8867 / 2027
Electromagnetism overview

Lesson 6 of 7

Circular motion in a magnetic field

  • I can describe and analyse charged-beam deflections in uniform electric and uniform magnetic fields.Syllabus

    Syllabus 8867, 10(k). Describe and analyse charged-beam deflections in uniform electric and uniform magnetic fields.

Make a guess. You are not marked.

Electrons and protons enter the same uniform magnetic field at the same speed, at right angles to it. Whose path is more tightly curved?

  1. The electrons', because their much smaller mass gives a smaller radius.
  2. Neither, because both carry the same size of charge and so feel equal forces.
  3. The protons', because they carry a positive charge.
Show the answer

The electrons', because their much smaller mass gives a smaller radius.

r = mv / (qB). Both have the same size of charge and speed, so the radius is proportional to mass. The electrons curve far more tightly, in the opposite sense.

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:

B|Q|v = mv2/r
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.

F = B|Q|v = 3.20 × 10-15 N
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 electron follows a true semicircleA uniform one-millitesla magnetic field points into the page only to the right of the ideal sharp boundary at x eighty. Position uses one thousand drawing units per metre in both directions. The electron enters at eighty,eighty-five travelling right and follows a right-hand semicircle of radius 113.875 drawing units about eighty,198.875. At the rightmost point 193.875,198.875, velocity is down and force left. At the exit eighty,312.75, velocity is left; its force immediately inside the field is up. Outside the field it continues left along the straight tangent with unchanged speed. Equal blue velocity arrows and equal purple force arrows are separate quantity scales. The radius is 0.113875 metre and the entry-to-exit separation is the diameter, 0.227750 metre.B = 1.00 mT into the pagevvvFFFrr = 0.113875 mEntry to exit: 0.227750 mDashed line: ideal field boundary

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.

Equal horizontal and vertical scales show the electron's true right-hand semicircle. Velocity is tangent and force inward at the marked positions. The field stops at the labelled ideal boundary, so the exit continues along a straight leftward tangent.

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):

Entry-to-exit vertical separation = 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.

Try it step by step

Work it out in steps: Momentum mv in kg m/s, then Radius r = mv / (qB) in m.

  1. Problem 1. A proton (mass 1.67 x 10-27 kg, charge 1.60 x 10-19 C) moves at 2.0 x 106 m/s at right angles to a uniform field of 0.50 T. Find the radius of its circular path.
  2. Problem 2. An electron (mass 9.11 x 10-31 kg, charge 1.60 x 10-19 C) moves at 3.0 x 107 m/s at right angles to a uniform field of 2.0 x 10-3 T. Find the radius of its circular path.
  3. Problem 3. An alpha particle (mass 6.64 x 10-27 kg, charge 3.20 x 10-19 C) moves at 1.5 x 107 m/s at right angles to a uniform field of 1.2 T. Find its momentum and the radius of its path.
Show the answer

Problem 1.

Momentum mv in kg m/s: mv = 1.67 x 10-27 x 2.0 x 106 = 3.34 x 10-21 kg m/s

Radius r = mv / (qB) in m: r = 3.34 x 10-21 / (1.60 x 10-19 x 0.50) = 0.0418 m

Problem 2.

Momentum mv in kg m/s: mv = 9.11 x 10-31 x 3.0 x 107 = 2.73 x 10-23 kg m/s

Radius r = mv / (qB) in m: r = 2.73 x 10-23 / (1.60 x 10-19 x 2.0 x 10-3) = 0.0854 m

Problem 3.

Momentum mv in kg m/s: mv = 6.64 x 10-27 x 1.5 x 107 = 9.96 x 10-20 kg m/s

Radius r = mv / (qB) in m: r = 9.96 x 10-20 / (3.20 x 10-19 x 1.2) = 0.259 m

Check your understanding

An electron enters at (0,0) moving right into a uniform into-page magnetic field occupying x > 0. Its circular radius is 0.113875 m, and other forces are negligible. What happens when it next leaves this field?

  1. It leaves 0.22775 m below entry, moving left at unchanged speed, then continues in a straight line.
  2. It leaves on a parabola, with its horizontal speed unchanged as it crosses the field.
  3. It leaves 0.113875 m below entry, moving downward.
Show the answer

It leaves 0.22775 m below entry, moving left at unchanged speed, then continues in a straight line.

The electron curves downward around the centre (0,-r). The right-hand semicircle ends at (0,-2r) with a leftward tangent. Magnetic force does no work, and outside the ideal field no force remains to keep turning it.