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Questions on charged particles in magnetic fields usually assume the field fills all of space, so the particle circles forever. Exam setters prefer a field that fills only part of space: a half-space beyond a boundary, or a strip of fixed width. Now you must find where the particle leaves, in which direction and after how long. This post gives one geometric method for both cases, with worked numbers for a proton.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "force on a moving charge in uniform magnetic and electric fields", and the JEE (Advanced) 2026 brochure lists "force on a moving charge and on a current-carrying wire in a uniform magnetic field". The NCERT Class 12 chapter Moving Charges and Magnetism (2026-27 reprint) covers the motion in Section 4.3. Bounded field regions are not named in the syllabus; they apply the same circle. Check the current year's documents.
What two facts do you need?
NCERT's Section 4.3 gives both. First, the magnetic force is perpendicular to the velocity, so it does no work and the speed never changes. Second, for velocity perpendicular to B, the path is a circle with
radius r = mv/(qB)
angular frequency ω = qB/m, independent of speed
So the time spent inside the field is simply the angle turned divided by ω: t = θ/ω = θm/(qB). Every question in this post is a geometry question about that angle θ.
For the numbers below, take a proton with m = 1.673 × 10⁻²⁷ kg and q = 1.602 × 10⁻¹⁹ C (CODATA values from NIST), speed 2.0 × 10⁶ m/s and B = 0.20 T, perpendicular to the plane of motion.
r = (1.673 × 10⁻²⁷ × 2.0 × 10⁶)/(1.602 × 10⁻¹⁹ × 0.20) ≈ 0.104 m
ω = qB/m ≈ 1.92 × 10⁷ rad/s
full period T = 2π/ω ≈ 3.28 × 10⁻⁷ s
Case 1: entering a half-space at right angles
The field fills everything beyond a straight boundary. The proton crosses the boundary at right angles. Its circle's centre lies on the boundary itself, so it completes a semicircle and leaves travelling straight back.
Exit point: one diameter, 2r ≈ 0.209 m, along the boundary from the entry point.
Time inside: half a period, about 1.64 × 10⁻⁷ s.
Case 2: entering a half-space at an angle
Now the proton crosses the boundary with its velocity at α = 30° to the boundary line. The chord joining entry and exit points lies along the boundary, and a circle meets a straight line symmetrically. So:
It always leaves at the same angle to the boundary that it entered with.
The chord length is 2r sin α = 2 × 0.104 × 0.5 ≈ 0.104 m.
Which arc does it travel? That depends on which way the force turns it. If the force bends it back towards the boundary straight away, it sweeps a short arc of 2α = 60°, taking T/6 ≈ 5.5 × 10⁻⁸ s. If the force first bends it away from the boundary, it sweeps the long arc, 360° − 60° = 300°, taking 5T/6 ≈ 2.7 × 10⁻⁷ s. An electron entering the same way turns the opposite sense, so it takes the other arc.
Case 3: a strip of finite width
The field fills a strip of width d = 5.0 cm, and the proton enters at right angles to the strip's edge. Since d is less than r (10.4 cm), it cannot turn back. It crosses the strip on an arc and leaves deflected by an angle φ.
Geometry: sin φ = d/r = 0.050/0.104 ≈ 0.479, so φ ≈ 28.6°.
Time inside: t = φ/ω = 0.499 rad/(1.92 × 10⁷ rad/s) ≈ 2.6 × 10⁻⁸ s.
Sideways shift at exit: r(1 − cos φ) ≈ 0.104 × (1 − 0.878) ≈ 1.3 cm.
What if the proton were slower? It turns back if r ≤ d, that is if v ≤ qBd/m = (1.602 × 10⁻¹⁹ × 0.20 × 0.050)/(1.673 × 10⁻²⁷) ≈ 9.6 × 10⁵ m/s. Below that speed it makes a semicircle and comes out on the side it went in.
Why do these questions catch students out?
Students can confuse electric and magnetic fields when problems look alike. Hernández and colleagues (2020) gave 322 introductory students two versions of a questionnaire with the same surface features, one about electric fields and one about magnetic fields, to compare their reasoning about charges at rest and in motion in uniform fields. In a magnetic field, unlike an electric one, the speed is constant and a charge at rest feels no force. Keep that difference in front of you.
Which mistakes should you avoid?
Using the time for a full circle. The particle is only inside for part of it.
Choosing the wrong arc in Case 2. Check the force direction at entry.
Measuring angles from the normal when the question gives them from the boundary, or the reverse. Draw both.
Changing the speed inside the field. Only the direction changes.
Forgetting the sign of q. An electron turns the opposite way and, with its tiny mass, on a much smaller circle.
For the radius and period in an unbounded field and the cyclotron, read charged particle motion and the cyclotron. For direction rules, see magnetic force and the right-hand rule.
A practice routine
Redo Case 2 with α = 45° and find both possible times.
Redo Case 3 with d = 8.0 cm.
Repeat Case 1 for an electron at the same speed and compare radius and time.
Draw every path to scale before you calculate anything.
The geometry is far easier when you can watch the path bend, and the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Start with step 1, and mark the circle's centre on your diagram before writing any equation.
