The big idea: The northern lights glow because charged particles streaming from the Sun get bent into spirals by Earth's magnetic field.
The force on a moving charge
- Any moving charge in a magnetic field feels a sideways force, F = qvB (when it moves at right angles to the field).
- This force is always perpendicular to the velocity — it pushes the charge sideways, never forwards or backwards.
- A constant sideways push bends the path into a circle.
Magnetic force on a moving charge
- Size: F = qvB (when v is perpendicular to B)
- Direction: perpendicular to the velocity v — always sideways
- A sideways push can't speed it up or slow it down, so it bends the path into a circle of radius r = mv/(qB)
Electric force on a charge (for contrast)
- Size: F = qE (in a field E)
- Direction: along the field — parallel to E
- A push along the motion changes the speed (it can speed up or slow the charge)
Spot it: Stationary charge → no magnetic force (you need v).
The force is sideways, so it can't change the speed — only the direction, curving the charge into a circle.
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In a magnetic field alone, the force qvB is always at right angles to the velocity. A force like that is a centripetal force: the charge moves in a circle.
- radius of the circular path (m)
- mass of the particle (kg)
- speed (m s⁻¹)
- charge (C)
- magnetic field strength (T)
Reading the circles
What the radius tells you: Rearranged, q/m = v/(Br): measure the radius at a known speed and field and you have the charge-to-mass ratio.
For ions of the same charge and speed, r ∝ m: heavier isotopes curve less, which is how isotopes are told apart.
At an angle: The general force is F = qvB sin θ, θ between v and B. Moving along the field (θ = 0): no force, no curving.
Worked example — radius of an electron's path
An electron (mass 9.11 × 10⁻³¹ kg) moves at 3.0 × 10⁶ m s⁻¹ at right angles to a 2.0 mT field. Find the radius of its path.
Solution
- Use the radius from qvB = mv²/r:
- Work it out:
Final answer
r = 8.5 mm. The electron keeps its speed; only its direction changes.
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The data booklet gives the magnetic force on a moving charge. For a charge moving at right angles to the field (the case the exam tests):
- magnetic force on the charge (N)
- size of the moving charge (C)
- speed of the charge (m s⁻¹)
- magnetic field strength (T, tesla)
What a velocity selector is: A velocity selector sends charges through crossed fields — an electric field E and a magnetic field B at right angles to each other.
The electric force qE pushes one way; the magnetic force qvB pushes the opposite way. Only charges of one exact speed feel zero net force and pass straight through.
A velocity selector has two charged parallel plates making a UNIFORM electric field E (evenly spaced arrows, from the + plate to the − plate). A magnetic field B is also applied across the same gap, at right angles to both E and the charge's motion.
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Inside the selector the electric force qE and the magnetic force qvB act in OPPOSITE directions on the moving charge. When qE = qvB they cancel, the net force is zero, and the charge travels in a straight line — undeflected.
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For the charge to go straight (undeflected) the two forces must be equal. Set them equal and the charge q cancels:
- selected speed — the speed that passes straight through (m s⁻¹)
- electric field strength between the plates (N C⁻¹ or V m⁻¹)
- magnetic field strength (T)
From qE = qvB the q cancels, leaving E = vB. Cover the one you want: v and B side by side → multiply (E = vB); E above B → divide (v = E ÷ B).
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A velocity selector has an electric field of E = 3.0 × 10⁴ N C⁻¹ and a magnetic field of B = 0.20 T, crossed at right angles. Find the speed of a charge that passes straight through undeflected.
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How this is tested — the velocity selector is a classic Paper 2 'crossed fields' question:
Paper 2 — the balance
- A particle passes undeflected through crossed fields.
- Find B (or E, or the speed v) from the balance qE = qvB.
Paper 2 — a different particle
- A different charge then enters the same fields.
- Draw or describe its path — it curves, because the magnetic force qvB changes.
The classic trap: Don't think the selected speed depends on the charge or mass — it doesn't. From qE = qvB the charge cancels, so v = E ÷ B only.
A different particle deflects: Set B by balancing one particle. A second particle of a different charge or mass moving at a different speed no longer satisfies qE = qvB, so one force wins and it curves.
Slower than v = E/B
- Magnetic force qvB is smaller (v is smaller)
- Electric force qE now wins
- The charge is deflected toward the − plate (the way qE points)
Faster than v = E/B
- Magnetic force qvB is larger (v is larger)
- Magnetic force now wins
- The charge is deflected the other way (the way qvB points)
An electron enters a velocity selector at 4.0 × 10⁶ m s⁻¹, where the electric field between the plates is E = 8.0 × 10⁵ N C⁻¹. (a) Find the magnetic field strength B that lets the electron pass through undeflected. (b) A second electron enters the same crossed fields moving faster than 4.0 × 10⁶ m s⁻¹. State whether it passes straight through, and which force wins.
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