Free D.4 Induction questions for IB Physics (SL & HL), HL only, part of D Fields. Each question has the answer and the marking points an examiner would look for.
Generate a full D.4 Induction worksheet →All D Fields questionsThe speed of rotation of a simple a.c. generator is multiplied by 2. Which statements are correct?
I. The frequency of the output is multiplied by 2.
II. The peak emf is multiplied by 2.
III. The peak flux linkage is multiplied by 2.
Answer: A
A coil of 100 turns and area 2 × 10−3 m2 is perpendicular to a magnetic field. The field changes uniformly from 0.1 T to 0.05 T in 0.02 s.
What is the magnitude of the average induced emf?
Answer: C
A metal rod of length 0.20 m slides at a constant 3.0 m s−1 along two parallel conducting rails in a uniform vertical magnetic field of 0.60 T. The rails are connected by a resistor of 1 Ω.
Calculate the induced emf.
Calculate the current in the circuit.
Calculate the force needed to keep the rod moving at constant speed.
Show that the mechanical power supplied equals the electrical power dissipated.
Outline how Lenz’s law applies to this system.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.360 V ε = BLv = 0.360 V ✓ | 1 |
| (b) | 0.360 A 0.360 A ✓ | 1 |
| (c) | 0.0432 N F = BIL ✓ = 0.0432 N ✓ | 2 |
| (d) | 0.130 W P = Fv = 0.130 W ✓ I2R = 0.130 W, the same ✓ | 2 |
| (e) | induced current opposes motion induced current flows in a direction to oppose the change producing it ✓ the force on the current-carrying rod opposes its motion, so work must be done ✓ | 2 |
A square loop of side s is made from copper wire of cross-sectional area A. The loop is held in a vertical plane and released from rest with its lower edge horizontal at a height h above the horizontal upper boundary of a region of uniform horizontal magnetic field B, perpendicular to the plane of the loop. The field region is deeper than s. Density of copper = 8960 kg m−3, resistivity of copper = 1.7 × 10−8 Ω m, g = 9.81 m s−2. Ignore air resistance and self-inductance.
For a certain value of h the loop falls at constant speed vT while it enters the field. Show that vT does not depend on s or A, and calculate vT when B = 0.10 T.
Calculate h.
B = 0.10 T, s = 5.0 cm and A = 1.0 mm2. Determine the charge that flows round the loop and the thermal energy generated while the loop enters the field.
The loop is instead released from a height 4h. Determine its acceleration just as it begins to enter the field, and describe its motion until it is completely inside the field and afterwards.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | v_T = 16ρ_mρ_e g/B² = 2.4 m s⁻¹ emf = Bsv, I = Bsv/R, magnetic force on lower edge = BIs = B2s2v/R ✓ constant speed: mg = B2s2vT/R ✓ m = ρmA(4s) and R = ρe(4s)/A ✓ vT = 16ρmρeg s2/(B2s2) = 16ρmρeg/B2 ✓ vT = 16 × 8960 × 1.7 × 10−8 × 9.81/0.010 = 2.4 m s−1 ✓ | 5 |
| (b) | 0.29 m h = vT2/2g = 0.29 m ✓ | 1 |
| (c) | 0.074 C; 8.8 × 10⁻⁴ J R = 1.7 × 10−8 × 0.20/1.0 × 10−6 = 3.4 × 10−3 Ω ✓ charge = (Bsv/R)(s/v) = Bs2/R = 0.074 C (independent of speed) ✓ m = 8960 × 1.0 × 10−6 × 0.20 = 1.79 × 10−3 kg ✓ constant speed: thermal energy = mgs = 8.8 × 10−4 J ✓ | 4 |
| (d) | g upwards (deceleration g) enters at 2vT, so the magnetic force is 2mg ✓ resultant force mg upwards: deceleration of magnitude g ✓ speed decreases towards vT (deceleration falls) while entering; once fully inside, no current, so it accelerates at g ✓ | 3 |
A square coil of 100 turns and side 10 cm is pulled at a constant speed of 1 m s−1 out of a region of uniform magnetic field of 0.3 T, perpendicular to the plane of the coil. The coil has a resistance of 5 Ω.
State Faraday’s law of electromagnetic induction.
Calculate the induced emf and current while one side of the coil is leaving the field.
Use Lenz’s law to explain why a force is needed to pull the coil at constant speed, and calculate this force.
Show that the work done by this force equals the energy dissipated in the coil.
Sketch a graph of the emf against time as the coil leaves the field completely, starting with the coil fully inside.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | emf = rate of change of flux linkage the induced emf equals the rate of change of magnetic flux linkage ✓ | 1 |
| (b) | 3.00 V; 0.600 A ε = NBav ✓ ε = 3.00 V ✓ I = 0.600 A ✓ | 3 |
| (c) | 1.80 N the induced current produces a force opposing the motion ✓ F = NBIa ✓ 1.80 N ✓ | 3 |
| (d) | Fv = I²R power = Fv = 1.80 W ✓ I2R = 1.80 W, equal ✓ | 2 |
| (e) | constant then zero zero emf while fully inside (no change of flux) ✓ constant emf while leaving, then zero once fully outside ✓ | 2 |
A rectangular coil of 200 turns, 5.0 cm × 4.0 cm, rotates at 50 revolutions per second in a uniform field of 0.12 T.
Calculate the maximum flux linkage.
Calculate the peak emf induced.
Explain, using Lenz’s law, why work must be done to rotate the coil when a current flows.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.048 Wb NBA = 200 × 0.12 × 2.0 × 10−3 ✓ 0.048 Wb ✓ | 2 |
| (b) | 15 V ε₀ = NBAω ✓ 15 V ✓ | 2 |
| (c) | induced current opposes motion the induced current produces a force/torque opposing the rotation ✓ so work must be done against it ✓ | 2 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in