iGCSE Physics (0625) 4.5.1 Electromagnetic induction Paper 4 -Exam Style Questions- New Syllabus
Question
Each coil position may be used once, more than once or not at all.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.5.2$ — The a.c. generator (Parts $\mathrm{(a)(i)}$, $\mathrm{(b)}$)
• Topic $4.5.1$ — Electromagnetic induction (Parts $\mathrm{(a)(ii)}$, $\mathrm{(a)(iii)}$)
▶️ Answer/Explanation
Correct Answer: To provide a continuous electrical connection between the rotating coil and the external circuit (lamp) without tangling the wires.
Detailed solution: Slip rings rotate with the coil while remaining in constant contact with the fixed carbon brushes. This setup allows the induced current to flow to the stationary external circuit. Crucially, it prevents the connecting wires from twisting and snapping as the coil turns, and ensures that each end of the coil is always connected to the same side of the external circuit, maintaining the alternating nature of the output.
Correct Answer: Rotation of the coil through a magnetic field induces a varying e.m.f. due to the changing magnetic flux linkage.
Detailed solution: As the coil rotates, its sides cut across the magnetic field lines between the $N$ and $S$ poles. This change in magnetic field through the coil induces an electromotive force ($e.m.f.$) according to Faraday’s Law. Because each side of the coil moves up and then down during every $180^{\circ}$ (half turn), the direction of the induced current reverses periodically, creating an alternating current ($a.c.$) in the complete circuit.
Correct Answer: Increase the speed of rotation and increase the strength of the magnetic field.
Detailed solution: To increase the maximum current, one must increase the magnitude of the induced $e.m.f.$. This can be achieved by increasing the rate at which magnetic field lines are cut (spinning the coil faster) or by using stronger magnets. Additionally, increasing the number of turns on the coil provides more conductors to cut the field, thereby increasing the total induced voltage and the resulting current.
Correct Answer: $2.5$ revolutions
Detailed solution: One complete revolution of the coil corresponds to one full cycle of the $a.c.$ wave on the graph (from zero, through a peak and trough, back to zero). From time $0$ to $B$ is $1$ revolution, and from $B$ to $C$ is another $1$ revolution. Point $E$ occurs after another half-cycle. Counting the waves: $1$ (at $B$) $+ 1$ (at $C$) $+ 0.5$ (at $E$) equals $2.5$ full revolutions.
Correct Answer:
Detailed solution: Induced $e.m.f.$ is maximum when the coil is horizontal because the sides are cutting the magnetic field lines at the fastest rate. Thus, $A$ and $C$ (peaks/troughs) match the horizontal coil diagrams. $e.m.f.$ is zero when the coil is vertical because the sides move parallel to the field lines and do not cut them. Therefore, $B$ and $D$ (zero points) match the vertical coil diagrams.