What you are looking at
A charged particle moves through a uniform
magnetic field that points straight out of (•) or
into (✕) the screen. The
green arrow is its velocity, the
gold arrow is the magnetic force on it, and the curve is the path it
traces. Notice the force is always
perpendicular to the velocity — it never speeds the
particle up or slows it down, it only turns it.
The Lorentz force
A magnetic field exerts a force on a moving charge given by the cross product:
F = q v × B
Its size is F = qvB sin θ (maximum when v is perpendicular to B), and its direction is given by the
right-hand rule, flipped for negative charges. Because the force is always sideways, it acts as a
centripetal force and bends the path into a
circle. Setting qvB = mv²/r and
solving gives the radius and period:
r = m v / (q B) T = 2π m / (q B) f = q B / (2π m)
A striking fact: the period and frequency do
not depend on the speed — a faster particle
simply travels a bigger circle in the same time. That constant "cyclotron frequency" is what makes the
cyclotron particle accelerator work.
What changes the orbit
A
stronger field or
larger charge tightens the circle (smaller r); a
heavier or
faster particle widens it. Flip the sign of the charge or reverse
the field and the particle curves the opposite way — this is exactly how a
mass spectrometer
separates ions, since heavier ones swing wider.
Crossed fields: the velocity selector
Switch on a perpendicular
electric field and the two forces compete. When the electric force
qE exactly balances the magnetic force qvB, the particle sails straight through — this happens only at one
special speed, v = E/B, so crossed fields act as a
velocity selector. Off that speed the path
becomes a drifting cycloid, with an overall
E×B drift at velocity E/B regardless of charge.
Things to try
Launch a positive charge and then a negative one in the same field to see them curl in opposite directions.
Crank up B to shrink the circle; raise the speed to grow it while the period stays put. Turn on the electric
field and hunt for the speed v = E/B where the path goes perfectly straight.