Consider a relativistic charged particle of rest mass mo 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 Fm has a constant magnitude. (b) Derive the equation of the path for a plane motion perpendicular

Question

Consider a relativistic charged particle of rest mass [latex]m_0[/latex] 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 [latex]\mathbf{F}_m[/latex] has a constant magnitude. (b) Derive the equation of the path for a plane motion perpendicular to a constant magnetic field B =Bk.

Accepted Answer

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

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

[latex]\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[/latex], (6.17a)

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

[latex]\mathbf{F}_m=q \mathbf{v} imes \mathbf{B}[/latex] (6.16)

[latex]\mathbf{p} \cdot \mathbf{B}=m \mathbf{v} \cdot \mathbf{B}=C ext {, a constant. }[/latex] (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 q 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, [latex]\left|\mathbf{F}_m ight|=q v_0 B \sin \langle\mathbf{v}, \mathbf{B} angle[/latex] is constant. Conversely, it follows from (6.16) that if [latex]\mathbf{F}_m[/latex] has a constant magnitude, [latex]\sin \langle\mathbf{v}, \mathbf{B} angle[/latex] is constant and hence the path is a circular helix .

The initial velocity [latex]\mathbf{v}_0[/latex] may be considered arbitrary. If the velocity is initially perpendicular to B, then, by (6.17b), [latex]\mathbf{p} \cdot \mathbf{B}=0[/latex] always, and the path is a circle in the plane perpendicular to B. If the initial velocity [latex]\mathbf{v}_0[/latex] is parallel to B, the constant force [latex]\mathbf{F}_m=\mathbf{0} ;[/latex] 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 [latex]d \mathbf{p} / d t=d(q \mathbf{x} imes \mathbf{B}) / d t[/latex]. Integration yields [latex]\mathbf{p}-\mathbf{q x} imes \mathbf{B}=\mathbf{A}[/latex], a constant vector. Let B = Bk and consider a plane motion perpendicular to B, so that [latex]\mathbf{x}=x \mathbf{i}+y \mathbf{j}[/latex] . Then [latex]\mathbf{p}=\left(A_1+q B y ight) \mathbf{i}+\left(A_2-q B x ight) \mathbf{j}[/latex], and with [latex]\mathbf{p} \cdot \mathbf{p}=|\mathbf{p}|^2=m^2 v_0^2[/latex], a constant, this yields the equation of a circular orbit of radius [latex]R \equiv m v_0 / q B:[/latex]

[latex]\left(x-\frac{A_2}{q B} ight)^2 + \left(y + \frac{A_1}{q B} ight)^2=R^2 .[/latex] (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 [latex]\omega=v_0 / R=q B / m=\left(q B / m_0 ight) \sqrt{1-\beta^2}[/latex]. This is known as the circular cyclotron frequency.

[latex]m=\gamma m_0=\frac{m_0}{\sqrt{1 - \beta^2}} \quad ext { with } \quad \beta \equiv \frac{\dot{s}}{c} ext {. }[/latex] (6.9)

Question 6.5: Consider a relativistic charged particle of rest mass mo mov...

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