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Q. 6.5

Consider a relativistic charged particle of rest mass  $m_0$  moving in a constant magnetic field of strength B. (a) Prove that the charge moves in a circular helix, a curve of constant curvature, and hence  $\mathbf{F}_m$  has a constant magnitude. (b) Derive the equation of the path for a plane motion perpendicular to a constant magnetic field B =Bk.

Verified Solution

Solution of (a). To determine the trajectory of a particle of charge q moving in a magnetic field of constant strength B, we recall Newton’s law in (5.34) and consider the relation

$\mathbf{F}(P, t)=\frac{d \mathbf{p}(P, t)}{d t}=\frac{d}{d t}[m(P) \mathbf{v}(P, t)]$                              (5.34)

$\frac{d}{d t}(\mathbf{p} \cdot \mathbf{B})=\frac{d \mathbf{p}}{d t} \cdot \mathbf{B}=\mathbf{F}_m \cdot \mathbf{B}=0$,                  (6.17a)

wherein (6.16) is the total force on q. Therefore , the component of the  momentum in the direction of B is constant:

$\mathbf{F}_m=q \mathbf{v} \times \mathbf{B}$                     (6.16)

$\mathbf{p} \cdot \mathbf{B}=m \mathbf{v} \cdot \mathbf{B}=C \text {, a constant. }$                (6.17b)

Since the magnitudes of p and B are constant, (6.l7b) implies that the angle between the fixed axis of B and the tangent to the space curve along which moves is constant everywhere along the path. Consequently, as described in  Example 1.14, the path is a circular helix, a space curve of constant curvature ; therefore, $\left|\mathbf{F}_m\right|=q v_0 B \sin \langle\mathbf{v}, \mathbf{B}\rangle$   is constant. Conversely, it follows from (6.16) that if  $\mathbf{F}_m$  has a constant magnitude, $\sin \langle\mathbf{v}, \mathbf{B}\rangle$  is constant and hence the path is a circular helix .

The initial velocity $\mathbf{v}_0$  may be considered arbitrary. If the velocity is initially perpendicular to B, then, by (6.17b), $\mathbf{p} \cdot \mathbf{B}=0$  always, and the path is a circle in the plane perpendicular to B. If the initial velocity  $\mathbf{v}_0$  is parallel to B, the  constant force $\mathbf{F}_m=\mathbf{0} ;$  the motion is uniform and the path is a straight line along the axis of B. The circle and the line are degenerate kinds of helices . In summary, the trajectory of a charged particle which is given an arbitrary initial velocity in a constant magnetic field is a circular helix .

Solution of (b). The path of a charge q in a plane motion perpendicular to the constant vector B is a circle. To describe this circle, we apply Newton’s law in (6.16) to write  $d \mathbf{p} / d t=d(q \mathbf{x} \times \mathbf{B}) / d t$.  Integration yields  $\mathbf{p}-\mathbf{q x} \times \mathbf{B}=\mathbf{A}$,  a constant vector. Let B = Bk and consider a plane motion perpendicular to B, so that $\mathbf{x}=x \mathbf{i}+y \mathbf{j}$ . Then  $\mathbf{p}=\left(A_1+q B y\right) \mathbf{i}+\left(A_2-q B x\right) \mathbf{j}$,  and with $\mathbf{p} \cdot \mathbf{p}=|\mathbf{p}|^2=m^2 v_0^2$,  a constant, this yields the equation of a circular orbit of radius  $R \equiv m v_0 / q B:$

$\left(x-\frac{A_2}{q B}\right)^2 + \left(y + \frac{A_1}{q B}\right)^2=R^2 .$                 (6.l7c)

We thus find with (6.9) that a charged relativistic particle in a uniform magnetic field moves on a circular orbit with angular speed  $\omega=v_0 / R=q B / m=\left(q B / m_0\right) \sqrt{1-\beta^2}$.   This is known as the circular cyclotron frequency.

$m=\gamma m_0=\frac{m_0}{\sqrt{1 – \beta^2}} \quad \text { with } \quad \beta \equiv \frac{\dot{s}}{c} \text {. }$                      (6.9)