## Chapter 6

## 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** =*B***k**.

## Step-by-Step

## 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 *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, \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** = *B***k** 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)