Understand how charged particles move in electric and magnetic fields, with plain English intuition, a clear diagram, the parabola versus circle idea, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.
Learn
Fire a charged particle into a field and it bends, but how it bends depends entirely on which field it meets. Drop it into an electric field and it curves like a thrown ball, tracing a smooth parabola. Send it into a magnetic field and it loops around in a perfect circle. Two fields, two shapes, and once you see why, the whole topic clicks into place.
An electric field bends it like gravity does
Picture two charged metal plates with a uniform electric field between them. A charge that wanders into that field feels a steady push in one fixed direction, always the same strength and always the same way.
That is exactly the situation a thrown ball is in. Gravity gives the ball a constant downward force, and the ball traces a parabola. Swap gravity for the electric force F=qE and you get the same shape. The particle keeps its sideways speed while the field steadily speeds it up in the other direction, so the path bends into a parabola.
A magnetic field bends it into a circle
A magnetic field plays by a different rule. The force on a moving charge, F=qvB, is always at right angles to the way the particle is moving. As the particle turns, the force turns with it, staying sideways the whole time.
A force that is always sideways to the motion is a centripetal force, the kind that holds something in a circle. So the particle loops around at a steady speed, and the radius of that circle is r=qBmv. A faster or heavier particle makes a bigger circle, while a stronger field or a bigger charge pulls the circle tighter.
On the left, the charge drifts in straight, then the electric force curves it down into a parabola, just like a projectile. On the right, the magnetic field (the crosses mean it points into the page) keeps bending the charge sideways, so it travels around a circle of radius r.
See it for yourself
Drop a few charges onto the field, then drag the sensor around to read the force a positive charge would feel at each point. The closer you move to a charge, the stronger the push.
The two fields feel similar but do opposite things to the energy of the particle. An electric field can speed a charge up or slow it down, because its force can point along the motion. A magnetic field never can, because its force is always sideways.
How to actually solve one
Most questions are a short, repeatable recipe.
To find the speed after accelerating through a voltage, set qV=21mv2 and solve for v=m2qV.
For a magnetic field, the radius of the circle is r=qBmv.
For the size of the magnetic force at any instant, use F=qvB.
In a uniform electric field, the force is F=qE and the path is a parabola, so treat it just like a projectile with this force in place of gravity.
Keep your powers of ten tidy. Electron and proton numbers are tiny, so write everything in scientific notation and combine the powers carefully.
Lock it in with active recall
Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
What path does a charge follow in a uniform electric field, and why?
A parabola. The constant force F=qE acts in one fixed direction, exactly like gravity on a thrown ball.
What path does a charge follow in a uniform magnetic field, and why?
A circle. The force F=qvB stays perpendicular to the velocity, acting as a centripetal force.
Write the radius of the circular path in a magnetic field.
r=qBmv — bigger for a faster or heavier particle, tighter for a stronger field or larger charge.
Why does a magnetic force never change a particle’s speed?
It is always perpendicular to the velocity, so it has no component along the motion and does no work. Only the direction changes.
A charge is accelerated from rest through a voltage V. Find its final speed.
Set qV=21mv2, so v=m2qV. The energy gained depends only on the voltage.
Recall · Electric Fields
Write the field between two parallel plates separated by d with voltage V.
E=dV — uniform between the plates, and the force on a charge is F=qE.
Recall · Magnetic Fields and Forces
When is the force on a moving charge in a magnetic field at its maximum?
When the velocity is perpendicular to the field, giving F=qvB. Parallel to the field there is no force.
See the recipe in action in the Worked Examples tab, then test yourself in Try It.
Worked examples
Worked Example 1Accelerating an electron through a voltage
An electron (m=9.1×10−31 kg, q=1.6×10−19 C) is accelerated from rest through a potential difference of 200 V. Find its final speed.
1
All the electrical energy the charge gains becomes kinetic energy. Set the work done by the field equal to the kinetic energy gained.
qV=21mv2
2
Rearrange to make the speed the subject.
v=m2qV
3
Put the numbers in.
v=9.1×10−312×1.6×10−19×200=7.03×1013
4
Take the square root for the final speed.
v=8.4×106 m/s
Answer
v=8.4×106 m/s
Worked Example 2That electron entering a magnetic field
The same electron then enters a magnetic field of 0.010 T at right angles to its motion. Find the radius of its circular path.
1
A magnetic force always acts at right angles to the velocity, so it bends the path into a circle. The magnetic force provides the centripetal force, which gives the radius formula.
r=qBmv
2
Put the numbers in, using the speed from the first example.
r=1.6×10−19×0.0109.1×10−31×8.4×106
3
Work out the top and the bottom separately.
r=1.6×10−217.64×10−24
4
Divide to get the radius.
r=4.8×10−3 m≈4.8 mm
Answer
r=4.8×10−3 m≈4.8 mm
Practice questions
Practice test
Try it yourself
6 questions, 11 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.A charged particle enters a uniform electric field at right angles to the field. Ignoring gravity, the path it follows is:
1mark
Need a hint?
A uniform electric field gives a constant force in one direction, just like gravity does to a thrown ball.
Show worked solution
A uniform electric field applies a constant force F=qE in a fixed direction. A constant sideways force on a moving particle bends its path into a parabola, exactly like a projectile under gravity. A circle only appears in a magnetic field, where the force keeps turning to stay at right angles to the velocity.
Q2.A charged particle moves in a circle in a uniform magnetic field. The magnetic force acting on it does no work on the particle because the force is always:
1mark
Need a hint?
Work needs a force component along the direction of motion. Think about the angle between the magnetic force and the velocity.
Show worked solution
The magnetic force F=qvB is always perpendicular to the velocity. Since work needs a force component along the motion, a force at right angles does zero work, so the speed never changes. The force only changes the direction, which is why the path is a circle at constant speed.
Q3.A particle of charge 2.0×10−19 C moves at 3.0×106 m/s through a magnetic field of 0.50 T, at right angles to the field. The magnetic force on it is closest to:
1mark
Need a hint?
Use F=qvB and multiply the three numbers together.
Show worked solution
F=qvB=2.0×10−19×3.0×106×0.50=3.0×10−13 N. Multiplying the powers of ten, 10−19×106=10−13, and the front numbers give 2.0×3.0×0.50=3.0.
Q4.Two particles enter the same magnetic field at the same speed, at right angles to the field. Particle X has twice the mass of particle Y but the same charge. Compared with the radius of Y, the radius of X is:
1mark
Need a hint?
Use r=qBmv and look at how the radius depends on the mass when everything else is fixed.
Show worked solution
From r=qBmv, the radius is directly proportional to the mass when v, q and B are unchanged. Doubling the mass doubles the radius, so X moves in a circle twice as large as Y.
Q5.A proton (m=1.6×10−27 kg, q=1.6×10−19 C) moves at 2.0×105 m/s at right angles to a magnetic field of 0.20 T. Find the radius of its circular path. Show your working.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
The magnetic force provides the centripetal force, so
r=qBmv=1.6×10−19×0.201.6×10−27×2.0×105.
The top is 3.2×10−22 and the bottom is 3.2×10−20, so
r=3.2×10−203.2×10−22=1.0×10−2 m=1.0 cm.
Q6.Electrons are accelerated between two parallel plates 50 mm apart with a potential difference of 250 V across them, in a vacuum. (a) Calculate the magnitude of the electric force on an electron between the plates (e=1.6×10−19 C). (b) The plate separation is then doubled to 100 mm with the same 250 V. Explain what happens to the kinetic energy an electron gains in crossing from one plate to the other.
4marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
(a) The field is E=dV=0.050250=5000 V/m, so the force is F=qE=1.6×10−19×5000=8.0×10−16 N.
(b) The kinetic energy gained is unchanged. The work done on the electron is ΔKE=qV, which depends only on the voltage, not the separation. Doubling d halves the field and the force, but the electron travels twice as far, so qV is the same.
VCAA 2025 Physics Exam, Section B Q6
Frequently asked questions
Why does an electric field give a parabola but a magnetic field gives a circle?
An electric field pushes a charge with a constant force in one fixed direction, so the path bends like a thrown ball into a parabola. A magnetic field pushes at right angles to the velocity, and that force keeps turning as the particle turns, so it traces a circle.
Why does a magnetic force never change a particle's speed?
The magnetic force is always perpendicular to the velocity, so it has no component along the motion. A force that is sideways to the motion does no work, so the kinetic energy and therefore the speed stay the same. Only the direction changes.
What does accelerating a charge through a voltage actually do?
The field does work on the charge equal to qV, and that energy turns into kinetic energy. Setting qV equal to half m v squared lets you find the speed the charge reaches, which is how electron guns in old television tubes worked.
