CIE iGCSE Co-Ordinated Science P4.2.1 Electrical charge Exam Style Questions Paper 4

(a)

Work done = force × distance = weight × height = \(mgh\).
\( W = 1200 \times 10 \times 15 = \mathbf{180\,000} \textbf{ J} \) (or \(1.8 \times 10^5 \text{ J}\)).
The work done equals the gravitational potential energy gained by the car.

(b)(i)

The magnetic field pattern of a solenoid resembles that of a bar magnet: field lines emerge from one end (north pole), curve around the outside of the solenoid, and re-enter at the other end (south pole).
Inside the solenoid, the field lines run parallel and are evenly spaced, indicating a uniform magnetic field.
The direction of the field (north pole end) is determined by the right-hand rule applied to the direction of current flow in the coil.

(b)(ii)

Using \( Q = It \): \( Q = 50 \times 30 = \mathbf{1500} \).
Unit: C (coulombs).
Charge is the product of current (in amperes) and time (in seconds), giving the total quantity of electric charge that has flowed.

(b)(iii)

First find the voltage: \( V = IR = 50 \times 5.0 = 250 \text{ V} \).
Then calculate power: \( P = IV = 50 \times 250 = \mathbf{12\,500} \textbf{ W} \).
Alternatively, \( P = I^2 R = 50^2 \times 5.0 = 2500 \times 5.0 = 12\,500 \text{ W} \).

(c)

One other advantage: the electromagnet can be switched on and off (by turning the current on or off).
This means the electromagnet can easily release the scrap metal by switching off the current, which a permanent magnet cannot do.
This makes it practical and controllable for lifting and dropping metal objects at a scrapyard.