Concept First Physics and Math Learning Centre In Singapore

1.

Describe an experiment to demonstrate electromagnetic induction. Explain the principles behind the phenomena. What variations in your experiment can you include to demonstrate the factors which affect the magnitude and the direction of the induced emf.

• When a magnet is pushed into the solenoid, there is a change in magnetic field lines linking the solenoid, which produces an induced e.m.f. This induced e.m.f. drives a current in the circuit, causing the pointer of the galvanometer deflects momentarily.

• When the magnet is stationary, there is no change in magnetic field lines linking the circuit. There is no induced e.m.f. and hence no current detected by the galvanometer.

• The experiment show that induced current (or induced emf) is produced in the coil due to the changing magnetic field in the solenoid. This process is called electromagnetic induction.

• The factors affecting the magnitude of the induced emf can be demonstrated as follows:

o   When the magnet moves at a faster speed in or out of the coil, the magnitude of the induced current is increased.

o   When a stronger magnet is used, the magnitude of the induced current is increased.

o   When the number of turns in the coil is increased, the magnitude of the induced current is increased.

2.

With the aid of a diagram, describe the structure of an a.c. generator and explain how it works.

### Structure

• An a.c. generator consists of a rectangular coil of wire connected to a pair of slip rings. Each slip ring is in contact with a carbon brush and the carbon brushes are connected to an external circuit.

• The rectangular coil is placed between the opposite poles of a magnet.

### How it works

• When a coil is rotated, the magnetic field flux through the coil changes and e.m.f is induced in the coil.

• The induced e.m.f generated drives a current through the external circuit.

3.

What are the functions of the slip rings and the carbon brush of an a.c. generator?

• The carbon brushes provide constant, sliding contact with the rotating slip rings.

• This ensures constant electrical contact between the coil and the external circuit. In this way, it allows the transfer of alternating e.m.f. induced in the rotating coil to the external circuit.

4.

With the aid of a diagram, describe the structure of a step-down transformer and explain how the application of an alternating voltage produces a lower output voltage.

### Structure

• The transformer consists of a primary coil and a secondary coil of wire wound round a laminated soft iron core.

• The primary coil is connected to an alternating current source and the secondary coil is connected to an external circuit.

### How it works

• The a.c. supply produces a changing magnetic field in the primary coil.

• The changing magnetic field is linked to the secondary coil via the soft iron core.

• The changing magnetic field will induce an alternating e.m.f in the secondary coil and hence an alternating current will flow.

### In a step-down transformer

• The number of turns in the secondary coil, Ns, is less than the number of turns in the primary coil, Np.

• The voltage across the secondary coil, Vs, is less than the voltage across the primary coil, Vp.

• Formula used in calculation:

5.

What is an ideal transformer?

• In an ideal transformer , all power supplied to the primary coil is transferred to the secondary coil i.e. there is no power lost (100% efficient).

• Pp = Ps à Vp Ip = Vs Is

6.

In the transmission of electricity from power stations to homes, why is it necessary to step up the voltage?

• The voltage is stepped up in the secondary coil so that the secondary current is decreased.

• In this way, the loss of power due to Joule Heating (P=I2R) in the cables can be reduced.

7.

In a step-down transformer, suggest why the wire used for the secondary coil is thicker than that used for the primary coil.

• As the voltage is stepped down in the secondary coil, the secondary current is increased and that caused a significant heating effect in the coil.

• Since the loss of power due to Joule Heating is P=I2R , a thicker wire has a lower resistance and the power lost can be reduced.