D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsMagnetic Force on Moving Charges and Circular Motion
D2.4 D2.5 D2.1About 75 minutes
Learning goals- I can use F = qvB sin θ and a hand rule to find the magnetic force on a moving charge.
- I can explain why a magnetic force changes a charge's direction but not its speed.
- I can derive and use r = mv/(qB) and T = 2πm/(qB) for charges moving in circles in a uniform magnetic field.
- I can explain velocity selectors, particle tracks in cloud and bubble chambers, and auroras.
Words to know
magnetic force on a moving charge
F = qvB sin θ, where θ is the angle between the velocity and the field. The force is perpendicular to both.
cyclotron period
The time for one circle of a charged particle in a uniform magnetic field, T = 2πm/(qB). It does not depend on the particle's speed.
velocity selector
A region of crossed electric and magnetic fields that lets through, undeflected, only particles with speed v = E/B.
charge-to-mass ratio (q/m)
A particle's charge divided by its mass. Thomson measured q/m for the electron in 1897 using electric and magnetic deflection.
cloud chamber / bubble chamber
Detectors in which charged particles leave visible trails of droplets or bubbles. In a magnetic field, the curvature of a trail shows the particle's charge sign and momentum.
The northern lights
On a clear winter night in northern Ontario, the Northwest Territories or Nunavut, the sky can ripple with green and purple curtains: the aurora borealis. Charged particles streaming from the Sun are steered by Earth's magnetic field, spiral along its field lines toward the polar regions and crash into the upper atmosphere, making the gases glow.
That steering is the magnetic force on a moving charge. It is behind particle accelerators, mass spectrometers and the detectors that discovered the positron. In this lesson you will calculate it and see why it makes charges move in circles.
From wires to single charges
A current is moving charge. In the last lesson, a length L of wire carrying current I felt F = ILB sin θ. If a charge q moves at speed v, it covers length L in time t = L/v, so the "current" times length is IL = (q/t)L = qv. Replace IL with qv:
F = qvB sin θ
- θ is the angle between the velocity and the field.
- No force on a charge at rest (v = 0) or moving parallel to B.
- Maximum force qvB when v is perpendicular to B.
Direction: use right-hand rule #3 with the thumb along the velocity of a positive charge, fingers along B; the palm pushes in the direction of the force. For a negative charge (an electron), use the left hand in the same way (as for electron flow in SPH3U), or use the right hand and reverse the answer.
Worked exampleExample 1: Force on a proton
A proton moves east at 2.0 × 10⁶ m/s through a 0.50 T field pointing straight down. Find the size and direction of the force.
- F = qvB sin 90° = (1.60 × 10⁻¹⁹)(2.0 × 10⁶)(0.50) = 1.6 × 10⁻¹³ N.
- Direction: right hand, thumb east (velocity), fingers down (field). The palm faces north.
1.6 × 10⁻¹³ N north. An electron moving the same way would be pushed south.
Why the path is a circle
The magnetic force is always perpendicular to the velocity. A force perpendicular to the motion does no work (W = Fd cos 90° = 0), so it cannot change the particle's speed or kinetic energy. It only changes the direction. A constant-size force always at right angles to the velocity is exactly the condition for uniform circular motion (from the unit Dynamics).
So for a charge moving perpendicular to a uniform field, the magnetic force is the centripetal force:
qvB = mv²/r → r = mv/(qB)
The period is T = 2πr/v = 2πm/(qB). Surprise: T does not depend on speed. A faster particle moves in a bigger circle and takes exactly the same time to go around. This fact is the key to the cyclotron (next lesson).
Notice that r = p/(qB), where p = mv is the momentum: the radius of the track measures momentum.
Worked exampleExample 2: Radius and period
The proton from Example 1 (2.0 × 10⁶ m/s, perpendicular to a 0.50 T field). Find the radius and period of its circular path.
- r = mv/(qB) = (1.67 × 10⁻²⁷)(2.0 × 10⁶)/((1.60 × 10⁻¹⁹)(0.50)) = 3.34 × 10⁻²¹/8.0 × 10⁻²⁰ = 0.042 m.
- T = 2πm/(qB) = 2π(1.67 × 10⁻²⁷)/(8.0 × 10⁻²⁰) = 1.3 × 10⁻⁷ s.
r ≈ 4.2 cm, T ≈ 1.3 × 10⁻⁷ s (about 7.6 million circles per second).
Worked exampleExample 3: An electron beam bent into a circle
Electrons are accelerated from rest through 200 V and enter a uniform 1.0 × 10⁻³ T field at right angles. Find the radius of their path.
- Speed from energy: v = √(2eΔV/m) = √(2(1.60 × 10⁻¹⁹)(200)/(9.11 × 10⁻³¹)) = 8.38 × 10⁶ m/s.
- r = mv/(eB) = (9.11 × 10⁻³¹)(8.38 × 10⁶)/((1.60 × 10⁻¹⁹)(1.0 × 10⁻³)) = 0.048 m.
r ≈ 4.8 cm. School "e/m" apparatus does exactly this, with a glowing gas that makes the circle visible. Measuring r, V and B gives e/m = 2ΔV/(B²r²) = 1.76 × 10¹¹ C/kg.
When v is not perpendicular to B
Split the velocity into a part along B (no force, so it stays constant) and a part across B (which makes a circle). Together they make a helix, like a stretched spring. Charged particles from the Sun spiral this way along Earth's field lines. Near the poles the field lines dip down into the atmosphere, which is why auroras appear in a ring around each magnetic pole, over northern Canada in the north. Earth's field also traps particles in the Van Allen radiation belts, which satellites must be designed to tolerate.
Velocity selectors
Put an electric field and a magnetic field at right angles to each other and to a beam. For a positive particle moving right, with E pointing down and B into the page, the electric force qE pushes down and the magnetic force qvB pushes up. They cancel only when
qE = qvB → v = E/B
Particles at exactly this speed go straight through a slit; faster ones curve one way and slower ones the other. The result does not depend on charge or mass. J. J. Thomson used crossed fields like this to measure the electron's speed and then its charge-to-mass ratio, and velocity selectors are the first stage of many mass spectrometers.
Worked exampleExample 4: A velocity selector
A velocity selector has E = 3.0 × 10⁴ N/C and B = 0.020 T. Which speed passes through? Would electrons at that speed also pass straight through?
- v = E/B = 3.0 × 10⁴/0.020 = 1.5 × 10⁶ m/s.
- For an electron, both forces reverse direction, but they still balance when qE = qvB.
1.5 × 10⁶ m/s, and yes: any charged particle at that speed passes undeflected, whatever its charge or mass.
Reading particle tracks (D2.5)
In a cloud chamber, alcohol vapour is cooled until it is ready to condense; a charged particle passing through ionizes the air, and droplets form along its path. In a bubble chamber, the trail is a line of tiny bubbles in a superheated liquid. With a magnetic field across the chamber:
- the direction of bending shows the sign of the charge;
- the radius shows the momentum (r = p/qB): straighter tracks mean more momentum;
- a particle that loses energy as it goes spirals inward;
- neutral particles leave no track, but their existence can be inferred from gaps and from conservation laws.
In 1932 Carl Anderson saw a track that bent like an electron's mirror image: the positron, the first antimatter particle discovered. Cloud chambers need dry ice (−78 °C) and isopropyl alcohol, so they are built only with an adult in a ventilated space, often at science centres. The activity below uses a simulation instead.
Just for funThe positron was announced in 1932 on the strength of a famous cloud-chamber photograph of a single curved track. Carl Anderson put a lead plate in his cloud chamber so he could tell which way the particle was moving (it loses energy and curves more tightly after the plate). One clever plate, one Nobel Prize.
Hands-on activityTrack analysis without dry ice
You need: a computer with a browser and a ruler (or the measuring tool in the simulation).
- Open a free charged-particle-in-a-magnetic-field simulation (search for "charged particle magnetic field simulation"; the Open Source Physics collection and several university physics departments offer one). Set B into the screen.
- Launch a proton. Measure the radius of its circle. Double its speed; measure again. Double B; measure again. Then launch an electron and an alpha particle at the same speed.
- Record r for each case and compare with r = mv/(qB).
- Now print or sketch the track diagram in this lesson. Using only the curvature, decide which tracks are positive, which are negative and which particle is slowing down.
What to look for: doubling v doubles r; doubling B halves r. The electron curves the opposite way to the proton, in a much smaller circle (its mass is 1/1836 of the proton's). An alpha particle (twice the charge, about four times the mass) has a radius about twice the proton's at the same speed. The time for one circle stays the same when you change the speed.
Your turn
1Find the force on an electron moving at 4.0 × 10⁶ m/s perpendicular to a 0.30 T field.
F = evB = (1.60 × 10⁻¹⁹)(4.0 × 10⁶)(0.30) = 1.9 × 10⁻¹³ N.
2An electron moves at 3.0 × 10⁶ m/s at 30° to a 0.010 T field. Find the force.
F = qvB sin θ = (1.60 × 10⁻¹⁹)(3.0 × 10⁶)(0.010)(0.50) = 2.4 × 10⁻¹⁵ N.
3A positive ion moves north through a field pointing east. Which way is the force? What if it were an electron?
Right hand: thumb north, fingers east: the palm faces down. Force on the ion is down; on an electron, up.
4Find the radius of the path of an electron moving at 5.93 × 10⁶ m/s (100 eV) perpendicular to a 1.0 × 10⁻³ T field.
r = mv/(eB) = (9.11 × 10⁻³¹)(5.93 × 10⁶)/((1.60 × 10⁻¹⁹)(1.0 × 10⁻³)) = 3.4 cm.
5Explain why the magnetic force cannot change a charged particle's kinetic energy.
The force is always perpendicular to the velocity, so it does no work (W = Fd cos 90° = 0). With no work done, the kinetic energy (and speed) stays constant; only the direction changes.
6Electrons in a 5.0 × 10⁻³ T field move in a circle of radius 2.0 cm. Find their speed.
v = qBr/m = (1.60 × 10⁻¹⁹)(5.0 × 10⁻³)(0.020)/(9.11 × 10⁻³¹) = 1.8 × 10⁷ m/s.
More practice: warm-up
1Write the formula for the magnetic force on a moving charge.
2What is the magnetic force on a charge at rest?
3What shape is the path of a charge moving perpendicular to a uniform magnetic field?
4Does the period of a charge circling in a magnetic field depend on its speed?
5Which hand do you use for the force on an electron?
The left hand (or the right hand, then reverse the answer).
More practice: on your own
1A proton and an alpha particle (charge 2e, mass 6.64 × 10⁻²⁷ kg) move at 1.0 × 10⁶ m/s perpendicular to a 0.50 T field. Find each radius.
Proton: r = (1.67 × 10⁻²⁷)(1.0 × 10⁶)/((1.60 × 10⁻¹⁹)(0.50)) = 2.1 cm. Alpha: r = (6.64 × 10⁻²⁷)(1.0 × 10⁶)/((3.20 × 10⁻¹⁹)(0.50)) = 4.2 cm (about twice the proton's).
2A cosmic-ray proton moving at 1.0 × 10⁷ m/s crosses Earth's field (about 5 × 10⁻⁵ T) at right angles. Find the radius of its path.
r = mv/(qB) = (1.67 × 10⁻²⁷)(1.0 × 10⁷)/((1.60 × 10⁻¹⁹)(5 × 10⁻⁵)) = 2.1 × 10³ m (about 2 km).
3What magnetic field would bend a 5.0 MeV proton into a circle of radius 1.0 m? (Use v = √(2Ek/m).)
Ek = 8.0 × 10⁻¹³ J; v = √(2(8.0 × 10⁻¹³)/1.67 × 10⁻²⁷) = 3.10 × 10⁷ m/s. B = mv/(qr) = (1.67 × 10⁻²⁷)(3.10 × 10⁷)/((1.60 × 10⁻¹⁹)(1.0)) = 0.32 T.
4In a cloud chamber photo with B out of the page, a track curves clockwise as seen in the photo. Is the particle positive or negative? Explain with a hand rule.
Look at the top of the clockwise circle, where the particle moves right and the centre is below. For a positive charge: right hand, thumb right, fingers out of the page: the palm faces down, toward the centre, which is what circular motion needs. So the particle is positive. (A negative particle would circle counterclockwise.)
5Show that the period T = 2πm/(qB) does not depend on speed, starting from r = mv/(qB).
T = 2πr/v = 2π(mv/qB)/v = 2πm/(qB). The v cancels: a faster particle travels a proportionally larger circle.
Challenge
1In an e/m apparatus, electrons accelerated through ΔV move in a circle of radius r in a field B. Show that e/m = 2ΔV/(B²r²), and find e/m if ΔV = 200 V, B = 1.0 × 10⁻³ T and r = 4.77 cm.
From energy: v² = 2eΔV/m. From the circle: v = eBr/m, so v² = e²B²r²/m². Set equal: 2eΔV/m = e²B²r²/m², so e/m = 2ΔV/(B²r²). Numbers: 2(200)/((1.0 × 10⁻³)²(0.0477)²) = 1.76 × 10¹¹ C/kg.
2A charged particle enters a uniform magnetic field with its velocity at 60° to the field lines. Describe its path and explain which part of the velocity sets the radius.
A helix around the field lines. The component along B (v cos 60°) is unaffected and carries the particle along the field. The component across B (v sin 60°) produces circular motion with radius r = m(v sin 60°)/(qB).
3A proton circles in a 0.50 T field with r = 4.2 cm. If it passes through a thin foil that removes 75% of its kinetic energy, what is its new radius?
Ek ∝ v², so a quarter of the Ek means half the speed (√0.25 = 0.5). r ∝ v, so the new radius is 2.1 cm. (This is how Anderson's lead plate showed the positron's direction of travel.)
Quiz
1. A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is
2. The magnetic force on a moving charge changes its
3. A proton circles in a uniform field. If its speed doubles, its radius
4. In a velocity selector with E = 2.0 × 10⁴ N/C and B = 0.10 T, undeflected particles have speed
5. An electron and a proton move at the same speed perpendicular to the same magnetic field. Compared with the proton, the electron
6. Auroras appear mostly near the polar regions because
Big idea (D2.4, D2.5): a magnetic field pushes on a moving charge with F = qvB sin θ, at right angles to its motion. That makes charges move in circles (r = mv/qB) at constant speed, which is used in velocity selectors, particle detectors and accelerators.
How to help: Have your teen act out the hand rule while explaining Example 1, and explain why the speed never changes. On a clear night with an aurora forecast (Natural Resources Canada's Space Weather Canada service and other agencies publish them), go look at the northern lights together and talk about where the particles came from.
Common difficulties:
- Using the right hand for electrons without reversing the answer.
- Thinking the magnetic force speeds particles up.
- Forgetting sin θ, or using the angle with the wrong line.
- Confusing r = mv/(qB) with the gravitational orbit formula.
Safety: real cloud chambers need dry ice and alcohol; build one only with an adult, with gloves, in a ventilated room, or visit a science centre.