IB DP Physics: Charged particles and conductors in electric and magnetic fields (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover D.3, Motion in electromagnetic fields, from the IB Physics guide, first assessment 2025. The content is the same at SL and HL. See the study guide or the revision notes. Course pages: IB Physics hub and checklist.

Use: e = 1.60 × 10^-19 C, electron mass 9.11 × 10^-31 kg, proton mass 1.67 × 10^-27 kg, atomic mass unit u = 1.66 × 10^-27 kg, μ0 = 4π × 10^-7 T m A^-1, g = 9.81 m s^-2.

Questions

1. A proton moves horizontally northwards through a uniform horizontal magnetic field that points east. State the direction of the magnetic force on the proton, and the direction it would be for an electron on the same path. [3]

2. An alpha particle (charge 3.2 × 10^-19 C, mass 6.64 × 10^-27 kg) moves at 1.8 × 10^6 m s^-1 through a uniform field of 0.50 T.

(a) Calculate the force on it when it moves at right angles to the field. [2] (b) Calculate the force when its velocity is at 40° to the field. [2] (c) For the case in (a), calculate the radius of its circular path. [3]

3. An electron starts from rest at one of two parallel plates 2.0 cm apart, with a potential difference of 400 V between them. Treat the field as uniform.

(a) Calculate the electric field strength. [1] (b) Calculate the acceleration of the electron. [2] (c) Calculate its speed on reaching the other plate. [2]

4. A charged particle moves in a circle in a uniform magnetic field. Explain why the path is circular and why the kinetic energy of the particle does not change. [3]

5. Singly charged ions of mass 20 u and 22 u, both moving at 2.0 × 10^5 m s^-1, enter a uniform field of 0.80 T at right angles and each completes a semicircle.

(a) Calculate the radius for the 20 u ions. [2] (b) Calculate the radius for the 22 u ions. [1] (c) Find the distance between where the two kinds of ion finish their semicircles. [2]

6. Charged particles pass through a region of crossed fields, E = 1.2 × 10^4 V m^-1 and B = 0.060 T, with the electric and magnetic forces opposing.

(a) Calculate the speed of particles that pass straight through. [2] (b) Explain whether a particle of twice the charge and the same speed also passes straight through. [2] (c) A particle enters at 2.5 × 10^5 m s^-1. State, with a reason, whether it is deflected towards the direction of the electric force or the magnetic force. [2]

7. A straight, horizontal wire of mass 4.0 g and length 0.20 m lies at right angles to a horizontal uniform field of 0.080 T. A current in the wire makes the magnetic force on it balance its weight. State the condition for this and calculate the current. [4]

8. Two long straight parallel wires are 3.0 cm apart in air. They carry 15 A and 20 A in opposite directions.

(a) Calculate the force per unit length on each wire. [2] (b) State whether the force is attractive or repulsive. [1] (c) Calculate the force on a 0.80 m length of one wire. [1] (d) The separation is doubled and the 15 A current is doubled. Determine the new force per unit length. [2]

9. Particles pass undeflected through crossed fields E = 2.0 × 10^2 V m^-1 and B = 1.0 × 10^-3 T. The electric field is then switched off, and the particles move on a circle of radius 1.14 mm in the same magnetic field. The field is into the page and the particles move to the right.

(a) Calculate the speed of the particles. [2] (b) Calculate the charge-to-mass ratio. [2] (c) The value for an electron is 1.76 × 10^11 C kg^-1. Comment on your result. [2] (d) The particles curve downwards on the page. Deduce the sign of their charge. [2]

10. Three long parallel wires lie in one plane, 5.0 cm apart in turn. The left and middle wires carry 10 A and the right wire carries 20 A, all in the same direction.

(a) Calculate the force per unit length on the middle wire due to the left wire, and state its direction. [2] (b) Calculate the force per unit length on the middle wire due to the right wire, and state its direction. [2] (c) Find the resultant force per unit length on the middle wire. [2]

11. A proton moves on a circular path of radius 0.15 m at right angles to a uniform field of 1.2 T.

(a) Calculate its speed. [2] (b) Calculate its kinetic energy in joules and in electronvolts. [3] (c) Calculate the time for one revolution. [2]

Answers

1. Right-hand rule: fingers along north, curl towards east [1], thumb points down, so the force on the proton is downward [1]. The electron is negative, so the force is reversed: upward [1]. [3] Examiner insight: Name the rule and the charge sign, so your reasoning can earn credit even if the direction is wrong.

2. (a) F = qvB sin 90° [1] = 3.2 × 10^-19 × 1.8 × 10^6 × 0.50 = 2.9 × 10^-13 N [1]. (b) F = qvB sin 40° [1] = 2.88 × 10^-13 × 0.643 = 1.9 × 10^-13 N [1]. (c) qvB = mv²/r, so r = mv/(qB) [1] = 6.64 × 10^-27 × 1.8 × 10^6 / (3.2 × 10^-19 × 0.50) [1] = 7.5 × 10^-2 m [1]. [7] Examiner insight: In (b), use the angle between velocity and field, and show the sin θ term.

3. (a) E = V/d = 400 / 0.020 = 2.0 × 10^4 V m^-1 [1]. (b) a = eE/m [1] = 1.60 × 10^-19 × 2.0 × 10^4 / 9.11 × 10^-31 = 3.5 × 10^15 m s^-2 [1]. (c) v² = 2ad [1] = 2 × 3.51 × 10^15 × 0.020, so v = 1.2 × 10^7 m s^-1 [1] (1.19 × 10^7). [5] Examiner insight: Carry unrounded values from (b) into (c).

4. The magnetic force is perpendicular to the velocity [1]. It therefore changes the direction but not the speed, so it acts as a centripetal force and the path is a circle [1]. The force does no work, so the kinetic energy is constant [1]. [3] Examiner insight: Link “perpendicular to velocity” to “no work” explicitly.

5. (a) r = mv/(qB) [1] = 20 × 1.66 × 10^-27 × 2.0 × 10^5 / (1.60 × 10^-19 × 0.80) = 5.2 × 10^-2 m [1]. (b) r = 22 × 1.66 × 10^-27 × 2.0 × 10^5 / (1.60 × 10^-19 × 0.80) = 5.7 × 10^-2 m [1]. (c) Each semicircle ends a diameter from the entry point, so separation = 2(r₂ − r₁) [1] = 2 × (5.706 − 5.188) × 10^-2 = 1.0 × 10^-2 m [1]. [5] Examiner insight: Use diameters, not radii.

6. (a) qE = qvB, so v = E/B [1] = 1.2 × 10^4 / 0.060 = 2.0 × 10^5 m s^-1 [1]. (b) Yes [1], because the charge cancels from qE = qvB, so the condition v = E/B does not depend on q [1]. (c) The magnetic force qvB is now larger than the electric force qE because v is greater than E/B [1], so the particle deflects in the direction of the magnetic force [1]. [6] Examiner insight: In (b), show the cancellation; a bare “yes” gives little to reward.

7. The magnetic force must equal the weight, with the magnetic force upward [1]. BIL = mg [1], so I = mg/(BL) = 4.0 × 10^-3 × 9.81 / (0.080 × 0.20) [1] = 2.5 A [1] (2.45 A). [4] Examiner insight: Convert grams to kilograms before substituting.

8. (a) F/L = μ0 I1 I2 / (2πr) [1] = 4π × 10^-7 × 15 × 20 / (2π × 0.030) = 2.0 × 10^-3 N m^-1 [1]. (b) Repulsive, because the currents are in opposite directions [1]. (c) F = 2.0 × 10^-3 × 0.80 = 1.6 × 10^-3 N [1]. (d) The force is proportional to I1 and to 1/r, so the factor is 2 × ½ = 1 [1]. The new value is 2.0 × 10^-3 N m^-1, unchanged [1]. [6] Examiner insight: In (d), proportional reasoning is quicker than recalculating.

9. (a) v = E/B [1] = 2.0 × 10^2 / 1.0 × 10^-3 = 2.0 × 10^5 m s^-1 [1]. (b) q/m = v/(Br) [1] = 2.0 × 10^5 / (1.0 × 10^-3 × 1.14 × 10^-3) = 1.75 × 10^11 C kg^-1 [1]. (c) It agrees with the electron value to within 1% [1], so the particles are probably electrons, though a ratio match alone is not proof [1]. (d) For a positive charge moving right in a field into the page, fingers right and curling into the page give an upward force [1]. The particles curve downward, so the charge is negative [1]. [8] Examiner insight: In (c), compare with the number and qualify the conclusion. In (d), state the rule used.

10. (a) F/L = 4π × 10^-7 × 10 × 10 / (2π × 0.050) [1] = 4.0 × 10^-4 N m^-1, attractive, so towards the left wire [1]. (b) F/L = 4π × 10^-7 × 10 × 20 / (2π × 0.050) [1] = 8.0 × 10^-4 N m^-1, attractive, so towards the right wire [1]. (c) The forces are in opposite directions, so subtract: 8.0 × 10^-4 − 4.0 × 10^-4 [1] = 4.0 × 10^-4 N m^-1 towards the 20 A wire [1]. [6] Examiner insight: Give a direction with a vector result. Both forces are attractive but point opposite ways.

11. (a) qvB = mv²/r, so v = qBr/m [1] = 1.60 × 10^-19 × 1.2 × 0.15 / 1.67 × 10^-27 = 1.7 × 10^7 m s^-1 [1]. (b) Ek = ½mv² [1] = 0.5 × 1.67 × 10^-27 × (1.7246 × 10^7)² = 2.5 × 10^-13 J [1]. Dividing by 1.60 × 10^-19 gives 1.6 × 10^6 eV [1]. (c) T = 2πr/v [1] = 2π × 0.15 / 1.7246 × 10^7 = 5.5 × 10^-8 s [1]. [7] Examiner insight: Convert joules to electronvolts by dividing by e, and keep the unrounded speed.

Where marks are usually lost

• Giving a direction without naming the rule or the sign of the charge.
• Leaving out sin θ when the angle is not 90°.
• Explaining constant kinetic energy by “the field is uniform” instead of “no work is done”.
• Adding the forces from two wires instead of finding the resultant with directions.

Next steps

• The study guide and the revision notes.
• See the IB Physics hub and the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. These questions cover D.3, Motion in electromagnetic fields.