Topic 8 of 9
Circular motion in a magnetic field
With perpendicular entry into a uniform magnetic field, a charged particle keeps its speed while its velocity turns. The magnetic force supplies the inward resultant required for circular motion.
For constant speed v on a circle of radius r, inward acceleration has magnitude v2/r. With magnetic force as the only significant force, its perpendicular-entry magnitude B|Q|v therefore gives:
r = mv/(|Q|B)
The magnetic force is the actual inward force; do not add a second force called centripetal force. At each point, velocity is tangent to the circle and magnetic force points towards its centre. The force does no work, so this model has constant speed, not constant horizontal velocity.
Worked magnetic beam
Use the radius to find the exit state
An electron enters at (0, 0), moving right at 2.00 × 107 m/s. A uniform 1.00 mT field into the page occupies x > 0, with an ideal sharp boundary at x = 0. Outside that region B is zero. Take right and up as positive.
Use me = 9.11 × 10-31 kg and |Q| = 1.60 × 10-19 C. Neglect electric fields, gravity, collisions, fringing and radiation, and use the classical model. Roughly, m of order 10-30 kg, v of order 107 m/s, |Q| of order 10-19 C and B of order 10-3 T suggest r of order 0.1 m.
r = [(9.11 × 10-31)(2.00 × 107)]
/[(1.60 × 10-19)(1.00 × 10-3)]
= 0.113875 m ≈ 0.114 m
The electron follows a true semicircle
The circular path uses equal position scales. Blue velocities are tangent; purple forces are inward. Force arrows at entry and exit refer to the field side of the boundary. Beyond it, the ideal magnetic force is zero and the path is straight.
The electron's initial force is down, so the circle centre is (0, -r). At the rightmost point (r, -r), velocity is down and force is left. The electron next reaches the boundary at (0, -2r):
= 0.227750 m ≈ 0.228 m
It leaves below entry, travelling left, still at 2.00 × 107 m/s. Immediately before exit, magnetic force is upward towards the centre. Outside the ideal field, with no further force, it travels straight along that leftward tangent.
Compare the electric and magnetic paths
In the uniform electric-field example, force has a fixed upward direction, horizontal velocity stays constant and the growing vertical component increases total speed. The path is a parabola within that field.
In the magnetic-only example, force continually turns with velocity and stays perpendicular to it. The path is circular at constant speed, while both horizontal and vertical velocity components change. Use the field type and initial direction to choose the motion model; a charged particle does not always follow one standard curve.