(i) A particle of rest mass \(m\) and charge \(q\) is injected at velocity \(\mathbf{u}\) into a constant pure magnetic field \(\mathbf{b}\) at right angles to the field lines. Use the Lorentz force law \((38.16)\) to establish that the particle will trace out a circle of radius \(m u \gamma(u) / q b\) with period \(2 \pi m \gamma(u) / q b\). [It was the y-factor in the period that necessitated the development of synchrotrons from cyclotrons, at whose energies the \(\gamma\) was still negligible.] (ii) If the particle is injected into the field with the same velocity but at an angle \(\theta \neq \pi / 2\) to the field lines, prove that the path is a helix, of smaller radius, but that the period for one complete cycle is the same as before.

1. Apply the Lorentz force law The Lorentz force acting on a charged particle moving in a magnetic field is given by \(\mathbf{F} = q(\mathbf{u} \times \mathbf{b})\). 2. Recognize the magnetic force is centripetal Since the particle's initial velocity is perpendicular to the magnetic field, the Lorentz force will always be perpendicular to the velocity. This means the force acts as a centripetal force, causing the particle to move in a circle. 3. Calculate the radius of the circle The centripetal force is given by \(\mathbf{F} = m\mathbf{a} = \frac{m\mathbf{u}^2}{r}\). Equate this with the Lorentz force: \(\frac{m\mathbf{u}^2}{r} = q(\mathbf{u} \times \mathbf{b})\). Use the fact that \(\mathbf{u} \times \mathbf{b} = ub\sin{\theta}\), and in this case, \(\theta = \pi/2\), so \(\sin{\theta} = 1\). Thus, we can write \(\frac{mu^2}{r} = qu- b\). Solving for the radius r, we get \(r = \frac{mu\gamma(u)}{qb}\). 4. Calculate the period of the motion To find the period, we can use the formula \(T = \frac{Distance}{Speed}\). Here, the distance is the circumference of the circle, \(2\pi r\), and the speed is constant and equal to u. Thus, \(T = \frac{2\pi r}{u}\). Substituting the expression for r, we get \(T = \frac{2\pi m\gamma(u)}{qb}\).

1. Decompose the initial velocity into parallel and perpendicular components We can decompose the initial velocity into components parallel (\(\mathbf{u}_{\parallel}\)) and perpendicular (\(\mathbf{u}_{\perp}\)) to the magnetic field. We have \(\mathbf{u}_{\parallel} = u\cos{\theta}\) and \(\mathbf{u}_{\perp} = u\sin{\theta}\). 2. Analyze the perpendicular motion The motion in the direction perpendicular to the magnetic field will be similar to the case in part (i). The magnetic force will be centripetal, and the radius of the helix will be given by \(r = \frac{mu_{\perp}\gamma(u)}{qb} = \frac{mu\sin{\theta} \gamma(u)}{qb}\). 3. Analyze the parallel motion The magnetic force acting in the direction parallel to the magnetic field is zero since the force is always perpendicular to the velocity. Therefore, the velocity in this direction will remain constant, and the helix will move with constant speed along the magnetic field lines. 4. Calculate the period of the helical motion Since there is no magnetic force acting in the parallel direction, the time it takes for one complete cycle of the helix will be the same as the time it takes for the perpendicular motion to complete one cycle. Thus, the period of the helical motion is the same as in part (i) and is given by \(T = \frac{2\pi m\gamma(u)}{qb}\). From this solution, we have shown that the particle traces out a circle with radius \(r = \frac{mu\gamma(u)}{qb}\) and period \(T = \frac{2\pi m\gamma(u)}{qb}\) when moving perpendicular to the magnetic field lines. When it starts with an angle \(\theta \neq \pi / 2\), it traces a helix with a smaller radius but with the same period.