Two motions at once
The last lesson fired charges along the field or straight across it. In general the velocity makes some angle \(\theta\) with \(\vec B\). Split it into two parts: \(v_\parallel = v\cos\theta\) along the field and \(v_\perp = v\sin\theta\) across it. The force \(q\,\vec v\times\vec B\) depends only on the part across the field, because the part along it gives a zero cross product.
The force is at right angles to \(\vec B\), so it never changes \(v_\parallel\). Along the field the charge simply drifts at the steady speed \(v\cos\theta\). Across the field it moves in a circle of radius \(mv\sin\theta/qB\), exactly as in the last lesson with \(v\sin\theta\) in place of \(v\). The two motions are independent, like the two motions of a projectile.
Watch the charge from a frame that moves along the field at \(v\cos\theta\). That frame moves at constant velocity, so the same laws hold in it. In it the charge has only the velocity \(v\sin\theta\) across the field, and it simply goes round a circle. Back in the ground frame, that circle is carried steadily along the field. A circle dragged along its own axis traces a helix, the shape of a spring.
Radius, period and pitch
The circling part has radius \(r = mv\sin\theta/qB\). Its period is still \(T = 2\pi m/qB\), because the period never depended on the speed. The distance the charge advances along the field during one turn is the pitch of the helix. It is the drift speed times the period.
Compare the three chips of Fig 3.1. At \(90°\) the path is a circle of radius \(0.209\,\text{m}\) with no drift. At \(60°\) the radius falls to \(0.181\,\text{m}\) and the proton advances \(0.656\,\text{m}\) per turn. At \(30°\) the radius is \(0.104\,\text{m}\) and the pitch \(1.14\,\text{m}\). In all three cases one turn takes \(6.56\times10^{-7}\,\text{s}\).
A proton moving at \(2\times10^6\,\text{m/s}\) enters a field of \(0.1\,\text{T}\) at \(60°\) to the field. Find the radius and the pitch of its helix.
Split the velocity: \(v_\perp = 2\times10^6\sin60° = 1.73\times10^6\,\text{m/s}\) and \(v_\parallel = 2\times10^6\cos60° = 10^6\,\text{m/s}\).
\(r = mv_\perp/qB\) and \(p = v_\parallel T\) with \(T = 2\pi m/qB\).
\(r = \dfrac{1.67\times10^{-27}\times1.73\times10^6}{1.6\times10^{-19}\times0.1} = 0.181\,\text{m}\). \(T = 6.56\times10^{-7}\,\text{s}\), so \(p = 10^6\times6.56\times10^{-7} = 0.656\,\text{m}\).
The proton circles on a radius of \(18\,\text{cm}\) and advances about \(66\,\text{cm}\) along the field per turn.
The same proton enters the \(0.1\,\text{T}\) field at \(30°\) to it. Find the pitch of its helix in metres.
A charge moves at an angle \(\theta\) to a uniform magnetic field. Which part of its velocity sets the radius of the helix, and which part carries it along the field?
When the pitch matches the circle
The ratio of pitch to circumference follows from the two formulas. Everything but the sine and the cosine cancels, so \(p/2\pi r = \cos\theta/\sin\theta = 1/\tan\theta\). The pitch equals the circumference only when \(\tan\theta = 1\), at \(45°\). A steeper angle gives a long, stretched helix; a flatter one gives a tightly wound coil.
\[ \dfrac{2\pi mv\cos\theta}{qB} = 2\pi\,\dfrac{mv\sin\theta}{qB} \]
Why. set the pitch equal to the circumference \(2\pi r\).
?
Your turn. cancel the common factors. Which line is it?
\[ \tan\theta = 1, \quad \theta = 45° \]
Read it. divide by \(\cos\theta\).
At what angle between the velocity and the field, in degrees, is the pitch of the helix equal to its circumference?
Add an electric field pointing along \(\vec B\). The magnetic force still acts only across the field, so the radius and the period of the circle do not change. The electric force acts along the field and accelerates the drift steadily. Each turn then takes the same time but covers more distance, so the helix stretches out more with every turn.
Tied to a field line
A charge in a field spirals round a field line and drifts along it. In a strong field the circles are small, so the charge seems tied to its line. If the field lines curve slowly, the charge follows them. This is how the Earth's magnetic field steers charged particles from the Sun.
A proton from the Sun moves at about \(400\,\text{km/s}\) across the Earth's field of about \(5\times10^{-5}\,\text{T}\) high above the atmosphere. Estimate the radius of its circle, in metres.
The Earth's field lines leave the southern hemisphere, arch out into space and come back into the northern hemisphere. Far out they run nearly parallel to the ground; near the poles they plunge steeply into it. A particle from the Sun caught on a line spirals along it towards the pole. There it strikes the thin upper air, about \(100\,\text{km}\) up, and makes the oxygen and nitrogen glow. The glow is the aurora.
Fig 3.2 explains why the aurora is a polar light. Near the equator the field lines run parallel to the ground, so a particle drifting along one never comes down. Only near the poles do the lines enter the atmosphere, and only there do the particles arrive. The field is also stronger there, and the helix tightens as it goes in.
Auroras are rare so far south, but a great solar storm can push them to lower latitudes. On the night of 10 to 11 May 2024, sky cameras at Hanle and Merak in Ladakh recorded a red auroral glow in the north. The storm was driven by several clouds of charged particles thrown out by the Sun one after another. It disturbed the Earth's field so strongly that the auroral glow spread far south of its usual polar zone.
Charged particles from the Sun reach the atmosphere mainly near the poles. Explain this using the helical motion of a charge in a magnetic field.
A charged particle with a velocity component along a field line spirals round the line and drifts along it, while the circling part keeps it close to the line. The Earth's field lines run from pole to pole, nearly parallel to the ground near the equator and plunging steeply into the ground near the poles. So particles guided along the lines can reach the upper atmosphere only at high latitudes. There they strike the air and make it glow as the aurora.
- Using \(v\sin\theta\) for the pitch.The pitch comes from the drift along the field, \(v\cos\theta\), times the period.
- Thinking the period changes with the angle.\(T = 2\pi m/qB\) contains neither \(v\) nor \(\theta\). Only the radius and the pitch change.
- Measuring \(\theta\) from the page or the ground instead of from \(\vec B\).\(\theta\) is always the angle between \(\vec v\) and \(\vec B\).