Form 6 Physics Practice Paper: Electric Fields, Circuits And Electromagnetic Induction (12 Questions, 100 Marks, 75 Minutes)

Form 6 Physics Practice Paper: Electric Fields, Circuits and Electromagnetic Induction (12 questions, 100 marks, 75 minutes)

2026-11-05

Form 6 Physics – 12 questions – 100 marks – 75 minutes

Form 6 Physics practice paper: 12 questions, 100 marks, 75 minutes on electric field, capacitance, circuits, magnetic force, magnetic flux, induced emf and transformer.

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Paper details　Form 6 Physics practice paper with 12 structured questions on electric fields, capacitance, circuit power, magnetic force and electromagnetic induction.

Section A: Electric Field, Capacitance and Circuits

Answer all parts, show all working and give units in your answers.

#	Question	Answer and marking notes
1	Two parallel metal plates are 2.0 cm apart and connected to a 500 V d.c. supply, giving a uniform electric field between them. (a) Find the electric field strength. (b) Find the force on a proton (charge 1.6 × 10^-19 C) between the plates. (c) Find the acceleration of the proton (mass 1.67 × 10^-27 kg).	(a) E = V / d = 500 / 0.020 = 2.5 × 10^4 V m^-1 (3 marks). (b) F = qE = 1.6 × 10^-19 × 2.5 × 10^4 = 4.0 × 10^-15 N (2 marks). (c) a = F / m = 4.0 × 10^-15 / 1.67 × 10^-27 = 2.4 × 10^12 m s^-2 (3 marks).
2	A 4.0 uF capacitor is charged to 12 V. (a) Find the charge stored. (b) Find the energy stored. (c) The supply is removed and the plate separation is doubled; find the new capacitance, voltage and stored energy.	(a) Q = CV = 4.0 × 10^-6 × 12 = 4.8 × 10^-5 C, i.e. 48 uC (2 marks). (b) U = 0.5 CV^2 = 0.5 × 4.0 × 10^-6 × 144 = 2.9 × 10^-4 J (3 marks). (c) Doubling the separation halves C to 2.0 uF; with the supply removed Q is unchanged, so V = Q/C doubles to 24 V and U = 0.5 QV doubles to 5.8 × 10^-4 J (3 marks).
3	A 12 V battery of negligible internal resistance is connected to a 4 ohm resistor in series with a parallel combination of 6 ohm and 3 ohm. (a) Find the equivalent resistance of the parallel part. (b) Find the total current. (c) Find the current in the 6 ohm and 3 ohm resistors. (d) Find the power dissipated in the 4 ohm resistor.	(a) 1/Rp = 1/6 + 1/3, so Rp = 2 ohm (2 marks). (b) Total resistance = 4 + 2 = 6 ohm, I = 12 / 6 = 2.0 A (2 marks). (c) Voltage across the parallel part = 2 × 2 = 4 V, so I(6 ohm) = 0.67 A and I(3 ohm) = 1.33 A (2 marks). (d) P = I^2 R = 4 × 4 = 16 W (2 marks).
4	An electric kettle is rated 1500 W at 220 V. (a) Find the resistance of its heating element. (b) Find the normal operating current. (c) If the supply falls to 200 V and the resistance stays the same, find the actual power.	(a) R = V^2 / P = 220^2 / 1500 = 32.3 ohm (3 marks). (b) I = P / V = 1500 / 220 = 6.8 A (2 marks). (c) P = V^2 / R = 200^2 / 32.3 = 1.2 × 10^3 W (3 marks).

Section B: Magnetic Force and Charged Particles

Answer all parts and state the formula used.

#	Question	Answer and marking notes
5	A straight wire 0.25 m long carries a current of 3.0 A in a uniform magnetic field of 0.40 T. (a) Find the magnetic force when the wire is perpendicular to the field. (b) Find the force when the wire makes an angle of 30 degrees with the field.	(a) F = BIl = 0.40 × 3.0 × 0.25 = 0.30 N (4 marks). (b) F = BIl sin30 = 0.30 × 0.5 = 0.15 N (4 marks).
6	A proton enters a uniform magnetic field of 0.50 T at 2.0 × 10^6 m s^-1, perpendicular to the field. (a) Find the radius of its circular path. (b) Find its cyclotron period.	(a) r = mv / (qB) = (1.67 × 10^-27 × 2.0 × 10^6) / (1.6 × 10^-19 × 0.50) = 4.2 × 10^-2 m (5 marks). (b) T = 2 pi m / (qB) = 2 pi × 1.67 × 10^-27 / (1.6 × 10^-19 × 0.50) = 1.3 × 10^-7 s (3 marks).
7	A coil of 200 turns has area 0.015 m^2 and resistance 6.0 ohm. A magnetic field perpendicular to the coil increases uniformly from 0.20 T to 0.60 T in 0.10 s. (a) Find the magnetic flux change per turn. (b) Find the induced emf. (c) Find the induced current.	(a) dPhi = A dB = 0.015 × 0.40 = 6.0 × 10^-3 Wb (2 marks). (b) emf = N dPhi / dt = 200 × 6.0 × 10^-3 / 0.10 = 12 V (4 marks). (c) I = emf / R = 12 / 6.0 = 2.0 A (2 marks).
8	An ideal transformer has its primary connected to 240 V a.c. with 1200 turns, and its secondary gives 12 V. (a) Find the number of secondary turns. (b) The secondary drives a load at 2.5 A; find the primary current.	(a) Ns / Np = Vs / Vp, so Ns = 1200 × 12 / 240 = 60 turns (4 marks). (b) Output power = 12 × 2.5 = 30 W; an ideal transformer has equal input power, so Ip = 30 / 240 = 0.125 A (4 marks).

Section C: Electromagnetic Induction, AC and Applications

Answer all parts and give numerical answers to suitable significant figures.

#	Question	Answer and marking notes
9	Two point charges of +2.0 uC each are 0.30 m apart (k = 8.99 × 10^9 N m^2 C^-2). (a) Find the electrostatic force between them and state whether it is attractive or repulsive. (b) Find the resultant electric field at the midpoint.	(a) F = kq1q2 / r^2 = 8.99 × 10^9 × (2.0 × 10^-6)^2 / 0.30^2 = 0.40 N, repulsive because the charges have the same sign (5 marks). (b) At the midpoint the two fields are equal in magnitude and opposite in direction, so the resultant field is zero (4 marks).
10	A 200 uF capacitor is charged to 10 V and then discharged through a 50 kohm resistor. (a) Find the time constant. (b) Find the voltage across the capacitor after 10 s. (c) Find the time for the voltage to fall to 1.0 V.	(a) tau = RC = 200 × 10^-6 × 50 × 10^3 = 10 s (3 marks). (b) V = V0 exp(-t/tau); at t = tau, V = 10 / e = 3.7 V (3 marks). (c) 1.0 = 10 exp(-t/10), so t = 10 ln10 = 23 s (3 marks).
11	A metal rod 0.40 m long moves at 5.0 m/s perpendicular to a uniform magnetic field of 0.25 T; its ends are connected across a 2.0 ohm resistor. (a) Find the induced emf. (b) Find the induced current. (c) Find the external force needed to keep the rod moving uniformly. (d) Show that the mechanical power equals the electrical power.	(a) emf = BLv = 0.25 × 0.40 × 5.0 = 0.50 V (3 marks). (b) I = emf / R = 0.50 / 2.0 = 0.25 A (2 marks). (c) F = BIl = 0.25 × 0.25 × 0.40 = 0.025 N (2 marks). (d) Mechanical power = Fv = 0.125 W; electrical power = I^2 R = 0.0625 × 2.0 = 0.125 W, equal (2 marks).
12	An electron is accelerated through a potential difference of 2000 V (charge 1.6 × 10^-19 C, mass 9.11 × 10^-31 kg). (a) Find the kinetic energy gained. (b) Find its final speed. (c) Explain whether a non-relativistic calculation is reasonable.	(a) KE = eV = 1.6 × 10^-19 × 2000 = 3.2 × 10^-16 J (3 marks). (b) v = sqrt(2KE / m) = sqrt(2 × 3.2 × 10^-16 / 9.11 × 10^-31) = 2.6 × 10^7 m s^-1 (4 marks). (c) v is about 0.09 c, far below the speed of light, so the non-relativistic result is in error by only about 1% and is reasonable (2 marks).

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