Electromagnetic Effects: Induction, Motors, Generators and Transformers

This lesson explains Electromagnetic Effects: Induction, Motors, Generators and Transformers for Cambridge IGCSE Physics 0625. It separates the Core requirements from the additional Supplement work for Extended candidates. Focus on the cause-and-effect explanation and the exact quantities being compared. Read the explanation once, then attempt the worked method from a blank page.

• electromagnetic induction: changing the magnetic field through a conductor can produce an e.m.f.
• the motor effect: a current-carrying conductor in a magnetic field can experience a force

The topic also includes magnetic fields caused by currents, a.c. generators, d.c. motors, transformers and electrical power transmission. Some requirements are Core and others are Supplement, so use the section labels rather than treating the whole page as one tier.

What magnetic field is produced by a current?

A current in a straight wire produces circular magnetic field lines around the wire. Reversing the current reverses the field direction.

A current in a solenoid produces a magnetic field similar to the field of a bar magnet. The field can be strengthened by:

• increasing the current
• increasing the number of turns per unit length
• placing a soft-iron core inside the coil

This effect is used in electromagnets, relays and other switching or lifting devices.

What is electromagnetic induction?

An e.m.f. is induced when a conductor cuts magnetic field lines or when the magnetic field through a circuit changes.

The induced e.m.f. can be increased by:

• moving the conductor or magnet faster
• using a stronger magnetic field
• increasing the number of turns on a coil

Reversing the relative motion or reversing the magnetic field reverses the induced e.m.f. Induction requires a change. A stationary magnet beside a stationary coil does not continuously induce an e.m.f.

How does an a.c. generator work?

An a.c. generator rotates a coil in a magnetic field. As the sides of the coil cut magnetic field lines, an e.m.f. is induced. The direction of the induced e.m.f. reverses every half-turn, so the output is alternating.

Slip rings and brushes connect the rotating coil to the external circuit. A graph of output voltage against time alternates between positive and negative values.

Increasing the speed of rotation, the field strength or the number of turns increases the output e.m.f.

What is the motor effect?

A current-carrying conductor placed in a magnetic field experiences a force when the current direction is not parallel to the field.

The force is increased by:

• increasing the current
• increasing the magnetic field strength
• increasing the length of conductor inside the field

Reversing either the current or the magnetic field reverses the force. Reversing both leaves the force direction unchanged.

Where the syllabus requires the direction, use the field direction and conventional current direction consistently. Conventional current is from positive to negative outside the power supply.

How does a d.c. motor keep turning?

A current-carrying coil in a magnetic field experiences opposite forces on its two sides, producing a turning effect.

The split-ring commutator reverses the current in the coil every half-turn. This keeps the turning effect in the same rotational direction. Brushes maintain electrical contact with the rotating commutator.

The turning effect can be increased by using a larger current, a stronger magnetic field, more turns or an appropriate coil area.

How does a transformer work?

A transformer changes the size of an alternating voltage.

1. Alternating current in the primary coil produces a changing magnetic field.
2. The soft-iron core carries the changing magnetic field to the secondary coil.
3. The changing field through the secondary coil induces an alternating e.m.f.

A transformer requires a changing magnetic field, so a steady direct current does not produce a continuous secondary e.m.f.

Core: the transformer voltage and turns equation

The voltage-turns relationship is Core content:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

where $V_p$ and $V_s$ are the primary and secondary voltages and $N_p$ and $N_s$ are the numbers of turns.

• A step-up transformer has $V_s > V_p$ and $N_s > N_p$.
• A step-down transformer has $V_s < V_p$ and $N_s < N_p$.

Core worked question

A transformer steps 240 V down to 12 V. The primary coil has 800 turns. Calculate the number of turns on the secondary coil. [3]

$\frac{240}{12} = \frac{800}{N_s}$

$N_s = \frac{800 \times 12}{240} = 40\text{ turns}$

Original marking guidance

• one mark for the transformer equation
• one mark for correct substitution or rearrangement
• one mark for 40 turns

Supplement: transformer power and transmission losses

For an ideal transformer:

$V_pI_p = V_sI_s$

The input and output powers are equal in the ideal model. A real transformer is not perfectly efficient because some energy is dissipated.

For a fixed transmitted power:

$P = IV$

Using a high transmission voltage allows a lower current. The power dissipated in a cable of resistance $R$ is:

$P_{\text{loss}} = I^2R$

A lower current therefore reduces heating losses in transmission cables.

The full chain is:

step up the voltage → reduce the current for the same power → reduce $I^2R$ heating → reduce energy dissipated in the cables

Supplement worked question

An ideal transformer supplies 24 V at 3.0 A from a 240 V primary supply. Calculate the primary current. [3]

$V_pI_p = V_sI_s$

$240I_p = 24 \times 3.0$

$I_p = 0.30\text{ A}$

Common mistakes

• Marking the turns-ratio equation as Supplement. The voltage-turns relationship is Core in the current syllabus.
• Using the ideal-power equation as Core. $V_pI_p = V_sI_s$ belongs to the Supplement.
• Saying a transformer works with steady d.c. A continuously changing magnetic field is required.
• Inverting the transformer ratio. Keep primary quantities on one side and secondary quantities on the other.
• Saying high voltage makes electricity travel faster. The purpose is to reduce current and therefore reduce $I^2R$ heating losses.
• Confusing slip rings with a split-ring commutator. Slip rings are associated with an a.c. generator; a split-ring commutator reverses coil current in a d.c. motor.
• Using electron flow when the question assumes conventional current. Unless stated otherwise, use conventional current direction.

Exam technique

For a device explanation, write a causal sequence rather than a list of component names. For a transformer calculation, identify whether the question needs the Core turns ratio or the Supplement ideal-power equation. For transmission, include both the current link and the $I^2R$ heating link.

How this is examined

Core and Extended papers can test magnetic fields produced by currents, induction, generators, the motor effect, d.c. motors, transformer operation and the voltage-turns equation at the depth specified by the syllabus. Extended papers can additionally require the ideal-transformer power relationship and a quantitative explanation of reduced cable losses. Practical questions may assess induction observations, motor variables or circuit and graph skills in an electromagnetic context.