Force on a moving charge
Charges that curve
- Fire a charged particle across a magnetic field and it curves into a circle.
- This is how particle accelerators bend beams and old TVs steered electrons.
- The force always pushes sideways.
Force on a moving charge
- $F = BQv\sin\theta$ — largest at right angles, zero when moving along the field.
- Direction: Fleming's left-hand rule (second finger = motion of a positive charge).
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.
Circular motion
- The force is always at right angles to $v$, so it does no work — speed stays constant.
- It acts as the centripetal force: $BQv = \dfrac{mv^{2}}{r}$, so $r = \dfrac{mv}{BQ}$, and $T = \dfrac{2\pi m}{BQ}$ (independent of speed).
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 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.
Hall effect and velocity selector
- A Hall probe uses $V_{\text{H}} = \dfrac{BI}{ntq}$ to measure a field $B$.
- A velocity selector (crossed $E$ and $B$) lets through only $v = \dfrac{E}{B}$.
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.
You've got it
- force on a moving charge: $F = BQv\sin\theta$ (max at right angles)
- it does no work → circular path, $r = \dfrac{mv}{BQ}$, period independent of speed
- velocity selector passes $v = \dfrac{E}{B}$