IGCSE Physics 4.6 Electromagnetic Effects MCQs – 30 Practice Questions with Answers (Cambridge 0625)
Free practice questions for Cambridge IGCSE Physics 0625, Section 4.6 Electromagnetic Effects, covering electromagnets and the motor effect, electromagnetic induction and generators, and transformers with power transmission. Questions marked [Supplement] are for students taking the Supplement (extended) papers, and every answer includes a worked explanation.
Formula recap: Transformer equation: Vs/Vp = Ns/Np. For an ideal (100% efficient) transformer: Vp × Ip = Vs × Is. Heating loss in transmission cables: P = I2R. Need the full revision list? See the IGCSE Physics Formula Sheet.
Electromagnets & the Motor Effect
Q1. Which change increases the strength of an electromagnet made from a coil wound on an iron core?
- (A) Decreasing the current in the coil
- (B) Replacing the iron core with a wooden core
- (C) Using fewer turns of wire
- (D) Increasing the current in the coil
Show Answer
Answer: D – The magnetic field around each turn is stronger when the current is larger, so the electromagnet becomes stronger. Fewer turns or a non-magnetic core would weaken it.
Q2. A solenoid carrying a current produces a magnetic field pattern that is most similar to the field of a…
- (A) single straight wire
- (B) horseshoe of wood
- (C) bar magnet
- (D) charged plastic rod
Show Answer
Answer: C – Outside a solenoid the field lines run from one end to the other just as they do for a bar magnet, and each end of the solenoid behaves like a magnetic pole.
Q3. Why is soft iron used as the core of an electromagnet rather than steel?
- (A) Soft iron is a better electrical conductor than steel
- (B) Soft iron is lighter than steel
- (C) Soft iron conducts heat away from the coil
- (D) Soft iron magnetises and demagnetises easily, so the magnetism switches off when the current stops
Show Answer
Answer: D – An electromagnet must lose its magnetism as soon as the current is switched off (for example in a scrapyard crane). Soft iron does this; steel stays magnetised.
Q4. In Fleming’s left-hand rule, the first finger, second finger and thumb point in the directions of…
- (A) Field, Current, Motion (force)
- (B) Current, Field, Motion (force)
- (C) Motion (force), Current, Field
- (D) Field, Motion (force), Current
Show Answer
Answer: A – First finger = magnetic Field, seCond finger = Current, thuMb = Motion (the direction of the force). The rule is used for the motor effect.
Q5. A straight wire carrying a current is placed at right angles to a magnetic field and experiences a force. Which change would NOT increase the size of this force?
- (A) Increasing the current in the wire
- (B) Using a stronger magnet
- (C) Rotating the wire until it is parallel to the field
- (D) Placing a longer length of wire in the field
Show Answer
Answer: C – The force is greatest when the wire is at right angles to the field and falls to zero when the wire is parallel to the field. Larger current, stronger field and longer wire in the field all increase the force.
Q6. What is the job of the split-ring commutator in a simple d.c. motor?
- (A) It reverses the connections to the coil every half turn so the coil keeps rotating in the same direction
- (B) It increases the current supplied to the coil
- (C) It turns the a.c. supply into a d.c. supply
- (D) It strengthens the magnetic field of the magnets
Show Answer
Answer: A – Each half turn, the commutator swaps which side of the coil is connected to which battery terminal. This reverses the current in the coil at exactly the right moment, so the forces keep pushing it round the same way.
Q7. Which set of changes would make a simple d.c. motor spin faster?
- (A) Fewer turns on the coil and a smaller current
- (B) More turns on the coil and a larger current
- (C) A weaker magnet and fewer turns on the coil
- (D) Removing the soft iron core from the coil
Show Answer
Answer: B – A larger force acts on the coil when the current is bigger, the coil has more turns, the field is stronger or the core concentrates the field. More turns and more current therefore give a faster motor.
Q8. How does a loudspeaker produce a sound wave?
- (A) A permanent magnet spins inside a coil and generates a.c.
- (B) The cone is heated by the current and expands
- (C) Sound waves shake a magnet, which induces a current in a coil
- (D) A coil attached to a paper cone sits in a magnetic field; the alternating current in the coil makes it (and the cone) vibrate backwards and forwards
Show Answer
Answer: D – This is the motor effect in action. The alternating current keeps reversing, so the force on the coil keeps reversing and the cone vibrates at the frequency of the signal, pushing the air to make sound. (Option C describes a moving-coil microphone instead.)
Q9. The north pole end of a solenoid is needed at the opposite end of the coil. What should be done?
- (A) Increase the current
- (B) Add an iron core
- (C) Reverse the connections to the power supply so the current flows the other way
- (D) Use thicker wire for the coil
Show Answer
Answer: C – The poles of a solenoid depend on the direction of the current. Reversing the current swaps the north and south ends. Changing the current size, core or wire only changes the strength, not the polarity.
Q10. A current-carrying wire lies along the direction of a magnetic field (parallel to the field lines). The force on the wire is…
- (A) maximum
- (B) zero
- (C) in the direction of the current
- (D) in the direction of the field
Show Answer
Answer: B – The motor-effect force only acts when the current crosses the field. A wire parallel to the field experiences no force; the force is greatest at right angles.
Electromagnetic Induction & Generators
Q11. A bar magnet is held stationary inside a coil connected to a sensitive meter. The meter shows…
- (A) zero, because an EMF is only induced while the magnetic field through the coil is changing
- (B) a reading that slowly increases
- (C) a steady reading
- (D) a reading that depends on the size of the magnet only
Show Answer
Answer: A – Induction needs a change: the magnet must move, the coil must move, or the field must change. A stationary magnet gives a constant field through the coil, so no EMF is induced.
Q12. Which change would NOT increase the EMF induced when a magnet is pushed into a coil?
- (A) Moving the magnet in faster
- (B) Using a coil with more turns
- (C) Using a stronger magnet
- (D) Holding the magnet still once it is inside the coil
Show Answer
Answer: D – The induced EMF depends on how quickly the field through the coil changes. Faster movement, more turns and a stronger magnet all increase it; holding the magnet still gives no change and no EMF.
Q13. A north pole is pushed into a coil and the meter needle kicks to the right. The magnet is then pulled out of the coil at the same speed. The needle…
- (A) kicks to the right by the same amount
- (B) kicks to the left by a similar amount, because the direction of the induced EMF reverses
- (C) stays at zero
- (D) kicks to the right by twice the amount
Show Answer
Answer: B – Reversing the movement reverses the direction of the change, so the induced EMF (and current) flows the opposite way. The size is similar because the speed is the same.
Q14. In a simple a.c. generator, the coil is connected to the external circuit by…
- (A) a split-ring commutator
- (B) an iron core
- (C) wires soldered directly to the coil
- (D) two slip rings and brushes
Show Answer
Answer: D – Slip rings rotate with the coil and keep each end of the coil connected to the same side of the external circuit, so the output alternates. A split-ring commutator would swap the connections and give d.c. instead.
Q15. As the coil of an a.c. generator rotates at a steady rate, the induced EMF is at its maximum when…
- (A) the plane of the coil is at right angles to the magnetic field
- (B) the coil is momentarily not cutting any field lines
- (C) the sides of the coil are cutting the field lines at the fastest rate (the plane of the coil is parallel to the field)
- (D) the coil has completed exactly one full turn
Show Answer
Answer: C – The EMF depends on the rate at which the coil cuts (or sweeps through) the flux. This rate is greatest when the coil’s sides move directly across the field lines, which happens when the plane of the coil is parallel to the field; the EMF is zero a quarter turn later.
Q16. Which change increases BOTH the peak EMF and the frequency of the output of a simple a.c. generator?
- (A) Rotating the coil faster
- (B) Adding more turns to the coil
- (C) Using a stronger magnet
- (D) Adding a soft iron core to the coil
Show Answer
Answer: A – Spinning faster makes the coil cut flux more quickly (bigger peak EMF) and completes more cycles per second (higher frequency). More turns, a stronger magnet or a core raise the peak EMF only, not the frequency.
Q17. [Supplement] A magnet’s north pole approaches the end of a coil. According to Lenz’s law, the induced current makes the near end of the coil behave as a…
- (A) south pole, attracting the magnet
- (B) north pole, repelling the approaching magnet and so opposing the change
- (C) neutral point with no field
- (D) north pole only after the magnet stops
Show Answer
Answer: B – Lenz’s law says the induced effect opposes the change that causes it. The approaching north pole is opposed by the coil presenting a north pole, which is why work must be done to push the magnet in – that work becomes electrical energy.
Q18. A moving-coil microphone works because…
- (A) sound waves vibrate a coil in a magnetic field, inducing an EMF in the coil
- (B) a current in a coil heats a diaphragm
- (C) a magnet is attracted to the sound waves
- (D) the coil acts as a transformer primary
Show Answer
Answer: A – The diaphragm and coil vibrate in the field of a permanent magnet. The movement induces an EMF in the coil that varies with the sound – electromagnetic induction, the reverse of a loudspeaker.
Q19. A wire is moved quickly at right angles through the field between the poles of a magnet. An EMF is induced because…
- (A) the wire becomes magnetised
- (B) the wire cuts the magnetic field lines, so the field through the circuit changes
- (C) friction between the wire and the air generates charge
- (D) the magnet heats the wire
Show Answer
Answer: B – Cutting field lines is the classic way to induce an EMF. The faster the cutting, the larger the EMF; if the circuit is complete, an induced current flows.
Q20. The coil of an a.c. generator is rotated twice as fast as before. The output changes so that…
- (A) only the frequency doubles
- (B) only the peak EMF doubles
- (C) both the frequency and the peak EMF (approximately) double
- (D) the output becomes d.c.
Show Answer
Answer: C – One rotation gives one full cycle of a.c., so doubling the rotation rate doubles the frequency. The flux is also cut twice as quickly, so the peak EMF roughly doubles as well.
Transformers & Power Transmission [mostly Supplement]
Q21. [Supplement] A transformer consists of a primary coil and a secondary coil. The two coils are wound on a core made of…
- (A) copper
- (B) plastic
- (C) solid steel
- (D) laminated soft iron
Show Answer
Answer: D – The iron core links the changing magnetic flux from the primary coil to the secondary coil. It is laminated (built of thin insulated sheets) to reduce energy-wasting eddy currents.
Q22. [Supplement] A transformer has 1200 turns on the primary coil and 60 turns on the secondary coil. The primary is connected to a 240 V a.c. supply. The secondary voltage is…
- (A) 4800 V
- (B) 240 V
- (C) 24 V
- (D) 12 V
Show Answer
Answer: D – Working: Vs/Vp = Ns/Np, so Vs = 240 × (60/1200) = 240 × 0.05 = 12 V. It is a step-down transformer.
Q23. [Supplement] Which transformer is a step-up transformer?
- (A) Np = 500 turns, Ns = 100 turns
- (B) Np = 100 turns, Ns = 500 turns
- (C) Np = 250 turns, Ns = 250 turns
- (D) Np = 1000 turns, Ns = 10 turns
Show Answer
Answer: B – The voltage steps up when the secondary coil has MORE turns than the primary: Vs/Vp = Ns/Np = 500/100 = 5, so the output voltage is five times the input.
Q24. [Supplement] A transformer will not work from a steady d.c. supply because…
- (A) d.c. cannot flow through a coil
- (B) d.c. supplies have too little power
- (C) the iron core repels d.c.
- (D) a steady current produces a constant magnetic field, so the flux through the secondary coil does not change and no EMF is induced
Show Answer
Answer: D – Transformers rely on electromagnetic induction, which needs a continually changing flux. A.c. changes direction all the time, but steady d.c. gives a constant field – an EMF appears in the secondary only at the moments of switch-on and switch-off.
Q25. [Supplement] An ideal transformer has a primary voltage of 240 V and a primary current of 0.50 A. The secondary voltage is 12 V. Assuming no energy losses, the secondary current is…
- (A) 0.025 A
- (B) 0.50 A
- (C) 10 A
- (D) 20 A
Show Answer
Answer: C – Working: input power = Vp × Ip = 240 × 0.50 = 120 W. For an ideal transformer, Vs × Is = 120 W, so Is = 120 / 12 = 10 A. Stepping the voltage down steps the current up.
Q26. [Supplement] Electricity is transmitted across the country at very high voltage (low current) because…
- (A) for the same power, a smaller current causes less heating loss (P = I2R) in the cables
- (B) high voltage travels faster through the cables
- (C) high voltage needs thinner insulation
- (D) generators cannot produce low voltages
Show Answer
Answer: A – Power transmitted = V × I, so raising V lets I be much smaller. The wasted power heating the cables is I2R, so halving the current cuts the loss to a quarter. Step-up transformers at the power station and step-down transformers near homes make this possible.
Q27. [Supplement] The iron core of a transformer is made from thin sheets (laminations) insulated from each other in order to…
- (A) make the core heavier
- (B) increase the eddy currents
- (C) reduce eddy currents in the core and so reduce energy loss as heat
- (D) allow d.c. to be used
Show Answer
Answer: C – The changing flux would induce swirling eddy currents in a solid core, wasting energy as heat. Laminations break up these current paths, improving efficiency. (Real transformers still get warm – no transformer is 100% efficient.)
Q28. [Supplement] A phone charger transformer reduces 230 V a.c. to 5.0 V. If the primary coil has 4600 turns, the number of turns on the secondary coil is…
- (A) 100 turns
- (B) 46 turns
- (C) 1000 turns
- (D) 211 600 turns
Show Answer
Answer: A – Working: Ns = Np × (Vs/Vp) = 4600 × (5.0/230) = 4600 × (1/46) = 100 turns.
Q29. [Supplement] An ideal step-down transformer supplies 120 W to a lamp at 12 V. Which row is correct for this transformer?
- (A) Secondary current = 10 A; primary voltage is higher than 12 V
- (B) Secondary current = 0.10 A; primary voltage is lower than 12 V
- (C) Secondary current = 1440 A; primary voltage is higher than 12 V
- (D) Secondary current = 10 A; primary voltage is lower than 12 V
Show Answer
Answer: A – Working: Is = P/Vs = 120/12 = 10 A. A step-DOWN transformer has a lower secondary voltage, so the primary voltage must be higher than 12 V (and the primary current smaller than 10 A, since power is conserved).
Q30. [Supplement] A transformer has an input power of 200 W and an output power of 180 W. Its efficiency and the main reason it is not 100% are…
- (A) 111%; eddy currents add extra power
- (B) 90%; energy is lost as heat, for example through eddy currents in the core and resistance of the coils
- (C) 90%; the missing energy is stored permanently in the core
- (D) 10%; most of the flux misses the secondary coil
Show Answer
Answer: B – Working: efficiency = (useful output power / input power) × 100% = (180/200) × 100% = 90%. The lost 20 W heats the transformer: eddy currents in the core (reduced by laminating it) and I2R heating in the coil windings.
Keep revising: Work through the rest of the topic pages on the Physics MCQs hub, and keep the key equations beside you with the IGCSE Physics Formula Sheet.
All questions are original practice questions written for QuaidianPoint – not Cambridge past-paper questions. QuaidianPoint is not affiliated with Cambridge Assessment International Education.