What is force on a moving charge?

A charge q moving with speed v at an angle θ to a magnetic field B experiences a force

SCSA WACE 2023-style practice: A proton (q=1.610^-19\ C, m=1.6710^-27\ kg) moves at 2.510^6\ m s^-1 perpendicular to a uniform magnetic field of 0.80\ T. (a) Calculate the magnetic force on the proton. (b) Calculate the radius of its circular path. (c) Explain why the speed of the proton does not change.

A 7 mark calculation rewards the force, the radius from circular motion and an energy argument. **(a) Force.** F=qvB=(1.610^-19)(2.510^6)(0.80)=3.210^-13\ N. **(b) Radius.** The magnetic force provides the centripetal force, qvB=mv^2r, so $r=(mv)/(qB)=(1.6710^-27)(2.510^6)(1.610^-19)(0.80)=3.310^-2\ m.$ **(c) Constant speed.** The magnetic force is always perpendicular to the velocity, so it does no work on the proton (W=Fdcos 90^=0). With no work done, the kinetic energy and hence the speed stay constant; only the direction changes. Markers reward F=qvB, r=mv/(qB), the value near 3.310^-2\ m and the perpendicular-force-does-no-work argument.

§-Syllabus dot point

WAPhysicsSyllabus dot point

How does a magnetic field exert a force on moving charges and currents?

Apply the magnetic force on a moving charge and on a current-carrying conductor, including the motor effect

A focused answer to the WACE Year 12 Physics Unit 3 dot point on magnetism and moving charges. The force on a moving charge and a current-carrying conductor, the right-hand rule, circular motion in a field and the motor effect.

Generated by Claude Opus 4.88 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

WACE wants you to predict the size and direction of the force a magnetic field exerts, both on a lone moving charge and on a wire carrying a current, and then to use that force to explain circular motion in a field and the operation of a simple DC motor. The key idea is that a magnetic force is always perpendicular to the velocity, so it changes direction but never speed.

Force on a moving charge

A charge $q$ moving with speed $v$ at an angle $\theta$ to a magnetic field $B$ experiences a force

F=qvB\sin\theta.

The force is greatest when the velocity is perpendicular to the field ( $\theta=90^\circ$ ) and zero when the charge moves along the field ( $\theta=0$ ). Because the force is always at right angles to $v$ , it does no work: it cannot change the kinetic energy, only the direction.

Direction comes from a hand rule. Using the right hand for conventional (positive) current, point the fingers in the direction of $v$ , curl them toward $B$ , and the thumb gives $F$ ; for a negative charge such as an electron the force is reversed. Many WACE students prefer the right-hand slap rule (fingers along $B$ , thumb along $v$ , palm pushes in the direction of $F$ for a positive charge).

Charges moving in circles

When a charge enters a uniform field at right angles, the constant perpendicular force supplies a centripetal force, so the charge follows a circular path. Setting the magnetic force equal to the centripetal force,

qvB=\frac{mv^2}{r}\quad\Rightarrow\quad r=\frac{mv}{qB}.

Faster or heavier particles curve in larger circles; stronger fields or larger charges tighten the radius. This is the principle behind mass spectrometers and the bending magnets in particle accelerators.

Force on a current-carrying conductor

A current is simply moving charge, so a straight wire of length $L$ carrying current $I$ in a field $B$ feels

F=BIL\sin\theta,

where $\theta$ is the angle between the current and the field. The direction follows the same right-hand rule, now with the fingers (or thumb) pointing along the conventional current. Two parallel wires therefore exert forces on each other: currents in the same direction attract, opposite currents repel.

The motor effect

Place a rectangular current loop in a magnetic field and the two sides carrying current across the field feel opposite forces, producing a turning effect (a torque) that rotates the coil. This is the motor effect. A split-ring commutator reverses the current every half turn so the torque keeps driving the coil the same way, giving continuous rotation. The torque is largest when the coil plane is parallel to the field and zero when the plane is perpendicular, which is why a real motor uses many turns and several coils to smooth the output.

Force and radius for a proton in a field

Proton moving across a 0.40 T field

A proton ( $q=1.60\times10^{-19}\ \text{C}$ , $m=1.67\times10^{-27}\ \text{kg}$ ) moves at $3.0\times10^{6}\ \text{m s}^{-1}$ perpendicular to a field of $0.40\ \text{T}$ .

Force on the proton:

F=qvB=(1.60\times10^{-19})(3.0\times10^{6})(0.40)=1.9\times10^{-13}\ \text{N}.

Radius of its circular path:

r=\frac{mv}{qB}=\frac{(1.67\times10^{-27})(3.0\times10^{6})}{(1.60\times10^{-19})(0.40)}=7.8\times10^{-2}\ \text{m},

about $7.8\ \text{cm}$ . The force sets the curvature; the speed and hence kinetic energy stay constant throughout the circle.

Reading the geometry

Exam questions live or die on the angle $\theta$ and the direction. Always identify whether the velocity or current is perpendicular to the field, draw the field, velocity and force on three mutually perpendicular axes, and state clearly which hand rule you used and whether the carrier is positive or negative.

Exam-style practice questions

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

WACE 20237 marksA proton (

q=1.6\times10^{-19}\ \text{C}

m=1.67\times10^{-27}\ \text{kg}

) moves at

2.5\times10^{6}\ \text{m s}^{-1}

perpendicular to a uniform magnetic field of

0.80\ \text{T}

. (a) Calculate the magnetic force on the proton. (b) Calculate the radius of its circular path. (c) Explain why the speed of the proton does not change.

Show worked answer →

A 7 mark calculation rewards the force, the radius from circular motion and an energy argument.

(a) Force: $F=qvB=(1.6\times10^{-19})(2.5\times10^{6})(0.80)=3.2\times10^{-13}\ \text{N}$ .
(b) Radius: The magnetic force provides the centripetal force, $qvB=\dfrac{mv^2}{r}$ , so
$r=\frac{mv}{qB}=\frac{(1.67\times10^{-27})(2.5\times10^{6})}{(1.6\times10^{-19})(0.80)}=3.3\times10^{-2}\ \text{m}.$
(c) Constant speed: The magnetic force is always perpendicular to the velocity, so it does no work on the proton ( $W=Fd\cos 90^\circ=0$ ). With no work done, the kinetic energy and hence the speed stay constant; only the direction changes.

Markers reward $F=qvB$ , $r=mv/(qB)$ , the value near $3.3\times10^{-2}\ \text{m}$ and the perpendicular-force-does-no-work argument.

WACE 20205 marksExplain why a stationary charge in a magnetic field feels no force, while the same charge moving across the field does, and explain how the direction of the force is found.

Show worked answer →

A 5 mark explanation needs the velocity dependence and the direction rule.

Velocity dependence: The magnetic force on a charge is $F=qvB\sin\theta$ , where $v$ is the charge's speed and $\theta$ is the angle between the velocity and the field. A stationary charge has $v=0$ , so $F=0$ ; it feels no magnetic force regardless of the field strength.
Moving across the field: When the charge moves with a component perpendicular to the field, $v\sin\theta$ is non-zero, so a force acts. The force is greatest when the velocity is perpendicular to the field ( $\theta=90^\circ$ ).
Direction: The force is perpendicular to both $v$ and $B$ , found with the right-hand rule (for a positive charge): fingers along the velocity, curl toward the field, thumb gives the force; reverse for a negative charge.

Markers reward $F=qvB\sin\theta$ with $v=0$ giving zero force, the perpendicular-component idea and a correct right-hand-rule statement with the sign caveat.