Force on Moving Charges and Velocity Selector

A current in a wire is really a flow of charged particles. So the force $F = BIL$F=BIL on a current-carrying conductor must ultimately be the sum of forces on the individual moving charges inside the wire. In this lesson we make that connection explicit, derive the force on a single moving charge $F = Bqv\sin\theta$F=Bqvsinθ, and explore two classic applications: circular motion in a magnetic field (with the all-important radius $r = mv/(qB)$r=mv/(qB) and period $T = 2\pi m/(qB)$T=2πm/(qB)), and the velocity selector (an OCR-specific experimental setup that is a favourite for long-answer questions).

Spec mapping (OCR H556 Module 6.3 — electromagnetism). This lesson covers the derivation of $F = Bqv\sin\theta$F=Bqvsinθ from $F = BIL$F=BIL via $I = nAvq$I=nAvq; the direction rule from Fleming's left-hand rule applied to a positive moving charge (reversed for electrons); circular motion of a charged particle in a uniform field with $r = mv/(qB)$r=mv/(qB) and $T = 2\pi m/(qB)$T=2πm/(qB) (independent of speed); the velocity selector $v = E/B$v=E/B from crossed electric and magnetic field force balance, independent of charge and mass; and the qualitative principles of the mass spectrometer combining a velocity selector with a curving-region magnetic field.

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Force on a Single Moving Charge

Consider a wire of cross-sectional area $A$A and length $L$L containing $n$n charge carriers per cubic metre, each of charge $q$q, drifting with velocity $v$v. The current is

$I = nAvq.$I=nAvq.

From the previous lesson, the force on the whole wire (perpendicular field) is

$F_{\text{wire}} = BIL = B(nAvq)L = (nAL)\,Bqv.$Fwire​=BIL=B(nAvq)L=(nAL)Bqv.

But $nAL$nAL is the total number of charge carriers in the wire. Dividing by $nAL$nAL, the force per charge is

$F = Bqv \quad \text{(velocity perpendicular to field).}$F=Bqv(velocity perpendicular to field).

If the velocity makes an angle $\theta$θ with the field, the formula is

$F = Bqv\sin\theta$F=Bqvsinθ

exactly as for wires. The force:

• Is perpendicular to both $\vec{B}$B and $\vec{v}$v.
• Is zero if $\vec{v}$v is parallel to $\vec{B}$B ($\sin\theta = 0$sinθ=0).
• Is maximum when $\vec{v}$v is perpendicular to $\vec{B}$B.
• Does no work on the charge, because $\vec{F} \cdot \vec{v} = 0$F⋅v=0 at all times.

The last point is crucial: magnetic forces never change the kinetic energy of a charged particle. They can change its direction but not its speed.

Direction — Fleming's Left-Hand Rule (and right-hand-slap rule)

For a positive charge moving in direction $\vec{v}$v, use Fleming's left-hand rule with "current" $=$= direction of motion. For a negative charge (electron), the force is in the opposite direction, because the equivalent conventional current is opposite to the electron velocity.

Right-hand-slap variant for positive charge: right hand flat, fingers along $\vec{B}$B, thumb along $\vec{v}$v, the palm pushes in the direction of $\vec{F}$F. For electrons, reverse the answer (or use the left hand).

flowchart TD
    A[Charge q moving with velocity v in field B] --> B[Is q positive?]
    B -->|yes| C[Fleming's left-hand rule with current along v]
    B -->|no| D[Fleming's left-hand rule with current opposite to v<br/>OR reverse the answer]
    C --> E[Force F perpendicular to both v and B]
    D --> E

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Worked Example 1 — Force on an Electron

An electron travels at $v = 1.5 \times 10^6$v=1.5×106 m s$^{-1}$−1 perpendicular to a magnetic field of $B = 0.20$B=0.20 T. Calculate the force on the electron.

$F = Bqv = (0.20)(1.60 \times 10^{-19})(1.5 \times 10^6) = 4.8 \times 10^{-14}\ \text{N}.$F=Bqv=(0.20)(1.60×10−19)(1.5×106)=4.8×10−14 N.

Tiny in absolute terms, but enormous per unit mass:

$a = \frac{F}{m_e} = \frac{4.8 \times 10^{-14}}{9.11 \times 10^{-31}} \approx 5.3 \times 10^{16}\ \text{m s}^{-2}.$a=me​F​=9.11×10−314.8×10−14​≈5.3×1016 m s−2.

About $5 \times 10^{15}\,g$5×1015g — five quadrillion times the acceleration due to gravity. The smallness of the electron mass is what makes magnetic forces dominant in cathode-ray tubes, beam-line physics, and synchrotron radiation.

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Circular Motion in a Magnetic Field

Because the magnetic force on a moving charge is always perpendicular to the velocity, it acts exactly like a centripetal force: the particle moves in a circle at constant speed (if the field is uniform and the particle enters perpendicular to it).

Setting magnetic force equal to centripetal force:

$Bqv = \frac{mv^2}{r}$Bqv=rmv2​

and rearranging,

$r = \frac{mv}{qB}.$r=qBmv​.

The radius of the circular path is proportional to the momentum $p = mv$p=mv and inversely proportional to $qB$qB. This is the principle behind:

• Cyclotrons and other circular particle accelerators.
• Mass spectrometers (heavier particles, larger arcs for the same speed and charge).
• Magnetic bottles in fusion research.
• The beautiful curving tracks visible in bubble chambers and cloud chambers.

The period of the circular motion is

$T = \frac{2\pi r}{v} = \frac{2\pi (mv/qB)}{v} = \frac{2\pi m}{qB}.$T=v2πr​=v2π(mv/qB)​=qB2πm​.

Notably, $T$T does not depend on $v$v. All particles of the same mass and charge in the same field take the same time to complete a circle. This property is what made the cyclotron possible — the oscillating accelerating voltage can be kept at a constant frequency (the cyclotron frequency $f_c = qB/(2\pi m)$fc​=qB/(2πm)).

Worked Example 2 — Radius of an Electron Path

An electron enters a uniform magnetic field of flux density $4.0$4.0 mT perpendicular to the field, with speed $v = 6.0 \times 10^6$v=6.0×106 m s$^{-1}$−1. Calculate the radius of the circular path and the period.

Radius.

$r = \frac{mv}{qB} = \frac{(9.11 \times 10^{-31})(6.0 \times 10^6)}{(1.60 \times 10^{-19})(4.0 \times 10^{-3})}$r=qBmv​=(1.60×10−19)(4.0×10−3)(9.11×10−31)(6.0×106)​

$r = \frac{5.466 \times 10^{-24}}{6.4 \times 10^{-22}} \approx 8.5 \times 10^{-3}\ \text{m} \approx 8.5\ \text{mm}.$r=6.4×10−225.466×10−24​≈8.5×10−3 m≈8.5 mm.

Period.

$T = \frac{2\pi m}{qB} = \frac{2\pi(9.11 \times 10^{-31})}{(1.60 \times 10^{-19})(4.0 \times 10^{-3})} \approx 8.9 \times 10^{-9}\ \text{s} \approx 8.9\ \text{ns}.$T=qB2πm​=(1.60×10−19)(4.0×10−3)2π(9.11×10−31)​≈8.9×10−9 s≈8.9 ns.

The electron completes a circle of radius $8.5$8.5 mm in about $9$9 ns — a cyclotron frequency of about $112$112 MHz (in the FM radio band). Note that if you halve $v$v, $r$r halves but $T$T stays the same.

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The Velocity Selector — OCR-Specific

A velocity selector is a beautifully simple device which uses crossed electric and magnetic fields to filter out particles of a specific speed, irrespective of their charge or mass. OCR H556 specifically requires candidates to understand how it works.

How It Works

Between two parallel plates there is a uniform electric field $E$E pointing (say) downwards. A magnetic field $B$B is applied perpendicular to $E$E and to the direction of travel (say, out of the page).

A positive charge $q$q moving horizontally to the right at speed $v$v experiences:

• An electric force $F_E = qE$FE​=qE downwards (along the field direction).
• A magnetic force $F_B = Bqv$FB​=Bqv upwards (by Fleming's left-hand rule: current right, field out, force up).

If the two forces are equal and opposite, the net force is zero and the particle passes through undeflected. The balance condition is

$qE = Bqv \quad \Rightarrow \quad v = \frac{E}{B}.$qE=Bqv⇒v=BE​.

This is the velocity-selection condition. Any particle with exactly this speed passes through the device in a straight line. Faster particles have a larger magnetic force and are deflected upwards; slower particles have a smaller magnetic force and are deflected downwards by the electric force. Only the selected velocity makes it through a slit at the exit.

Three important features:

1. The selected velocity depends only on $E$E and $B$B, not on $q$q or $m$m. It works for any charged particle.
2. The sign of $q$q does not matter either: a negative charge has both forces reversed, so they still cancel at the same speed.
3. Combined with a subsequent pure magnetic-field region (where $r = mv/(qB)$r=mv/(qB)), a velocity selector is the first stage of a mass spectrometer: select a velocity, then measure the radius to find $m/q$m/q.

Worked Example 3 — Velocity Selector

A velocity selector uses $B = 50$B=50 mT and parallel plates $d = 2.0$d=2.0 cm apart with $V = 1000$V=1000 V across them. What speed of particles will pass through undeflected?

$E = \frac{V}{d} = \frac{1000}{0.020} = 5.0 \times 10^4\ \text{V m}^{-1}$E=dV​=0.0201000​=5.0×104 V m−1

$v = \frac{E}{B} = \frac{5.0 \times 10^4}{0.050} = 1.0 \times 10^6\ \text{m s}^{-1}.$v=BE​=0.0505.0×104​=1.0×106 m s−1.

Any particle — electron, proton, helium ion, Rb$^+$+ ion — travelling at exactly $10^6$106 m s$^{-1}$−1 in the right direction will pass through. All others are deflected.

Exam tip. The velocity selector is a classic 6-mark OCR synoptic question. Make sure you can (a) derive $v = E/B$v=E/B from force balance, (b) explain why it selects velocity not momentum, and (c) explain why it is independent of charge and mass.

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The Mass Spectrometer — Application

After the velocity selector, the beam enters a pure magnetic field region where it curves in a circle of radius

$r = \frac{mv}{qB}.$r=qBmv​.

The two magnetic fields (in the selector and in the curving region) may or may not be the same. Measuring $r$r for a particle of known charge and selected velocity gives $m$m directly:

$m = \frac{qBr}{v}.$m=vqBr​.

This is how chemists routinely measure the masses of ions with precision better than one part in $10^6$106 — enough to distinguish isotopes of the same element.

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Synoptic Links

• Module 6.3 (magnetic fields and forces, previous lesson). $F = Bqv$F=Bqv derives from $F = BIL$F=BIL by recognising $I = nAvq$I=nAvq and dividing by the number of carriers.
• Module 5.2 (circular motion). The centripetal-force equation $F = mv^2/r$F=mv2/r provides the right-hand-side; magnetic force provides the left-hand-side. The combination is universal in particle physics.
• Module 6.2 (electric fields). The electric force $F = qE$F=qE on a charge in the velocity selector is the same force introduced in Module 6.2; the velocity selector is the cleanest synoptic exam moment in Module 6.
• Module 6.4 (nuclear and particle physics). Cyclotron operation, the LHC, and beta-decay tracking all use $r = mv/(qB)$r=mv/(qB). A muon's $r$r vs an electron's $r$r at the same momentum is a $207$207:$1$1 ratio — the way muons were discovered in 1936.
• Chemistry — mass spectrometry. Every chemistry laboratory's mass spectrometer applies $r = mv/(qB)$r=mv/(qB) (or its modern equivalents) to identify molecules by their mass-to-charge ratio.
• Astrophysics — aurorae and Van Allen belts. Charged particles from the solar wind spiral along Earth's magnetic field lines (helical motion: circular motion perpendicular to $\vec{B}$B combined with free drift along $\vec{B}$B); when they hit the upper atmosphere they create aurorae.

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