A charge moving in a magnetic field
A magnetic field pushes only on a moving charge. The push (force) is always at right angles to the velocity. A sideways push can turn the particle, but it cannot make it faster or slower. So the speed stays the same and the path is a circle.
The magnetic force qvB is the force that points to the centre of the circle. Set it equal to the needed centre-seeking force mv²/r:
qvB = mv²/r, so r = mv/(qB)
Time for one full circle: T = 2πr/v = 2πm/(qB). The speed v cancels out. A faster particle makes a bigger circle, but it needs the same time to go round it.
Mass spectrometer: sorting ions by mass
A mass spectrometer measures the mass of atoms or molecules. It has three jobs.
- Make ions. The sample is given an electric charge q.
- Speed them up. An ion falls through a voltage V and gains energy qV. So ½mv² = qV and v = √(2qV/m).
- Bend them. The ion enters a magnetic field B through a slit and moves in a half-circle. It hits a plate at distance d = 2r.
Put v into r = mv/qB:
r = (1/B) √(2mV/q), so m = qB²r²/(2V)
For the same q, V and B, the radius depends on mass: r ∝ √m. A heavy ion lands farther from the slit. Two isotopes of the same element (same charge, a little different mass) land in two separate marks. The plate tells you the mass, and the size of each mark tells you how much of each isotope there was.
Cyclotron: speeding up particles in small steps
To give a particle a lot of energy with a small voltage, a cyclotron uses the same voltage again and again.
- Two hollow D-shaped metal boxes (dees) sit in a strong magnetic field B, with a small gap between them.
- An alternating voltage is applied across the gap. The inside of a dee has no electric field, so the particle only turns there.
- The particle crosses the gap, gets pushed by the electric field (energy +qV), enters a dee and makes a half-circle. When it is back at the gap, the voltage has just reversed, so it gets pushed again.
Each push makes the circle bigger, because r ∝ v. The path is a spiral that grows outwards.
Key point: the time for each half-turn is πm/(qB), the same for small and large circles. So one fixed frequency of the voltage works for the whole journey:
f = qB/(2πm) (the cyclotron frequency)
Energy and limits of a cyclotron
The particle leaves at the edge of the dee, where r = R. Then v = qBR/m, and the final kinetic energy is
KE = q²B²R²/(2m)
More energy needs a bigger B or a bigger R. It does not depend on the gap voltage; a smaller voltage only means more turns.
Limits: (1) At speeds near the speed of light the mass grows, so the timing slips out of step. (2) Electrons are too light, so they reach such speeds quickly; cyclotrons are used for protons and heavier ions. (3) A very big magnet is costly.
Try it: a practical
Try it: In the 3D, choose Circle. Predict what happens to r if you double the speed, then check. Predict what happens to the time for one turn. Then choose Spectrometer and set the heavy mass to 4. Predict how far its mark is compared to the light ion (answer: twice as far, because √4 = 2). At home: tie a ball on a string and whirl it. A longer string or a faster ball is like a bigger r. The string is the force that bends the path, just like B.
Key formulas and definitions
- r = mv / (qB)
- T = 2πm / (qB), f = qB / (2πm)
- Mass spectrometer: ½mv² = qV, r = (1/B)√(2mV/q), d = 2r
- Cyclotron: f = qB/(2πm); final KE = q²B²R²/(2m)
- Units: B in tesla (T), v in m/s, m in kg, q in coulomb (C), V in volt (V)
Worked examples
1. A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) enters a field of 0.50 T at 3.0 × 10⁶ m/s at right angles to the field. Find the radius of its path.
r = mv/(qB) = (1.67 × 10⁻²⁷ × 3.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.50) = 5.01 × 10⁻²¹ / 8.0 × 10⁻²⁰ = 0.063 m = 6.3 cm.
2. If the speed of the particle in the example above is doubled, what happens to the radius and to the time for one circle?
r ∝ v, so r doubles to 12.6 cm. T = 2πm/qB does not contain v, so the time for one circle stays the same.
3. A proton moves in a field of 1.0 T. Find the cyclotron frequency.
f = qB/(2πm) = (1.6 × 10⁻¹⁹ × 1.0)/(2π × 1.67 × 10⁻²⁷) = 1.52 × 10⁷ Hz, about 15 MHz.
4. In a mass spectrometer, singly charged neon-20 ions land 20.0 cm from the slit. Where do neon-22 ions land (same V and B)?
d ∝ √m. d₂ = 20.0 × √(22/20) = 20.0 × 1.0488 = 20.98 cm, about 21.0 cm. The two isotopes are about 1 cm apart.
5. A proton and an alpha particle (mass 4 times, charge 2 times) are accelerated through the same V and enter the same B. Compare their radii.
r = (1/B)√(2mV/q), so r ∝ √(m/q). For the alpha: √(4/2) = √2 times the proton's. rα : rp = 1.41 : 1.
6. A cyclotron has dees of radius 0.50 m and a field of 1.5 T. Find the energy of protons that leave it, in MeV.
v = qBR/m = (1.6 × 10⁻¹⁹ × 1.5 × 0.50)/(1.67 × 10⁻²⁷) = 7.2 × 10⁷ m/s. KE = ½mv² = 4.3 × 10⁻¹² J. Divide by 1.6 × 10⁻¹³ J/MeV: about 27 MeV.
Common mistakes
- Thinking the magnetic field speeds the particle up. It only turns it. The speed gains come from the electric field at the gap.
- Saying the time for a circle is longer for a faster particle. T = 2πm/qB does not depend on v.
- Forgetting that d = 2r in a mass spectrometer. The ion lands at the other end of the half-circle.
- Believing the final energy of a cyclotron depends on the gap voltage. It depends on q, B, R and m only.