Force on a moving charge
| English | Chinese | Pinyin |
|---|---|---|
| centripetal force | 向心力 | xiàng xīn lì |
| velocity selector | 速度选择器 | sù dù xuǎn zé qì |
| Hall voltage | 霍尔电压 | huò ěr diàn yā |
| Hall probe | 霍尔探头 | huò ěr tàn tóu |
A force that can never speed anything up
- A magnetic force on a moving charge is always at right angles to the velocity. Always, without exception.
- A force perpendicular to the motion does no work, so it cannot change the particle's speed. It can only change its direction.
- A constant-magnitude force that only ever turns you is exactly the recipe for a circle, which is why every particle accelerator is round.
- This lesson is $F = BQv\sin\theta$, the circular motion it produces, and the two instruments built on it.
The force on a moving charge
- $\theta$ is the angle between the velocity and the field. The force is greatest at right angles and zero when the charge moves along the field.
- The direction comes from Fleming's left-hand rule, with the second finger along the motion of a positive charge. For a negative charge, such as an electron, the force is the opposite way.
- A charge at rest feels no force at all: $v = 0$ makes $F = 0$.
A $2.0\ \text{C}$ charge moves at $3.0\ \dfrac{\text{m}}{\text{s}}$ at right angles to a $0.50\ \text{T}$ field. What is the force?
$F = BQv = 0.50 \times 2.0 \times 3.0 = 3.0\ \text{N}$.
For motion at right angles, the magnetic force on a moving charge is $F = BQ$____.
$F = BQv$ — proportional to the charge's speed.
Why the path is a circle
- The force is always perpendicular to $v$, so it does no work, and the speed is constant. Only the direction changes.
- A constant-magnitude force always pointing at right angles to the motion is a centripetal force 向心力, so the path is a circle:
- The radius is proportional to the momentum and inversely proportional to the charge and the field. A stronger field bends the path more tightly.
The force turns the velocity without ever adding to it
Force on a moving charge
F = BQv
The magnetic force on a charge is proportional to its speed (for a fixed field and charge).
The magnetic force does no work on a charge moving in a circle.
The force is always perpendicular to the velocity, so it changes direction but not speed — no work done.
The radius of a charge's circular path in a magnetic field is:
Setting $BQv = \dfrac{mv^{2}}{r}$ gives $r = \dfrac{mv}{BQ}$ — bigger momentum, bigger circle.
The period is independent of speed
- Substituting $v = 2\pi r/T$ into $r = mv/(BQ)$ gives:
- The speed has cancelled. A faster particle travels a bigger circle in exactly the same time.
- That surprising fact is what makes the cyclotron possible: one fixed accelerating frequency works for the particle throughout, however fast it becomes.
The time for one circular orbit does not depend on the speed.
$T = \dfrac{2\pi m}{BQ}$ — a faster particle just goes round a bigger circle in the same time.
A charged particle moves in a circle in a magnetic field. Which quantities stay constant? Select all that apply.
The force does no work, so speed and kinetic energy cannot change; only the direction does, which is what makes the path circular.
Worked example: bending an electron beam
- An electron of speed $2.0 \times 10^7\ \text{m/s}$ enters a field of $1.5\ \text{mT}$ at right angles. Find the radius of its path. ($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$, $e = 1.60 \times 10^{-19}\ \text{C}$.)
- What if the speed doubles? The radius doubles, but the period does not change, since $T = 2\pi m/(BQ)$ contains no $v$.
- And the kinetic energy? Unchanged by the magnetic force, which does no work. Only an electric field can change the speed.
An electron at 2.0e7 m/s enters a 1.5 mT field at right angles. What is the radius of its path, in metres? (m = 9.11e-31 kg, e = 1.60e-19 C)
r = mv/(BQ) = (9.11e-31 x 2.0e7)/(1.5e-3 x 1.60e-19) = 0.076 m. Doubling the speed doubles r but leaves the period unchanged.
The velocity selector
- Cross an electric field with a magnetic field so their forces on a moving charge are opposite.
- The electric force $QE$ does not depend on speed; the magnetic force $BQv$ does. So they balance for exactly one speed:
- Particles at that speed pass straight through the slit. Faster ones are pushed one way and slower ones the other, whatever their charge or mass. That is a velocity selector 速度选择器.
One speed goes straight; everything else is deflected
A velocity selector (crossed E and B fields) lets through only particles with speed:
The forces balance ($qE = qvB$) only at $v = \dfrac{E}{B}$; others are deflected.
In a velocity selector the electric and magnetic forces balance for particles of speed v = ____.
QE = BQv, and the charge cancels, so only one speed passes straight through whatever the particle's charge or mass.
The Hall effect
- Pass a current through a flat slab in a perpendicular field. The magnetic force pushes the moving charges to one face, which builds up until the electric field it creates just balances the magnetic force.
- The steady p.d. across the slab is the Hall voltage 霍尔电压:
- Since everything except $B$ is fixed for a given slab, $V_{\text{H}} \propto B$, which is how a Hall probe 霍尔探头 measures a magnetic field.
Why can a Hall probe be used to measure magnetic flux density?
Charges are pushed to one face until the electric field they build up balances the magnetic force, leaving a steady p.d. proportional to B.
Marks that slip away
- The magnetic force does no work, so the speed and kinetic energy are unchanged. Only the direction changes.
- The period is independent of speed; the radius is not. Check which the question asks for.
- $\theta$ is between the velocity and the field, and a charge at rest feels no force.
- For a negative charge the force is opposite to the left-hand-rule direction for the conventional current. Say so when you use it.
You've got it
- $F = BQv\sin\theta$: greatest at right angles, zero along the field, zero for a charge at rest, and perpendicular to both $v$ and $B$
- because it does no work, the speed is constant and the path is a circle with $r = \dfrac{mv}{BQ}$
- the period $T = \dfrac{2\pi m}{BQ}$ is independent of speed, which is what makes a cyclotron work
- a velocity selector passes only $v = E/B$; a Hall probe uses $V_{\text{H}} = BI/(ntq)$, proportional to $B$, to measure a field