one where there is no energy loss. All the energy supplied by the primary coil is transferred to the secondary. The power in the primary coil equals the power in the secondary coil. It is 100% efficient.
In practice the core becomes hot and this causes energy to be lost.
Example 2
​An ideal transformer has 300 turns on the primary coil and 8100 turns on the secondary. The input voltage to the primary is 9.0V. Calculate the voltage across the secondary coil.
Example 3
​A transformer is used to change the mains power supply in order to run a lamp which is rated at 50W, 12V.
​​(i) Calculate the number of turns in the primary coil of the transformer. Â
​(ii) Calculate the current in the primary coil.
Example 4
​An ideal 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
​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.
Examples 1​
(a) Arrow vertically upwards.      (b) Arrow INTO plane of paper (90o to plane).
Examples 2: Vs  = 243V ​
Examples 3: Np = 5600            Ip = 0.21A Â
​
Example 4:Â Â (C)Â Â 6VÂ Â 0.36A
Example 5:Â Â Â 0.1A
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