Coaching Centre
Charged particles in magnetic fields
The magnetic force on a moving charge, F = qvB sin θ, the right-hand palm rule, why a charge in a uniform magnetic field moves in a circle of radius r = mv/qB, helical paths, velocity selectors and mass spectrometers, and how magnetic and electric fields compare.
On a clear winter night in southern Tasmania the sky can glow green and red: the aurora australis. Charged particles streaming from the Sun are caught by Earth's magnetic field, spiral along the field lines towards the poles and crash into the upper atmosphere. The same force bends the electron beams in the Australian Synchrotron round a 216 m ring.
The force on a moving charge
A magnetic field has no effect at all on a charge that is sitting still. Set the charge moving, and it feels a force that depends on how fast it is moving and on the angle between its velocity and the field:
$$F = qv_{\perp}B = qvB\sin\theta$$
Here θ is the angle between the velocity v and the field B, and $v\sin\theta$ is the component of the velocity perpendicular to the field. Only that component counts. A charge moving parallel to the field lines feels no force at all; one moving at right angles feels the maximum force, qvB. The unit of B is the tesla (T): a field of 1 T exerts 1 N on a charge of 1 C moving at 1 m s−1 at right angles to it.
Which way? The right-hand palm rule
Hold your right hand flat. Point your fingers in the direction of the field (north to south), your thumb in the direction a positive charge is moving, and your palm faces the direction of the force. For a negative charge such as an electron, work out the force for a positive charge and reverse it (or point your thumb opposite to the electron's velocity). The force is always at right angles to both v and B.
Diagrams often show fields coming straight out of or going into the page. A dot (·) is the tip of an arrow coming towards you, a field out of the page; a cross (×) is the tail feathers of an arrow going away, a field into the page.
Circular motion
Because the magnetic force is always perpendicular to the velocity, it can change the particle's direction but never its speed. It does no work, so the kinetic energy stays constant. A force of constant size that is always at right angles to the velocity is exactly what produces uniform circular motion. The magnetic force supplies the centripetal force:
$$qvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB}$$
Faster or heavier particles move in bigger circles; more charge or a stronger field makes the circle tighter. The time for one revolution is
$$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$$
which does not depend on the speed at all: a faster particle travels a proportionally bigger circle in the same time. That surprising result is what makes a cyclotron work.
A proton moving at 3.0 × 106 m s−1 enters a uniform 0.25 T magnetic field at right angles. Find the force on it, the radius of its path and its period.
$F = qvB = (1.602 \times 10^{-19})(3.0 \times 10^{6})(0.25) = 1.2 \times 10^{-13}$ N.
$r = \dfrac{mv}{qB} = \dfrac{(1.673 \times 10^{-27})(3.0 \times 10^{6})}{(1.602 \times 10^{-19})(0.25)} = 0.13$ m.
$T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.673 \times 10^{-27})}{(1.602 \times 10^{-19})(0.25)} = 2.6 \times 10^{-7}$ s.
An electron at the same speed would move in a circle $m_p/m_e \approx 1840$ times smaller, radius about 0.068 mm, and curve the opposite way.
Entering at an angle: helices
If the velocity is at an angle θ to the field, split it into components. The perpendicular component, $v\sin\theta$, produces circular motion; the parallel component, $v\cos\theta$, is unaffected. Together they make a helix, a spiral wound around the field lines. This is how charged particles from the solar wind spiral along Earth's field towards the polar regions to produce aurorae.
- Proton, 2.0 × 106 m s−1, 100 mT. Check the radius (20.9 cm) and period with r = mv/qB and T = 2πm/qB.
- Double the speed. What happens to the radius, and to the period? Explain why the period doesn't change.
- Switch to a deuteron (a proton plus a neutron) and raise B to 200 mT. Then try an alpha particle at 200 mT. Why do they land almost in the same place? (Hint: compare q/m.)
- Choose the electron, then flip the field out of the page. Which way does it curve now, and why?
- Velocity selector: set E = 200 kV m−1 and B = 100 mT. Predict the speed that passes straight through, then check it. What happens to faster protons?
Velocity selectors and mass spectrometers
A velocity selector uses crossed electric and magnetic fields. The electric force qE is the same for every particle, while the magnetic force qvB depends on speed. For one particular speed the two balance, and those particles go straight through a slit:
$$qE = qvB \quad\Rightarrow\quad v = \frac{E}{B}$$
Notice that this works for any charge and any mass. A mass spectrometer sends ions of a known speed from a velocity selector into a second magnetic field. Each ion follows a semicircle of radius $r = mv/qB$, so ions of different mass (for the same charge) land at different distances, $2r$, from the entrance slit. Chemists use this to identify compounds (Chemistry Module 8), and geologists use it to date rocks by measuring isotope ratios.
Singly charged carbon-12 and carbon-14 ions leave a velocity selector with E = 4.0 × 104 V m−1 and B = 0.20 T, then enter a deflection field of 0.50 T. How far apart do they land? (1 u = 1.661 × 10−27 kg)
$v = \dfrac{E}{B} = \dfrac{4.0 \times 10^{4}}{0.20} = 2.0 \times 10^{5}$ m s−1 for both.
$r_{12} = \dfrac{(12 \times 1.661 \times 10^{-27})(2.0 \times 10^{5})}{(1.602 \times 10^{-19})(0.50)} = 0.0498$ m; $r_{14} = \dfrac{14}{12} r_{12} = 0.0581$ m.
They land at 2r from the slit, so the gap is $2(0.0581 - 0.0498) = 0.017$ m, about 1.7 cm.
Comparing electric and magnetic fields
| Uniform electric field | Uniform magnetic field | |
|---|---|---|
| Force on a stationary charge | Yes, F = qE | None |
| Direction of force | Along the field (+) or against it (−) | Perpendicular to both v and B |
| Depends on speed? | No | Yes, F = qvB sin θ |
| Work done on the charge | Yes, qV: speed changes | None: speed stays constant |
| Path when fired perpendicular to the field | Parabola | Circle (helix if at an angle) |
Magnetic circular motion is one more example of the uniform circular motion you studied in Module 5. In each case some real force provides the centripetal force: friction for a car on a roundabout, tension for a ball on a string, gravity for a satellite, and qvB for a charged particle. The equations $F = mv^2/r$ and $T = 2\pi r/v$ apply to all of them.
Test yourself
An electron moves parallel to the field lines of a uniform magnetic field. The magnetic force on it is
θ = 0°, so sin θ = 0.
A proton moves to the right across the page through a magnetic field directed into the page. The force on it is
Fingers into the page, thumb to the right: the palm faces up the page.
The speed of a charged particle moving in a circle in a uniform magnetic field is doubled. Its period
T = 2πm/qB has no v in it.
Which statement about a charged particle moving in a uniform magnetic field is correct?
The force is perpendicular to the velocity, so it changes direction (velocity) but not speed.
In a velocity selector E = 3.0 × 105 V m−1 and B = 0.15 T. Particles pass undeflected if their speed is
v = E/B = 3.0 × 105 ÷ 0.15 = 2.0 × 106 m s−1, for any charge.
Label the mass spectrometer.
Which field does this describe?
The magnetic force on a moving charge is always to its velocity, so it changes the direction but not the . In a uniform field the particle moves in a , and if it enters at an angle it follows a . Doubling the field strength the radius. The unit of magnetic field strength is the .
Compare the motion of an electron fired at right angles into a uniform electric field with the motion of an electron fired at right angles into a uniform magnetic field.
In both fields the electron experiences a force proportional to its charge, and the force deflects it from a straight line. The differences come from the direction of the force. In a uniform electric field the force F = qE has the same size and direction everywhere (towards the positive plate) and does not depend on the electron's velocity. The electron therefore has a constant acceleration at right angles to its initial velocity, and it follows a parabola, like a projectile. The field does work on it (W = qV), so its speed increases. In a uniform magnetic field the force F = qvB is always perpendicular to the electron's velocity, so its direction changes as the electron turns. It acts as a centripetal force, producing uniform circular motion of radius r = mv/qB. Because the force is always perpendicular to the motion it does no work, so the electron's speed and kinetic energy stay constant.
An electron travelling at 1.5 × 107 m s−1 enters a uniform magnetic field of 2.0 mT at right angles. Calculate the radius of its path and explain why the radius would double if the speed doubled.
The magnetic force provides the centripetal force: qvB = mv²/r, so r = mv/qB = (9.109 × 10−31 × 1.5 × 107) ÷ (1.602 × 10−19 × 2.0 × 10−3) = 4.3 × 10−2 m. If the speed doubles, the magnetic force qvB doubles, but the centripetal force needed for the same radius, mv²/r, would quadruple. The force can only supply this if r doubles, which is what r = mv/qB shows: r is directly proportional to v.