Pick your class, read the steps, then try them right in the lab —
change one slider at a time and watch the rays and graphs respond.
Class 12 — The magnetic force on a moving charge
Concept: a magnetic field pushes a moving charge sideways with a force F = qvB sinθ, always at right angles to
the velocity. It does no work, so the speed stays the same and the charge goes round a circle of radius r = mv ÷ qB.
- Open Charges in B & E fields with a proton, v = 2 × 106 m/s, B = 50 mT and the angle at 90°.
The radius is r = mv ÷ qB = 417.6 mm, the period T = 2πm ÷ qB = 1.31 µs and the frequency 762 kHz.
- Double the speed to 4 × 106 m/s: the radius doubles to 835.3 mm, but the period is still 1.31 µs.
A faster particle goes round a bigger circle in the same time.
- Double the field to 100 mT instead: the radius halves to 208.8 mm and the period falls to 656 ns.
- Choose an electron: r = 227.4 µm, 1836 times smaller than the proton's, and it goes round the other way because its
charge is negative. Its period is only 0.71 ns.
Takeaway: r = mv ÷ qB = p ÷ qB and T = 2πm ÷ qB: the period does not depend on the speed.
Class 12 — Helical motion
Concept: only the part of the velocity at right angles to B is turned. The part along B is unchanged, so a charge that
enters at an angle moves in a helix.
- Keep the proton and set the angle between v and B to 40°. The part across the field is v sin 40° = 1.286 × 106 m/s,
so the radius shrinks to 268.5 mm (r = m v sinθ ÷ qB).
- The part along the field is v cos 40° = 1.532 × 106 m/s and the period is still 1.31 µs, so the pitch is
v cosθ × T = 2.01 m.
- At 90° the pitch is zero (a circle); as the angle falls towards 30° the helix stretches out along the field.
Takeaway: a helix = circular motion across the field + steady drift along it. This is how charged particles spiral
along the Earth's field lines towards the poles.
Class 12 — The velocity selector
Concept: crossed electric and magnetic fields push a charge in opposite directions. The forces cancel at one speed
only: qE = qvB, so v = E ÷ B.
- Choose Crossed E and B with E = 50 kV/m and B = 50 mT. v = 50 000 ÷ 0.05 = 1 × 106 m/s, and at this speed
both forces are 8.01 × 10−15 N. The particle goes straight through the slit.
- Raise the speed to 1.4 × 106 m/s: the magnetic force wins and the particle hits a plate. At 0.7 × 106 m/s the electric
force wins and it is deflected the other way.
- Speeds within about 2% of 106 m/s (0.98 or 1.02) still pass the slit; the right-hand graph shows how narrow the range is.
- Switch to an electron: both forces reverse and the electron passes at the same speed, because neither the mass nor the sign of
the charge appears in v = E ÷ B.
- Double E to 100 kV/m: the selected speed doubles to 2 × 106 m/s.
Takeaway: a velocity selector passes particles of one speed, v = E ÷ B, whatever their mass or charge.
Class 12 — The mass spectrometer
Concept: ions that leave a velocity selector all have the same speed, so in a second field B2 they follow semicircles
with r = mv ÷ qB2. Heavier ions follow bigger circles and land further from the slit.
- Open The mass spectrometer with neon. The selector gives v = E ÷ B1 = 1 × 106 m/s. With B2 = 0.5 T the
²⁰Ne+ ions land 828.9 mm from the slit and the ²²Ne+ ions at 911.8 mm: the lines are 82.9 mm apart.
- Double B2 to 1 T: both distances halve (to 414.4 mm and 455.9 mm), so the lines move closer together.
A weak B2 spreads them out, which makes similar masses easier to tell apart.
- Choose the carbon isotopes: ¹²C, ¹³C and ¹⁴C land at 497.5 mm, 539.1 mm and 580.6 mm. The distance is
proportional to the mass (2r = 2mv ÷ qB2), so equal mass differences give equal gaps (about 41.5 mm).
Takeaway: 2r ∝ m: the position of a line on the plate measures the mass of the ion.
Class 12 — The cyclotron
Concept: inside a dee a particle goes round a semicircle in a time that does not depend on its speed, so an alternating
voltage of frequency f = qB ÷ 2πm can push it forward every time it crosses the gap.
- Open The cyclotron with a proton, B = 1 T and V = 300 kV. The resonance frequency is f = qB ÷ 2πm = 15.24 MHz. The
dees have a radius of 0.5 m, so the final energy is KE = q2B2R2 ÷ 2m = 11.97 MeV.
- Each crossing gives qV = 300 keV, so the proton crosses the gap 39 times (19.5 revolutions) in about 1.28 µs. Watch the circles grow
as √n.
- Double the voltage to 600 kV: the energy is still 11.97 MeV but it needs only 19 crossings. The voltage sets how fast the particle
gains energy, not how much it ends up with.
- Double the field to 2 T: the frequency doubles to 30.49 MHz and the energy rises fourfold to 47.9 MeV (KE ∝ B2).
- Choose an alpha particle at 1 T: f = 7.67 MHz (half the proton's, because q ÷ m is halved) and KEmax = 12.05 MeV.
Takeaway: f = qB ÷ 2πm and KEmax = q2B2R2 ÷ 2m, independent of V.
Class 10 — The magnetic field of a current
Concept: a wire carrying a current makes a magnetic field round it. The field lines are circles, and their direction is
given by the right-hand thumb rule: thumb along the current, fingers curl along the field.
- Open Fields of currents with a long straight wire carrying 5 A, and look at a point 4 cm away. The field is
B = μ0I ÷ 2πr = 25 µT, about half of the Earth's field.
- Move to 8 cm: the field halves to 12.5 µT. Now keep 8 cm and double the current to 10 A: it goes back to 25 µT.
- The lines get further apart as you go away from the wire: the field is weaker there.
- Switch to the solenoid: a coil of many turns makes a strong, straight field inside and a weak one outside, like a bar magnet.
With 800 turns per metre and 5 A the field inside is 5 mT, about 100 times the Earth's.
Takeaway: the field of a straight wire is proportional to the current and falls off as 1 ÷ r; a solenoid is a strong electromagnet.
Class 12 — The Biot–Savart law and a current loop
Concept: the Biot–Savart law adds up the field of every small piece of a current. For a circular loop it gives the field on
the axis: B = μ0NIR2 ÷ 2(R2 + x2)3/2.
- Choose A circular current loop with R = 5 cm, I = 5 A and N = 1. At the centre B = μ0NI ÷ 2R = 62.83 µT. The
explanation checks this by adding up 48 pieces of the loop numerically.
- Make N = 10 turns: B = 628.3 µT, ten times as much. Double the radius to 10 cm (N = 1): B = 31.4 µT, half as much.
- Move the probe along the axis to x = 5 cm (one radius): B = 22.2 µT, only 35% of the central value. The graph is
strongest at the centre and falls on both sides.
- The field lines look like those of a bar magnet: the loop is a magnetic dipole with moment m = NIA = 0.0393 A·m2.
Takeaway: B = μ0NI ÷ 2R at the centre of a loop, and the loop acts like a small magnet.
Class 12 — Ampère's law and the solenoid
Concept: Ampère's circuital law, ∮B·dl = μ0I, gives the field inside a long solenoid: B = μ0nI.
- Choose A solenoid: n = 800 turns per metre, I = 5 A, length 20 cm. The long-solenoid value is μ0nI = 5.03 mT.
- The solenoid is only 20 cm long and 4 cm across, so at the centre the field is slightly smaller: 4.93 mT (98.1%). At the end of the
axis (x = 10 cm) it is 2.5 mT, about half.
- Double the turns per metre to 1600: the field inside doubles (9.86 mT here). Change the radius: the field does not depend on it.
- Look at the field lines: straight and parallel inside, and spreading out and returning round the outside, like a bar magnet.
Takeaway: B = μ0nI inside a long solenoid; it depends on n and I only. An iron core multiplies it by the relative permeability.
Class 10 — The force on a current and the motor effect
Concept: a wire carrying a current in a magnetic field feels a force at right angles to both. Fleming's left-hand rule gives
its direction: first finger = field, second finger = current, thumb = force.
- Open Forces on currents with the wire: B = 0.5 T, I = 4 A, L = 20 cm, at 90° to the field. F = BIL sin 90° = 0.5 × 4 × 0.2 = 0.4 N,
pointing into the page.
- Turn the wire to 30°: F = 0.4 × sin 30° = 0.2 N. Turn it to 0°, parallel to the field: there is no force at all.
- Switch to the coil with the commutator on. The two long sides are pushed in opposite directions, which turns the coil, and the
commutator reverses the current every half turn so it keeps turning: this is the electric motor.
Takeaway: F = BIL sinθ: greatest at right angles to the field, zero along it. A motor turns this force into rotation.
Class 12 — Torque on a current loop and the galvanometer
Concept: a coil of N turns and area A with current I in a field B has a magnetic moment m = NIA, and feels a torque
τ = mB sinθ, where θ is the angle between m and B.
- Choose A coil in a field: N = 20, 6 cm × 4 cm (A = 24 cm2), I = 2 A, B = 0.4 T. Then m = NIA = 0.096 A·m2 and the greatest torque is
mB = 38.4 mN·m, when the plane of the coil is parallel to the field.
- Press Play without the commutator: the coil swings about the position where its plane is perpendicular to the field and slowly
settles there, where the torque is zero. For small swings T = 2π√(J ÷ mB) = 453 ms.
- Tick the commutator: the torque never changes sign and the coil spins faster and faster until the drag balances it. This is a DC motor.
- In a moving-coil galvanometer a spring balances this torque, so the deflection is proportional to the current: NIAB = kφ.
Takeaway: τ = NIAB sinθ = m × B: a loop turns until its magnetic moment lines up with the field.
Class 12 — The force between parallel wires and the ampere
Concept: each of two parallel wires sits in the field of the other, so they push or pull each other. Currents in the same
direction attract; opposite currents repel.
- Choose Two parallel wires with I1 = I2 = 5 A, d = 4 cm. The field of wire 1 at wire 2 is 25 µT, so the force per metre is
F ÷ L = μ0I1I2 ÷ 2πd = 125 µN/m, and the wires attract.
- Double the distance to 8 cm: the force halves to 62.5 µN/m.
- Set I2 to −5 A (the opposite way): the force is still 125 µN/m but now the wires repel.
- With I1 = I2 = 1 A and d = 1 m the force would be 2 × 10−7 N/m: this is the definition of the ampere.
Takeaway: F ÷ L = μ0I1I2 ÷ 2πd, and the two forces are equal and opposite.