The Workbook

\(\vec F = q\vec v \times \vec B\), magnitude \(qvB\sin\theta\), perpendicular to both \(\vec v\) and \(\vec B\). \(F = 1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 = 1.6 \times 10^{-13}\) N.

1. \(\vec v\) is to the right (\(\hat i\)) and \(\vec B\) into the page (\(-\hat k\)): \(\hat i \times (-\hat k) = \hat j\). For a positive charge the force is up the page.

2. The electron's charge is negative, so its force is opposite: down the page.

\(\vec F = q\vec v \times \vec B\) is always perpendicular to \(\vec v\), so \(\vec F \cdot \vec v = 0\) and the power delivered is zero. The force changes only the direction of motion; the speed and the kinetic energy stay constant.

(i) \(\sin\theta = 0\), so \(F = 0\): the charge moves on undisturbed. (ii) \(v = 0\), so \(F = 0\): a magnetic field does not act on a stationary charge.

\(\vec v \times \vec B = 10^{4} \times 0.1\,(3\hat i + 4\hat j) \times \hat k = 10^{3}(-3\hat j + 4\hat i)\). \(\vec F = 2 \times 10^{-6} \times 10^{3}(4\hat i - 3\hat j) = (8\hat i - 6\hat j) \times 10^{-3}\) N, magnitude \(10^{-2}\) N.

The electric force \(qE\) and the magnetic force \(qvB\) are arranged to be opposite. They cancel when \(qE = qvB\), that is \(v = \dfrac{E}{B}\), whatever the charge or mass. \(v = \dfrac{10^{5}}{0.2} = 5 \times 10^{5}\) m/s.

The force \(qvB\) is the centripetal force: \(qvB = \dfrac{mv^2}{r}\), so \(r = \dfrac{mv}{qB}\). \(T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}\). Proton: \(r = \dfrac{1.67 \times 10^{-27} \times 10^{6}}{1.6 \times 10^{-19} \times 0.1} \approx 0.1\) m; \(T = \dfrac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-20}} \approx 6.6 \times 10^{-7}\) s.

1. \(r = \dfrac{mv}{qB} = \dfrac{9.1 \times 10^{-31} \times 10^{7}}{1.6 \times 10^{-19} \times 10^{-3}} \approx 5.7 \times 10^{-2}\) m \(= 5.7\) cm.

2. The path is a semicircle, so \(d = 2r \approx 11.4\) cm.

3. Half a period: \(t = \dfrac{\pi m}{qB} = \dfrac{\pi \times 9.1 \times 10^{-31}}{1.6 \times 10^{-22}} \approx 1.8 \times 10^{-8}\) s.

\(r = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB}\). (i) Same \(K\): \(\dfrac{r_\alpha}{r_p} = \dfrac{\sqrt{4m}/2e}{\sqrt{m}/e} = 1\): equal radii. (ii) Same momentum \(p\): \(r = \dfrac{p}{qB}\), so \(\dfrac{r_p}{r_\alpha} = \dfrac{2e}{e} = 2\).

\(f = \dfrac{1}{T} = \dfrac{qB}{2\pi m} = \dfrac{1.6 \times 10^{-19} \times 10^{-3}}{2\pi \times 9.1 \times 10^{-31}} \approx 2.8 \times 10^{7}\) Hz. It does not depend on the speed: a faster electron moves on a bigger circle in the same time.

1. \(r \propto \dfrac{m}{q}\) at the same speed. The deuteron has twice the mass and the same charge: \(\dfrac{r_d}{r_p} = 2\).

2. \(T \propto \dfrac{m}{q}\) too: \(\dfrac{T_d}{T_p} = 2\).

\(p = qBr = 1.6 \times 10^{-19} \times 0.01 \times 0.02 = 3.2 \times 10^{-23}\) kg m/s. \(v = \dfrac{p}{m} = \dfrac{3.2 \times 10^{-23}}{9.1 \times 10^{-31}} \approx 3.5 \times 10^{7}\) m/s.

\(L = mvr\) and \(mv = qBr\), so \(L = qBr^2 = 1.6 \times 10^{-19} \times 0.01 \times (0.02)^2 = 6.4 \times 10^{-25}\) kg m\(^2\)/s.

\(v_\perp = 2 \times 10^{6}\sin 60^\circ = 1.73 \times 10^{6}\) m/s, \(v_\parallel = 10^{6}\) m/s. \(r = \dfrac{mv_\perp}{qB} = \dfrac{1.67 \times 10^{-27} \times 1.73 \times 10^{6}}{1.6 \times 10^{-20}} \approx 0.18\) m. \(T = 6.6 \times 10^{-7}\) s (Q7), so the pitch \(p = v_\parallel T \approx 0.66\) m.

1. \(v\sin\theta\), perpendicular to \(\vec B\), feels the force and produces the circle. \(v\cos\theta\), along \(\vec B\), feels no force and carries the particle steadily along the field.

2. \(r = \dfrac{mv\sin\theta}{qB}\) and \(p = v\cos\theta \times T = \dfrac{2\pi mv\cos\theta}{qB}\).

Pitch \(= 2\pi r\) means \(\dfrac{2\pi mv\cos\theta}{qB} = \dfrac{2\pi mv\sin\theta}{qB}\), so \(\tan\theta = 1\) and \(\theta = 45^\circ\).

A particle with a velocity component along the field spirals around a field line and drifts along it. The Earth's field lines run from pole to pole and dip into the surface near the poles. The spiralling particles are guided there and strike the upper atmosphere, giving the aurora.

1. The proton gains energy only while crossing the gap. It crosses once every half revolution, alternately in opposite directions, so the gap voltage must reverse every half period. The oscillator must therefore run at the frequency of revolution, \(\dfrac{qB}{2\pi m}\), which is the same for every radius.

2. \(f = \dfrac{1.6 \times 10^{-19} \times 1.5}{2\pi \times 1.67 \times 10^{-27}} \approx 2.3 \times 10^{7}\) Hz, about \(23\) MHz.

At the edge \(v = \dfrac{qBR}{m}\), so \(K = \dfrac{q^2B^2R^2}{2m} = \dfrac{(1.6 \times 10^{-19})^2 \times 2.25 \times 0.25}{2 \times 1.67 \times 10^{-27}} \approx 4.3 \times 10^{-12}\) J \(\approx 27\) MeV.

Electrons are light and reach speeds close to \(c\) after only a few million volts. Their mass then grows with speed, so the revolution frequency \(\dfrac{qB}{2\pi m}\) falls. The electrons arrive at the gap out of step with the fixed oscillator. They stop gaining energy.

Two gap crossings per revolution, each giving \(qV = 50\) keV: \(100\) keV per revolution. Revolutions \(= \dfrac{27 \times 10^{6}}{10^{5}} = 270\).

\(v = \dfrac{E}{B} = \dfrac{2 \times 10^{4}}{0.01} = 2 \times 10^{6}\) m/s. For a faster particle the magnetic force \(qvB\) exceeds the electric force \(qE\), so it is deflected in the direction of the magnetic force.

\(r = \dfrac{mv}{qB}\). For \(20\) u: \(\dfrac{20 \times 1.66 \times 10^{-27} \times 2 \times 10^{6}}{1.6 \times 10^{-19} \times 1} = 0.415\) m. For \(22\) u: \(0.457\) m. The two isotopes land about \(8\) cm apart after a semicircle.

\(r = \dfrac{mv}{qB}\), so \(T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}\). The speed has cancelled: a faster charge moves on a proportionally larger circle and takes the same time. Only \(\dfrac{m}{q}\) and \(B\) fix the period.