Free D.3 Motion in electromagnetic fields questions for IB Physics (SL & HL), part of D Fields. Each question has the answer and the marking points an examiner would look for.
Generate a full D.3 Motion in Electromagnetic Fields worksheet →All D Fields questionsA wire of length 0.3 m carries a current of 3 A in a uniform magnetic field of 0.05 T. The wire makes an angle of 30° with the field.
What is the force on the wire?
Answer: A
Electrons are accelerated from rest through a potential difference of 2000 V and then enter the region between two parallel plates, 4 cm apart and 5 cm long, moving parallel to the plates. The potential difference between the plates is 400 V.
Show that the speed of the electrons entering the plates is about 2.7 × 107 m s−1.
Calculate the electric field strength between the plates and the acceleration of an electron.
Determine the vertical deflection of an electron as it leaves the plates.
Describe the shape of the path between the plates and after it leaves them.
A magnetic field is applied so that the electrons pass through undeflected. Calculate the magnetic flux density and state its direction relative to the electric field.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 2.65 × 107 m s−1 eV = ½mv2 ✓ v = 2.65 × 107 m s−1 ✓ | 2 |
| (b) | 10000 V m−1; 1.76 × 1015 m s−2 E = Vd = 10000 V m−1 ✓ a = eEm = 1.76 × 1015 m s−2 ✓ | 2 |
| (c) | 3.13 mm time between plates = Lv = 1.89 × 10−9 s ✓ y = ½at2 ✓ 3.13 mm ✓ | 3 |
| (d) | parabola then straight line parabolic between the plates (constant force perpendicular to initial velocity) ✓ straight line after leaving the field ✓ | 2 |
| (e) | 3.8 × 10−4 T eE = evB ✓ B = Ev = 3.8 × 10−4 T ✓ perpendicular to both the electric field and the velocity ✓ | 3 |
Ions of charge +e and mass 3.8 × 10−26 kg are accelerated through 2.0 kV and enter a magnetic field of 0.25 T at right angles.
Calculate the speed of the ions.
Calculate the radius of their path.
State why the kinetic energy of the ions does not change in the field.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.3 × 105 m s−1 qV = ½mv2 ✓ 1.30 × 105 m s−1 ✓ | 2 |
| (b) | 0.12 m r = mv/qB ✓ 0.12 m ✓ | 2 |
| (c) | force perpendicular to velocity the magnetic force is perpendicular to the velocity, so no work is done ✓ | 1 |
Deuterons are accelerated from rest through 5000 V and then enter a uniform magnetic field of 0.3 T perpendicular to their velocity.
Calculate the speed of the particles as they enter the field.
Explain why the particles move in a circle.
Calculate the radius of the path.
Show that the time for one revolution does not depend on the speed, and calculate it.
Alpha particles accelerated through the same potential difference enter the same field. Determine the ratio of their radius to that of the deuterons.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 6.92 × 105 m s−1 qV = ½mv2 ✓ v = 6.92 × 105 m s−1 ✓ | 2 |
| (b) | force perpendicular to velocity the magnetic force is always perpendicular to the velocity ✓ so it changes the direction but not the speed, providing the centripetal force ✓ | 2 |
| (c) | 4.82 cm r = mvqB ✓ 4.82 cm ✓ | 2 |
| (d) | 4.38 × 10−7 s T = 2πrv ✓ = 2πmqB, independent of v ✓ 4.38 × 10−7 s ✓ | 3 |
| (e) | 1.00 r = 1B√2mVq ✓ ratio = 1.00 ✓ | 2 |
An electron enters the region between two horizontal plates at 3.0 × 107 m s−1, moving parallel to the plates. The field between the plates is 1.0 × 104 V m−1 and the plates are 6.0 cm long.
Calculate the time the electron spends between the plates.
Calculate the vertical deflection of the electron.
State the shape of the path between the plates.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 2.0 × 10−9 s 0.060/3.0 × 107 = 2.0 × 10−9 s ✓ | 1 |
| (b) | 3.5 mm a = eE/m = 1.76 × 1015 m s−2 ✓ s = ½at2 ✓ = 3.5 × 10−3 m ✓ | 3 |
| (c) | parabola parabolic ✓ | 1 |
An electron travelling at 2.0 × 107 m s−1 enters a uniform magnetic field of 1.5 × 10−3 T at right angles to the field.
Calculate the radius of its path.
Explain why the speed of the electron does not change.
Calculate the time for one complete orbit.
An electric field is added so that the electron passes straight through. Calculate the electric field strength.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 7.6 cm r = mv/qB ✓ = 0.076 m ✓ | 2 |
| (b) | force ⟂ velocity the magnetic force is always perpendicular to the velocity ✓ so no work is done on the electron ✓ | 2 |
| (c) | 2.4 × 10−8 s T = 2πm/qB ✓ = 2.4 × 10−8 s ✓ | 2 |
| (d) | 3.0 × 104 V m−1 qE = qvB ✓ E = vB = 3.0 × 104 V m−1 ✓ | 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