State Faraday’s law of electromagnetic induction - Leaving Cert Physics - Question 9 - 2011

When the magnet is moved towards the coil, the needle on the sensitive meter will deflect, indicating that a current is generated in the coil. The direction of the deflection will be influenced by the direction of the movement of the magnet.

When the magnet is stationary in the coil, there will be no movement of the needle on the meter. This is because there is no change in the magnetic flux when the magnet is not moving.

The first observation is due to the induced emf generated by the change in the magnetic field as the magnet moves, leading to a flow of current in the coil. The second observation occurs because when the magnet is stationary, there is no change in magnetic flux, and thus no induced current in the coil.

Changing the speed of the magnet will affect the rate of change of the magnetic field acting on the coil. If the magnet moves faster, it will result in a greater induced emf and a higher deflection of the needle, indicating more current. Conversely, if the magnet moves slower, there will be less induced emf and a lower deflection of the needle.

A.c. stands for alternating current, which is an electric current that periodically reverses direction, as opposed to direct current (d.c.) that flows in one direction only.

The transformer consists of two coils (primary and secondary) wound around a magnetic core. The primary coil is connected to the input voltage source, while the secondary coil provides the output voltage. The core helps in transferring magnetic flux between the coils.

Using the transformer voltage equation, we have:

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

where:

• $V_p$ is the primary voltage (230 V)
• $N_p$ is the number of turns in the primary coil (200)
• $N_s$ is the number of turns in the secondary coil (600)

Rearranging the equation to solve for $V_s$ gives us:

$V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{600}{200}$

Calculating this yields:

$V_s = 690 \text{ V}$