What is the period is independent of speed?

Substituting v = 2π rT into the radius equation and simplifying gives the period:

What is helical motion?

When a charge enters at an angle to the field, only the perpendicular component v_ = vsinθ drives the circular motion, while the parallel component v_ = vcosθ carries the charge steadily along a field line. The result is a helix (a spiral) whose radius is set by v_ and whose pitch (the distance advanced per turn) is set by v_. This is exactly how charged particles spiral along Earth's magnetic field lines in the Van Allen belts and funnel toward the poles to create aurorae.

TASC TCE 2019-style practice: A beam of electrons moving at about 3 × 10^7\ m s^-1 enters a magnetic field of flux density 11.5\ mT at right angles to the field. Determine the force on the electrons, the radius of their circular path, and the frequency of the electron orbit.

The magnetic force on a moving charge (v perpendicular to B) is F = qvB: $F = (1.610^-19)(3.010^7)(11.510^-3) = 5.52 × 10^-14\ N.$ This force is centripetal, so qvB = mv^2r gives r = mvqB: $r = (9.1110^-31)(3.010^7)(1.610^-19)(11.510^-3) = 1.49 × 10^-2\ m.$ Frequency: f = v2π r = 3.010^72π (1.4910^-2) = 3.21 × 10^8\ Hz. So F ≈ 5.5 × 10^-14\ N, r ≈ 1.5\ cm and f ≈ 3.2 × 10^8\ Hz. Markers want F = qvB, r = mvqB and f = v2π r.

TASC TCE 2023-style practice: In the Large Hadron Collider protons move in large circles controlled by magnetic fields. At full energy a proton has a momentum of 3.73 × 10^-15\ kg m s^-1 and is kept in a circle of radius 4.23\ km. Calculate the magnetic flux density required to keep the protons in this circle.

For circular motion the magnetic force provides the centripetal force: qvB = mv^2r, which rearranges to B = mvqr = pqr, since momentum p = mv. $B = (p)/(qr) = 3.7310^-15(1.610^-19)(4230) = 3.7310^-156.76810^-16 = 5.51\ T.$ A flux density of about 5.5\ T is needed, which is why the LHC uses superconducting magnets. Markers want B = pqr with the radius in metres.

§-Syllabus dot point

TASPhysicsSyllabus dot point

Why do magnetic fields make charged particles move in circles?

Analyse the circular motion of charged particles in uniform magnetic fields and its applications.

The force on a moving charge as a centripetal force, the radius and period of circular motion in a magnetic field, and applications such as the mass spectrometer and velocity selector.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

This dot point applies the magnetic force on a charge to the special and very useful case of circular motion, and to the instruments that exploit it.

The magnetic force as a centripetal force

The force on a charge $q$ moving at speed $v$ through a field $B$ is $F = qvB\sin\theta$ , maximum when the velocity is perpendicular to the field. This force is always perpendicular to the velocity, so it never speeds the charge up or slows it down; it only changes the direction. A force of constant size always perpendicular to the motion is exactly what produces uniform circular motion.

Setting the magnetic force equal to the centripetal force:

qvB = \frac{mv^2}{r}

and solving for the radius gives:

r = \frac{mv}{qB}

Faster or heavier particles curve in larger circles; stronger fields and larger charges curve them tighter.

The period is independent of speed

Substituting $v = \dfrac{2\pi r}{T}$ into the radius equation and simplifying gives the period:

T = \frac{2\pi m}{qB}

Remarkably, the speed cancels: all particles of the same charge-to-mass ratio circle in the same time regardless of how fast they move. Faster particles travel larger circles but take the same time per loop. This constant period is what lets a cyclotron drive particles with an alternating voltage of fixed frequency.

Helical motion

When a charge enters at an angle to the field, only the perpendicular component $v_\perp = v\sin\theta$ drives the circular motion, while the parallel component $v_\parallel = v\cos\theta$ carries the charge steadily along a field line. The result is a helix (a spiral) whose radius is set by $v_\perp$ and whose pitch (the distance advanced per turn) is set by $v_\parallel$ . This is exactly how charged particles spiral along Earth's magnetic field lines in the Van Allen belts and funnel toward the poles to create aurorae.

Applications

In a mass spectrometer, ions are first accelerated to a known speed, then sent into a uniform magnetic field. Heavier ions follow a larger radius $r = \dfrac{mv}{qB}$ , so the field separates ions by mass, allowing the composition of a sample to be measured.

A velocity selector crosses an electric field with a magnetic field. A charge passes straight through only if the electric force balances the magnetic force, $qE = qvB$ , which happens at one particular speed $v = \dfrac{E}{B}$ , no matter the charge or mass. This delivers a beam of single, known speed for the spectrometer.

Worked example: radius of a proton's path

A proton ( $q = 1.6\times10^{-19}\ \text{C}$ , $m = 1.67\times10^{-27}\ \text{kg}$ ) moves at $2.0\times10^{6}\ \text{m s}^{-1}$ perpendicular to a $0.50\ \text{T}$ magnetic field. Find the radius of its circular path.

Step 1: Identify why the proton travels in a circle

Because the proton's velocity is perpendicular to the field, the magnetic force $F = qvB$ is always perpendicular to the proton's velocity. A force that is always perpendicular to the motion does no work and does not change the speed; it continuously deflects the proton sideways, which is exactly the condition for uniform circular motion. We can therefore set the magnetic force equal to the centripetal force:

qvB = \frac{mv^2}{r}

Step 2: Rearrange for the radius

Cancelling one factor of $v$ from both sides and solving for $r$ shows that the radius depends on the proton's momentum and the field strength. A heavier or faster proton traces a larger circle; a stronger field curves the path more tightly:

r = \frac{mv}{qB}

Step 3: Substitute values and evaluate

With $m = 1.67\times10^{-27}\ \text{kg}$ , $v = 2.0\times10^{6}\ \text{m s}^{-1}$ , $q = 1.6\times10^{-19}\ \text{C}$ , and $B = 0.50\ \text{T}$ :

r = \frac{(1.67\times10^{-27})(2.0\times10^{6})}{(1.6\times10^{-19})(0.50)} = \frac{3.34\times10^{-21}}{8.0\times10^{-20}} = 0.042\ \text{m}

Final answer: The proton circles with a radius of $0.042\ \text{m}$ (about $4.2\ \text{cm}$ ).

In the exam, recognise circular motion in a magnetic field whenever a charge moves across a uniform field. Set $qvB = \dfrac{mv^2}{r}$ to find the radius, and remember the period is independent of speed. For combined fields, balance the electric and magnetic forces to find the selected velocity, and resolve velocity into perpendicular and parallel parts when the entry is at an angle.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20195 marksA beam of electrons moving at about

3 \times 10^7\ \text{m s}^{-1}

enters a magnetic field of flux density

11.5\ \text{mT}

at right angles to the field. Determine the force on the electrons, the radius of their circular path, and the frequency of the electron orbit.

Show worked answer →

The magnetic force on a moving charge ( $v$ perpendicular to $B$ ) is $F = qvB$ :

F = (1.6\times10^{-19})(3.0\times10^7)(11.5\times10^{-3}) = 5.52 \times 10^{-14}\ \text{N}.

This force is centripetal, so $qvB = \dfrac{mv^2}{r}$ gives $r = \dfrac{mv}{qB}$ :

r = \frac{(9.11\times10^{-31})(3.0\times10^7)}{(1.6\times10^{-19})(11.5\times10^{-3})} = 1.49 \times 10^{-2}\ \text{m}.

Frequency: $f = \dfrac{v}{2\pi r} = \dfrac{3.0\times10^7}{2\pi (1.49\times10^{-2})} = 3.21 \times 10^8\ \text{Hz}$ .

So $F \approx 5.5 \times 10^{-14}\ \text{N}$ , $r \approx 1.5\ \text{cm}$ and $f \approx 3.2 \times 10^8\ \text{Hz}$ . Markers want $F = qvB$ , $r = \dfrac{mv}{qB}$ and $f = \dfrac{v}{2\pi r}$ .

TCE 20232 marksIn the Large Hadron Collider protons move in large circles controlled by magnetic fields. At full energy a proton has a momentum of

3.73 \times 10^{-15}\ \text{kg m s}^{-1}

and is kept in a circle of radius

4.23\ \text{km}

. Calculate the magnetic flux density required to keep the protons in this circle.

Show worked answer →

For circular motion the magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{r}$ , which rearranges to $B = \dfrac{mv}{qr} = \dfrac{p}{qr}$ , since momentum $p = mv$ .

B = \frac{p}{qr} = \frac{3.73\times10^{-15}}{(1.6\times10^{-19})(4230)} = \frac{3.73\times10^{-15}}{6.768\times10^{-16}} = 5.51\ \text{T}.

A flux density of about $5.5\ \text{T}$ is needed, which is why the LHC uses superconducting magnets. Markers want $B = \dfrac{p}{qr}$ with the radius in metres.

TCE 20222 marksCharged particles are trapped in the Earth's Van Allen Belts. Calculate the radius of the circular motion of a proton moving at

60^\circ

to a field of

3.7\ \text{T}

at a speed of

2.0 \times 10^8\ \text{m s}^{-1}

Show worked answer →

Only the velocity component perpendicular to the field causes circular motion. The perpendicular speed is $v_\perp = v\sin 60^\circ = (2.0\times10^8)(0.866) = 1.732 \times 10^8\ \text{m s}^{-1}$ .

Radius, using the proton mass $1.67 \times 10^{-27}\ \text{kg}$ :

r = \frac{m v_\perp}{qB} = \frac{(1.67\times10^{-27})(1.732\times10^8)}{(1.6\times10^{-19})(3.7)} = 0.489\ \text{m}.

The proton circles with a radius of about $0.49\ \text{m}$ ; the parallel component carries it along the field line, giving a helix. Markers want the perpendicular velocity component and $r = \dfrac{m v_\perp}{qB}$ .