Along the field: nothing happens
Suppose a charge is fired exactly along a uniform field. Then \(\vec v\) is parallel to \(\vec B\), \(\sin\theta = 0\), and the force is zero. With no force, the velocity never changes, and the charge stays parallel to the field for ever. It crosses the field in a straight line at constant speed, as if the field were not there.
Across the field: a circle
Now fire the charge at right angles to a uniform field. The force \(qvB\) is perpendicular to the velocity. The last lesson showed that the speed never changes, so the size of the force \(qvB\) never changes either. A force of constant size, always at right angles to the motion, is the centripetal force of uniform circular motion. The chapter Dynamics of Circular Motion studied it. So the charge moves in a circle.
\[ qvB = \dfrac{mv^2}{r} \]
Why. the magnetic force supplies the whole centripetal force; \(\sin\theta = 1\) because \(\vec v\perp\vec B\).
\[ qB = \dfrac{mv}{r} \]
Why. divide both sides by \(v\), which is not zero.
\[ r = \dfrac{mv}{qB} \]
Read it. solve for the radius.
A heavier or faster charge has more momentum to turn, so it swings out on a wider circle. A larger charge or a stronger field gives a larger force, so the circle is tighter. The sense of rotation follows from the right-hand rule. In a field into the page, a positive charge goes round anticlockwise, and an electron clockwise.
The period hides a surprise
One turn covers \(2\pi r\) at speed \(v\), so the period is \(T = 2\pi r/v\). Put \(r = mv/qB\) into it. The speed cancels, and \(T = 2\pi m/qB\). A fast charge runs a large circle and a slow one a small circle, but both take the same time for one turn. The frequency \(f = qB/2\pi m\) is often called the cyclotron frequency.
A proton moves at \(10^6\,\text{m/s}\) at right angles to a field of \(0.1\,\text{T}\). Find the radius and the period of its circle.
\(r = \dfrac{mv}{qB} = \dfrac{1.67\times10^{-27}\times10^6}{1.6\times10^{-19}\times0.1} = 0.104\,\text{m}\), about \(10\,\text{cm}\).
\(T = \dfrac{2\pi m}{qB} = \dfrac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-20}} = 6.56\times10^{-7}\,\text{s}\).
The proton makes about \(1.5\) million turns every second, and at twice the speed it would still make the same number.
Values used: \(e = 1.6\times10^{-19}\,\text{C}\), \(m_p = 1.67\times10^{-27}\,\text{kg}\), and for the alpha particle below \(m = 4m_p\), \(q = 2e\).
Try the speeds \(0.4\), \(0.8\) and \(1.2\times10^6\,\text{m/s}\) for the proton. The radius is \(4.2\), \(8.4\) and \(12.5\,\text{cm}\), in proportion to the speed. Yet the proton spends \(3.28\times10^{-7}\,\text{s}\) in the field every time, half a period. The alpha particle has four times the mass and twice the charge, so its \(m/q\) is twice the proton's. Its circle is twice as wide and its period twice as long.
An electron circles in a field of \(10^{-3}\,\text{T}\). Find its frequency of revolution, in units of \(10^7\,\text{Hz}\).
The speed of a charge circling in a uniform field is doubled. What happens to the radius and to the period?
Radius from momentum or energy
The top of \(mv/qB\) is the momentum \(p\), so \(r = p/qB\). Kinetic energy is \(K = p^2/2m\), so \(p = \sqrt{2mK}\) and \(r = \sqrt{2mK}/qB\). For a charge accelerated from rest through a potential difference \(V\), \(K = qV\), and then \(r = \sqrt{2mV/q}/B\). The chapter Electric Potential, Energy and the Dipole gives the energy \(qV\).
These forms settle comparisons quickly. A proton has mass \(m\) and charge \(e\); an alpha particle has \(4m\) and \(2e\). With equal kinetic energies, \(\sqrt{2(4m)K}/2e = \sqrt{2mK}/e\), so the radii are equal. With equal momenta, \(r = p/qB\) is halved by the alpha particle's double charge: \(r_p : r_\alpha = 2 : 1\).
An electron moves in a circle of radius \(2\,\text{cm}\) in a field of \(0.01\,\text{T}\). Find its momentum, its speed and its angular momentum about the centre.
Read the momentum straight from the circle: \(p = qBr\). Then \(v = p/m\) and \(L = mvr = pr\).
\(p = eBr\) and \(L = (eBr)\,r = eBr^2\).
\(p = 1.6\times10^{-19}\times0.01\times0.02 = 3.2\times10^{-23}\,\text{kg m/s}\). \(v = 3.2\times10^{-23}/9.1\times10^{-31} = 3.5\times10^7\,\text{m/s}\). \(L = 3.2\times10^{-23}\times0.02 = 6.4\times10^{-25}\,\text{kg m}^2/\text{s}\).
The electron carries \(3.2\times10^{-23}\,\text{kg m/s}\) of momentum at \(3.5\times10^7\,\text{m/s}\), with \(L = 6.4\times10^{-25}\,\text{kg m}^2/\text{s}\).
A proton and a deuteron (one proton and one neutron, charge \(e\), mass \(2m_p\)) enter the same field at the same speed. Compare the radii and the periods, deuteron to proton.
A field with an edge
Real fields fill a limited region. A charge that enters at right angles to the edge starts on a circle of radius \(r\), centred on the edge line. If the region is deeper than \(r\), the charge completes a half circle and comes back out through the same edge. It leaves \(2r\) from where it entered, moving the opposite way, after half a period, \(\pi m/qB\).
If the field is only a strip of width \(d\) narrower than \(r\), the charge reaches the far edge first. It leaves turned through an angle \(\theta\) with \(\sin\theta = d/r\). The time inside is the same fraction of a period: \(t = (\theta/2\pi)\,T = \theta m/qB\).
The dashed line in Fig 2.1 joins the centre of the circle to the exit point. It makes the angle \(\theta\) with the edge line, and its horizontal reach is \(r\sin\theta = d\). That is where \(\sin\theta = d/r\) comes from. The velocity at the exit is perpendicular to this radius, so it too is turned through \(\theta\).
An electron moving at \(10^7\,\text{m/s}\) enters a wide region of field \(10^{-3}\,\text{T}\), at right angles to its straight edge. Find the radius of its path, the distance between the entry and exit points, and the time spent in the field.
\(r = \dfrac{m_ev}{eB} = \dfrac{9.1\times10^{-31}\times10^7}{1.6\times10^{-19}\times10^{-3}} = 5.7\times10^{-2}\,\text{m} = 5.7\,\text{cm}\).
The path is a half circle, so the electron leaves \(2r = 11.4\,\text{cm}\) from where it entered.
The time is half a period: \(t = \dfrac{\pi m_e}{eB} = \dfrac{\pi\times9.1\times10^{-31}}{1.6\times10^{-22}} = 1.8\times10^{-8}\,\text{s}\).
The proton of the first example (\(r = 10.4\,\text{cm}\), \(T = 6.56\times10^{-7}\,\text{s}\)) crosses a strip of field \(5.2\,\text{cm}\) wide. For how long is it in the field, in units of \(10^{-8}\,\text{s}\)?
Four particles move at the same speed at right angles to the same field. Put them in order of the radius of their circles, smallest first.
- Thinking a faster charge goes round faster, in turns per second.\(T = 2\pi m/qB\) has no \(v\). A faster charge runs a larger circle in the same time.
- Using \(\sqrt2\) for the proton and alpha particle at equal energy.\(r = \sqrt{2mK}/qB\): four times the mass under the root gives 2, and twice the charge cancels it. The radii are equal.
- Taking the time in a strip as half a period whatever its width.Only a strip wider than \(r\) gives a half circle. A narrower strip gives \(t = \theta m/qB\) with \(\sin\theta = d/r\).