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Circular paths in a magnetic field

Key idea: For motion perpendicular to a uniform magnetic field, magnetic force supplies the centripetal force at constant speed. The circular radius is mv/(B|q|).

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  • GCE A-Level H2 Physics 2027
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A beam of electrons fired into a uniform magnetic field, at right angles to it, bends round into a circle. The magnetic force never changes the speed of the electrons, only their direction, and that is exactly what a centripetal force does. Combining the magnetic force with circular motion gives the radius of the circle.

A circular path

A charged particle moving at right angles to a uniform magnetic field feels a force BQv of constant size, always at right angles to its velocity. A force like that keeps the speed constant and makes the particle move in a circle. The magnetic force provides the centripetal force:

BQv = mv²/r ⇒ r = mv/BQ = p/BQ

where p = mv is the momentum. This classical relation applies when the speed is much smaller than the speed of light. Faster or heavier particles move in larger circles; a stronger field or a larger charge makes the circle smaller. The radius is a distance, so use the size of the charge. The sign of the charge decides which way the path curves. For a proton or electron, the charge magnitude is e = 1.60 × 10⁻¹⁹ C.

Charged-particle paths in a uniform magnetic fieldA positive charge moves left at the top of a circle in a field into the page; force points down towards the centre. The second panel projects a three-dimensional helix around a rightward field axis.velocity perpendicular to BB into page (×)+vF toward centrespeed constant; radius r = mv/(|q|B)perpendicular and parallel componentsB axis; parallel motion →+radius uses v⊥; pitch is axial advance per turn
Charged-particle paths in a uniform magnetic fieldA positive charge moves left at the top of a circle in a field into the page; force points down towards the centre. The second panel projects a three-dimensional helix around a rightward field axis.velocity perpendicular to BB into page (×)+vF toward centrespeed constant; radius r = mv/(|q|B)perpendicular and parallel componentsB axis; parallel motion →+radius uses v⊥; pitch is axial advance per turn
Schematic, not to scale. Perpendicular entry gives a circle; both non-zero perpendicular and parallel components give a helix. The right panel is a projection of a three-dimensional path. Only the perpendicular component sets its radius.

The particle moves in a circle only while it is in the field. When it leaves, there is no more force, and it carries on in a straight line along the direction it had at the edge of the field.

Check your understanding 1 After the field ends

A particle reaches the edge of a magnetic field with its velocity pointing diagonally upwards and right. There are no forces outside the field. Which path does it follow next?

Path outside the field
Show answer

The particle keeps its exit velocity: a straight line diagonally upwards and right, tangent to the circle at the exit. Removing the force removes the acceleration, not the velocity.

Worked example 1

Radius from momentum

Problem

A proton with momentum 4.0 × 10⁻²⁰ kg m s⁻¹ enters a uniform magnetic field of 0.50 T at right angles to it. Find the radius of its path.

Show full solution

  1. Choose the relation

    Method

    Use r = p/BQ.

    Reason

    The momentum is given, so there is no need to find the speed first.

    Working

    Answer

    r = p/BQ

  2. Radius

    Working

    Answer

    r = (4.0 × 10⁻²⁰)/((0.50)(1.60 × 10⁻¹⁹)) = 0.50 m

The graph shows how the radius depends on the field for protons moving at 3.0 × 10⁶ m s⁻¹. Since r ∝ 1/B, equal increases in B do not reduce the radius by equal amounts.

Magnetic path radius at fixed proton speed

An inverse radius-versus-field curve. At 0.20 T the radius is 0.157 m; at 0.40 T it is 0.0783 m. The speed and particle type are fixed.

Scroll across the figure to read all labels.

An inverse radius-versus-field curve. At 0.20 T the radius is 0.157 m; at 0.40 T it is 0.0783 m. The speed and particle type are fixed.An inverse radius-versus-field curve. At 0.20 T the radius is 0.157 m; at 0.40 T it is 0.0783 m. The speed and particle type are fixed.
Calculated for protons entering perpendicular to uniform B at 3.0 × 10⁶ m s⁻¹, using m = 1.67 × 10⁻²⁷ kg and |q| = 1.60 × 10⁻¹⁹ C. Other forces and relativistic effects are neglected.
Open full-size graph
What to noticeDoubling B halves r at fixed momentum. Equal increases in B do not give equal decreases in r.
View figure data
Figure data: Magnetic path radius at fixed proton speed
Magnetic flux density, B (T)Proton radius: r = mv/(|q|B)
0.2000.15656
0.2050.15274
0.2100.14911
0.2150.14564
0.2200.14233
0.2250.13917
0.2300.13614
0.2350.13324
0.2400.13047
0.2450.12781
0.2500.12525
0.2550.12279
0.2600.12043
0.2650.11816
0.2700.11597
0.2750.11386
0.2800.11183
0.2850.10987
0.2900.10797
0.2950.10614
0.3000.10437
0.3050.10266
0.3100.10101
0.3150.09940
0.3200.09785
0.3250.09635
0.3300.09489
0.3350.09347
0.3400.09210
0.3450.09076
0.3500.08946
0.3550.08820
0.3600.08698
0.3650.08579
0.3700.08463
0.3750.08350
0.3800.08240
0.3850.08133
0.3900.08029
0.3950.07927
0.4000.07828
0.4050.07731
0.4100.07637
0.4150.07545
0.4200.07455
0.4250.07368
0.4300.07282
0.4350.07198
0.4400.07116
0.4450.07037
0.4500.06958
0.4550.06882
0.4600.06807
0.4650.06734
0.4700.06662
0.4750.06592
0.4800.06523
0.4850.06456
0.4900.06390
0.4950.06326
0.5000.06263
0.5050.06200
0.5100.06140
0.5150.06080
0.5200.06022
0.5250.05964
0.5300.05908
0.5350.05853
0.5400.05799
0.5450.05745
0.5500.05693
0.5550.05642
0.5600.05592
0.5650.05542
0.5700.05493
0.5750.05446
0.5800.05399
0.5850.05353
0.5900.05307
0.5950.05263
0.6000.05219
0.6050.05176
0.6100.05133
0.6150.05091
0.6200.05050
0.6250.05010
0.6300.04970
0.6350.04931
0.6400.04893
0.6450.04855
0.6500.04817
0.6550.04781
0.6600.04744
0.6650.04709
0.6700.04674
0.6750.04639
0.6800.04605
0.6850.04571
0.6900.04538
0.6950.04505
0.7000.04473
0.7050.04441
0.7100.04410
0.7150.04379
0.7200.04349
0.7250.04319
0.7300.04289
0.7350.04260
0.7400.04231
0.7450.04203
0.7500.04175
0.7550.04147
0.7600.04120
0.7650.04093
0.7700.04067
0.7750.04040
0.7800.04014
0.7850.03989
0.7900.03964
0.7950.03939
0.8000.03914

Guided practice 2

An electron beam

About 4 min

Problem

Electrons moving at 2.0 × 10⁷ m s⁻¹ enter a uniform magnetic field of 5.0 mT at right angles to it. Convert the field to tesla, find the momentum of one electron, then find the radius of its path. The mass of an electron is 9.11 × 10⁻³¹ kg.

Find the radius

Show the steps Optional working
Unit: T
Unit: kg m s⁻¹
Unit: m

Hints

Hint 1

units

5.0 mT = 5.0 × 10⁻³ T. Use the size of the electron’s charge.

Show full solution

  1. Radius

    Working

    Answer

    B = 5.0 × 10⁻³ T and p = mv = (9.11 × 10⁻³¹)(2.0 × 10⁷) = 1.82 × 10⁻²³ kg m s⁻¹.

    r = mv/BQ = ((9.11 × 10⁻³¹)(2.0 × 10⁷))/((5.0 × 10⁻³)(1.60 × 10⁻¹⁹)) = 0.023 m

Try it yourself 3

An alpha particle

Minimal support

Problem

An alpha particle, of mass 6.68 × 10⁻²⁷ kg and charge + 3.20 × 10⁻¹⁹ C, moves to the right at 1.0 × 10⁶ m s⁻¹ into a uniform magnetic field of 0.20 T directed into the page. Find the radius of its path, and the direction in which it starts to curve.

Find the radius and direction

Unit: m
It starts to curve

Hints

Hint 1

charge

The alpha particle carries two units of charge, which is already included in 3.20 × 10⁻¹⁹ C.

Show solution step by step

  1. Radius

    Reason

    The magnetic force provides the centripetal force.

    Working

    Answer

    r = ((6.68 × 10⁻²⁷)(1.0 × 10⁶))/((0.20)(3.20 × 10⁻¹⁹)) = 0.10 m

  2. Direction

    Working

    Answer

    First finger into the page, second finger to the right: the thumb points up the page, so the path curves upwards.

Time for one revolution

One revolution is a distance 2π r at speed v, so the time it takes is

T = (2π r)/v = (2π m)/BQ

The speed cancels. A faster particle of the same kind moves on a larger circle, but takes the same time to go round. For a proton in a 0.20 T field, T = 3.3 × 10⁻⁷ s for speeds much smaller than the speed of light.

Check your understanding 2 Twice as fast

A proton moves in a circle in a uniform 0.20 T field. Its mass is 1.67 × 10⁻²⁷ kg. Compare it with a proton moving at twice the speed in the same field, with both speeds much smaller than the speed of light. Find the ratio of the second radius to the first, the ratio of their revolution times, and the time for one revolution.

Unit: s
Show answer

The radius ratio is 2. The time ratio is 1: T = 2π r/v, and both r and v double. Substituting r = mv/(BQ) gives T = 2π m/(BQ), so

T = (2π(1.67 × 10⁻²⁷))/((0.20)(1.60 × 10⁻¹⁹)) = 3.3 × 10⁻⁷ s for either proton.

Exam-style question 1

Ions with equal kinetic energy

5 marks

Examination question

Two singly charged positive ions enter the same uniform magnetic field at right angles to it, with equal kinetic energies. Ion Y has four times the mass of ion X.

(a) Show that the radius of an ion’s path is r = mv/BQ. [2 marks]

(b) Find the ratio (radius of Y's path)/(radius of X's path). [3 marks]

Write your answer before viewing the mark scheme

Show solution step by step

  1. (a) Radius

    2 marks

    Method

    Equate the magnetic force to the centripetal force.

    Reason

    The magnetic force is the only force, and it points to the centre of the circle.

    Working

    Answer

    BQv = mv²/r, so r = mv/BQ

  2. Momentum at equal kinetic energy

    2 marks

    Reason

    The speeds are not equal, so write mv in terms of the kinetic energy Eₖ.

    Working

    Answer

    Eₖ = p²/2m, so mv = p = square root of 2mEₖ and r = (square root of 2mEₖ)/BQ ∝ square root of m

  3. (b) Ratio

    1 mark

    Working

    Answer

    rY/rX = square root of 4 = 2

Common mistakes

  • Adding the magnetic force and the centripetal force as two separate forces; they are the same force.
  • Assuming r ∝ m when the particles have equal kinetic energy instead of equal speed.
  • Giving a negative charge a negative radius; the sign of the charge changes only the direction of curving.
  • Continuing the circle after the particle has left the field.
  • Using B in mT without converting to tesla.

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