Magnetic Force on a Moving Charge: The Right-Hand Rule Explained

Point a compass anywhere near a wire carrying current and the needle swings to align with the magnetic field the current creates — that much feels intuitive. But turn the question around: fire a charged particle through a magnetic field, and the force it feels is not along the field, not along its own velocity, but at right angles to both of them simultaneously — a direction neither one actually points in. There is no everyday intuition for that. The only way through it is a hand gesture physicists have relied on for over a century: the right-hand rule.

This page is an interactive right-hand rule simulator and magnetic force calculator, built around three tools rather than one: a vector explorer where you rotate the field yourself and watch the force respond (including the moment it vanishes entirely), a velocity selector where you watch two competing forces balance directly, and a mass spectrometer where you control the one variable — mass — that actually separates real ions.

What Is the Formula for Magnetic Force on a Moving Charge?

The force a magnetic field exerts on a moving charge is given by the Lorentz force law:

$\vec{F} = q\vec{v} \times \vec{B}$

Because this is a cross product, the force has three properties — each a little counter-intuitive — that trip up nearly every student the first time:

• Zero force on a stationary charge. If $v = 0$, then $F = 0$: magnetic fields only push on charges that are already moving. That's the biggest difference from electric force, which acts on a charge whether it's moving or not (see our electric field post for that comparison).
• Force perpendicular to velocity. The cross product $\vec{v} \times \vec{B}$ is always perpendicular to $\vec{v}$, so the magnetic force can never speed up or slow down a charge — only change its direction. It plays the same role a string plays on a whirling ball, except nothing is physically touching the particle at all; see how magnets pull with no contact needed.
• Magnitude depends on the angle. The full magnitude is $F = qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. When the charge moves perpendicular to the field ($\theta = 90°$, the case this page focuses on), $\sin\theta = 1$ and the force simplifies to $F = qvB$ — the maximum force that speed and field can produce together.

What Is the Right-Hand Rule for Magnetic Force?

To find the direction of $\vec{v} \times \vec{B}$, and therefore the force on a positive charge:

1. Point your right-hand fingers in the direction of the velocity $\vec{v}$.
2. Curl your fingers toward the direction of the field $\vec{B}$.
3. Your thumb now points in the direction of the force $\vec{F}$.

For a negative charge, the force points opposite to what your thumb indicates. Flip the direction, or use your left hand instead.

What Is Fleming's Left-Hand Rule?

Many curricula (NCERT, JEE, NEET) teach the related Fleming's left-hand rule for the force on a current-carrying conductor, using the left hand with three mutually perpendicular fingers: First finger = Field, seCond finger = Current, thuMb = Motion (force). It answers the same underlying question as the right-hand rule above — just with a different hand. Conventional current is defined as the flow of positive charge, so Fleming's left-hand rule and "right-hand rule, then flip for negative charge" describe identical physics, just taught with different hand gestures in different classrooms.

Interactive Right-Hand Rule Simulator

Choose the charge's sign, and drag the sliders for speed, mass, field strength, and — the one to actually experiment with — the angle between velocity and field. The green arrow is velocity, fixed pointing right; the indigo arrow is the field, and you control where it points. Watch the amber symbol in the corner: a dot (out of the page) or a cross (into the page) shows the force's direction, sized by its magnitude. Sweep the angle through 0° and 180° and the symbol disappears — a charge moving parallel to the field feels no magnetic force at all, a fact a circle-only diagram can never show.

This diagram is deliberately not a moving orbit — it isolates the direction-finding problem the right-hand rule actually solves. The special case where the field sits locked at 90° to the velocity (the default above) is exactly the configuration that produces circular motion, covered next.

Why Does a Charged Particle Move in a Circle in a Magnetic Field?

Because the magnetic force is always perpendicular to velocity, it can never do any work on the particle ($W = \vec{F} \cdot \vec{d}$, and a force perpendicular to displacement does zero work). With no work done, the particle's speed never changes — only its direction, continuously. A constant-magnitude force always perpendicular to a constant-speed velocity is the definition of uniform circular motion.

What Is the Formula for the Radius of Circular Motion in a Magnetic Field?

Setting the magnetic force equal to the centripetal force required for a circle of radius $r$:

$qvB = \frac{mv^2}{r} \quad\Longrightarrow\quad r = \frac{mv}{qB}$

This is the cyclotron radius — sometimes called the radius of gyration. Rearranged, it also gives the cyclotron angular frequency and period:

$\omega = \frac{qB}{m} \qquad\qquad T = \frac{2\pi m}{qB}$

Does the Period of Circular Motion Depend on Speed?

Look closely at the period formula: $T = 2\pi m / (qB)$ contains no $v$ at all. A faster particle traces a bigger circle, but takes exactly the same time to complete one full loop as a slower particle of the same charge and mass in the same field. This is the property that makes the cyclotron particle accelerator possible: a fixed oscillating voltage, switching at the one frequency $\omega = qB/m$, keeps accelerating a particle no matter how large its orbit has grown, because that frequency never has to change. Set the vector explorer above to θ = 90° (its default) and adjust the speed slider: its "if B were perpendicular to v" readout shows the radius growing while the period stays fixed — the same result, read off the special case rather than an animated orbit.

What Is a Velocity Selector and How Does It Work?

Real particle beams rarely arrive at one clean, known speed on their own — they need filtering first. A velocity selector handles that, using perpendicular electric and magnetic fields to let only one speed through undeflected.

With the field pointing up and the magnetic field out of the page, a particle entering horizontally feels two competing sideways forces: an electric force $qE$ and a magnetic force $qvB$. At most speeds these forces are unequal and the particle deflects — hitting a wall before it ever reaches the far side. At exactly one speed they cancel:

$qE = qvB \quad\Longrightarrow\quad v = \frac{E}{B}$

Notice what happens for a negative charge: both forces flip sign together, so the balance condition doesn't change at all. A velocity selector filters the same speed regardless of whether the beam is positive or negative.

Watch the two force arrows rather than the trajectory alone: Fₑ (green, constant) versus Fᵇ (amber, growing with speed). Drag the speed slider until they're the same length — that's v = E/B, and the path turns green. Anything faster gets bent one way by the now-dominant magnetic force; anything slower gets bent the other way by the now-dominant electric force. Only particles moving at the selected speed make it through — everything else hits a wall before it clears the region.

How Does a Mass Spectrometer Separate Isotopes?

Once a beam has been filtered to a single known speed, it can be sent into a region with only a magnetic field, where the cyclotron radius formula from earlier, $r = mv/(qB)$, takes over. Since $v$, $q$ and $B$ are now identical for every ion in the beam, radius depends on nothing but mass. Heavier ions sweep wider semicircles and land further from where they entered; lighter ions sweep tighter ones. Reading the landing position tells you the mass — no chemistry involved, just geometry.

This isn't a hypothetical setup. In 1919, Francis Aston built this instrument and used it to show that ordinary neon isn't one substance but two: atoms of mass 20 and mass 22, in roughly a 9:1 ratio, chemically identical and impossible to separate by any chemical means, only by mass. It was the first direct proof that a non-radioactive element could have isotopes.

The two neon traces are fixed real-world references — Aston's actual isotopes, 10% apart in mass. Drag the sample ion's mass slider and watch its own path (dashed, purple) move continuously between and beyond them: radius scales directly with mass, which is the entire principle a real instrument relies on to identify an unknown sample by matching its landing position against known references.

Worked Examples for Physics Exams

Example 1: Finding the radius and period

A particle of charge $q = 2$ C and mass $m = 0.5$ kg moves at $v = 10$ m/s perpendicular to a $B = 2$ T field. Find the radius and period of its circular path.

$r = mv/(qB) = (0.5 \times 10)/(2 \times 2) = 1.25$ m. $\omega = qB/m = (2\times2)/0.5 = 8$ rad/s, so $T = 2\pi/8 \approx 0.785$ s. Verify in the simulator: these are the default settings — the status readout shows r = 1.25 m and T = 0.785 s.

Example 2: Finding the field strength

A proton ($q = 1.6 \times 10^{-19}$ C, $m = 1.67 \times 10^{-27}$ kg) moving at $v = 2.0 \times 10^{6}$ m/s is bent into a circle of radius $r = 0.10$ m. What magnetic field is needed?

Rearranging $r = mv/(qB)$ for $B$: $B = mv/(qr) = (1.67\times10^{-27} \times 2.0\times10^{6}) / (1.6\times10^{-19} \times 0.10) \approx 0.21$ T, a field easily produced by a lab electromagnet.

Example 3: A velocity selector

A velocity selector uses $E = 3000$ V/m and $B = 0.5$ T. What speed passes straight through?

$v = E/B = 3000/0.5 = 6000$ m/s. Any charge, positive or negative, moving at this speed exits undeflected; every other speed is filtered out.

Example 4: A real mass spectrometer separating neon isotopes

Singly ionised neon ions ($q = 1.6\times10^{-19}$ C) exit a velocity selector at $v = 1.0\times10^{5}$ m/s and enter a $B = 0.2$ T analyser field. Neon-20 has a mass of $3.32\times10^{-26}$ kg; neon-22 has a mass of $3.65\times10^{-26}$ kg. Find both radii and the separation between where they land.

$r_{20} = m_{20}v/(qB) = (3.32\times10^{-26} \times 1.0\times10^{5})/(1.6\times10^{-19} \times 0.2) \approx 0.104$ m. $r_{22} = (3.65\times10^{-26} \times 1.0\times10^{5})/(1.6\times10^{-19} \times 0.2) \approx 0.114$ m. Separation $= 2(r_{22}-r_{20}) \approx 0.021$ m, close to 2 cm, easily resolved by a detector array — the same method Aston used to tell the two isotopes apart.