DSE Physics Practice Paper: Electricity And Magnetism (12 Questions, 50 Marks, 75 Minutes)

DSE Physics Practice Paper: Electricity and Magnetism (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Physics: Electricity and Magnetism

DSE Physics: 12 original questions, 50 marks, 75 minutes on charges, circuits, magnetic force, induction and the photoelectric effect.

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Paper details　Level: S6 | Subject: Physics | Time: 75 minutes | Marks: 50 | Questions: 12

Section A: Short Questions (4 marks each)

Formulae you may use (you must still substitute values yourself)

Coulomb law: F = kQ₁Q₂/r², k = 9.0 × 10⁹ N·m²/C² | Ohm law: V = IR | Power: P = IV = I²R = V²/R | Series: R = R₁ + R₂; parallel: 1/R = 1/R₁ + 1/R₂ | Closed circuit: I = ε/(R + r), terminal voltage V = ε − Ir | Magnetic force: F = BIL (wire), F = qvB (moving charge) | Circular radius: r = mv/(qB) | Transformer: V₁/V₂ = N₁/N₂ and, for an ideal transformer, V₁I₁ = V₂I₂ | Capacitor: Q = CV, E = ½CV² | a.c.: V₀ = √2 V_rms, P = V_rms²/R | Photon energy: E = hc/λ (hc = 1 240 eV·nm) | Photoelectric effect: K_max = hc/λ − φ | Radioactive decay: A = A₀ (1/2)^(t/T½) | Faraday law: ε = NΔΦ/Δt. In every answer, write the formula, the substitution and the unit, and state the direction where it matters.

Answer all questions. Show the formula, substitution and units.

#	Question	Answer and marking notes
1	Two point charges of 2.0 × 10⁻⁶ C and 4.0 × 10⁻⁶ C are 0.30 m apart. Use Coulomb’s law to find the electrostatic force between them, and state whether it is attractive or repulsive. (4 marks)	F = kQ₁Q₂/r² = (9.0 × 10⁹ × 2.0 × 10⁻⁶ × 4.0 × 10⁻⁶)/0.30² = 0.072/0.09 = 0.80 N. Both charges are positive, so the force is repulsive.
2	A 12 Ω heater carries a current of 0.50 A for 5.0 minutes. Find (a) the potential difference across it, (b) its power, (c) the electrical energy used in joules. (4 marks)	V = IR = 0.50 × 12 = 6.0 V; P = I²R = 0.25 × 12 = 3.0 W; E = Pt = 3.0 × 300 = 900 J.
3	Two resistors of 6.0 Ω and 3.0 Ω are connected (a) in parallel, (b) in series. Find the equivalent resistance in each case. (4 marks)	Parallel: 1/R = 1/6.0 + 1/3.0 = 0.50 S, so R = 2.0 Ω. Series: R = 6.0 + 3.0 = 9.0 Ω.
4	A battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to an external resistor of 5.5 Ω. Find (a) the current, (b) the terminal voltage, (c) the power dissipated inside the battery each second. (4 marks)	I = ε/(R + r) = 12/6.0 = 2.0 A; V = ε − Ir = 12 − 1.0 = 11.0 V; P = I²r = 4.0 × 0.50 = 2.0 W.
5	An electron enters a uniform magnetic field of flux density 0.20 T at 3.0 × 10⁶ m/s, perpendicular to the field. Find (a) the magnetic force on the electron, (b) the radius of its circular path. (Electron charge 1.6 × 10⁻¹⁹ C, mass 9.1 × 10⁻³¹ kg) (4 marks)	F = qvB = 1.6 × 10⁻¹⁹ × 3.0 × 10⁶ × 0.20 = 9.6 × 10⁻¹⁴ N; r = mv/(qB) = (9.1 × 10⁻³¹ × 3.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.20) ≈ 8.5 × 10⁻⁵ m.
6	An ideal transformer has its primary connected to 240 V a.c. and delivers 12 V at 2.0 A from the secondary. Find (a) the turns ratio, (b) the output power, (c) the primary current. (4 marks)	Turns ratio 240/12 = 20:1; output power = 12 × 2.0 = 24 W; for an ideal transformer input power equals output power, so I₁ = 24/240 = 0.10 A.

Section B: Application Questions (4 marks each)

Answer all questions. Give units and directions where applicable.

#	Question	Answer and marking notes
7	A straight wire 0.40 m long carries a current of 2.5 A and is placed perpendicular to a magnetic field of flux density 0.60 T. Find the force on the wire and state how the force changes if the current is reversed. (4 marks)	F = BIL = 0.60 × 2.5 × 0.40 = 0.60 N; reversing the current reverses the direction of the force but its magnitude stays 0.60 N.
8	A capacitor of capacitance 4.7 μF is connected to a 12 V d.c. supply. Find (a) the charge stored, (b) the energy stored. (4 marks)	Q = CV = 4.7 × 10⁻⁶ × 12 ≈ 5.6 × 10⁻⁵ C; E = ½CV² = 0.5 × 4.7 × 10⁻⁶ × 144 ≈ 3.4 × 10⁻⁴ J.
9	The mains supply has an r.m.s. voltage of 220 V at 50 Hz and is connected to a 44 Ω heater. Find (a) the r.m.s. current, (b) the average power, (c) the peak voltage. (4 marks)	I = V/R = 220/44 = 5.0 A; P = V²/R = 48 400/44 = 1 100 W; V₀ = 220√2 ≈ 311 V.

Section C: Long Questions (14 marks in total)

Show full working; write explanations in complete sentences.

#	Question	Answer and marking notes
10	Monochromatic light of wavelength 400 nm falls on a metal surface of work function 2.0 eV. Find (a) the photon energy in eV, (b) the maximum kinetic energy of the photoelectrons, (c) the threshold wavelength of the metal. (4 marks)	E = hc/λ = 1 240/400 = 3.1 eV; maximum kinetic energy = 3.1 − 2.0 = 1.1 eV; threshold wavelength = 1 240/2.0 = 620 nm.
11	A radioactive isotope has a half-life of 6.0 hours. A sample has an initial activity of 800 Bq. Find (a) the activity after 24 hours, (b) the percentage of undecayed nuclei remaining at that time. (4 marks)	24 hours is 4 half-lives; activity = 800/2⁴ = 50 Bq; fraction remaining = (1/2)⁴ = 6.25 %.
12	A coil of 200 turns, each of area 0.015 m², lies in a uniform magnetic field perpendicular to its plane. The flux density falls uniformly from 0.80 T to zero in 0.20 s. Find (a) the flux change per turn, (b) the average induced e.m.f. in the coil; (c) use Lenz’s law to explain the direction of the induced current and relate it to conservation of energy. (6 marks)	ΔΦ = BΔA = 0.80 × 0.015 = 0.012 Wb; average e.m.f. = NΔΦ/Δt = 200 × 0.012/0.20 = 12 V. By Lenz’s law the induced current creates a field that opposes the decrease in flux, so the induced field is in the same direction as the original field; the right-hand rule then gives the current direction. Work must be done against the opposing effect, which is why energy is conserved rather than created.

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