Trajectory of Charged Particle in Cross Field

Introduction

Mutually perpendicular electric and magnetic field is generally referred to as Cross field. In this document we work out the trajectory of a charged particle in a cross field. The charged particle is stationary in the beginning.

Problem

Let a charged particle with charge $q$ be located at the origin with the magnetic field $\vec{B} = B \vec{k}$ and electric field $\vec{E} = E \vec{j}$ . We are interested in working out the trajectory of the charged particle.

The total force on the charged particle is

\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})

From Newton’s second law:

\frac{d}{d t} \vec{P} = \vec{F}

\frac{d \vec{P}}{d t} = q (\vec{E} + \vec{v} \times \vec{B})

Taking the component along x axis.

\begin{aligned} \frac{d \vec{P}}{d t} \cdot \vec{i} & = q (\vec{E} + \vec{v} \times \vec{B}) \cdot \vec{i} \\ \frac{d}{d t} (\vec{P} \cdot \vec{i}) & = q (\vec{E} + \vec{v} \times \vec{B}) \cdot \vec{i} \\ \frac{d}{d t} (P_{x}) & = q \vec{E} \cdot \vec{i} + q (\vec{v} \times \vec{B}) \cdot \vec{i} \\ m \frac{d v_{x}}{d t} & = 0 + q (\vec{v} \times \vec{B}) \cdot \vec{i} \\ m \frac{d v_{x}}{d t} & = - q (\vec{v} \cdot (\vec{B} \times \vec{i})) \\ m \frac{d v_{x}}{d t} & = - q (\vec{v} \cdot B \vec{j}) \end{aligned}

\begin{aligned} (1) & m \frac{d v_{x}}{d t} & = - q B v_{y} \end{aligned}

Similarly taking the y component we get

\begin{aligned} \frac{d \vec{P}}{d t} \cdot \vec{j} & = q (\vec{E} + \vec{v} \times \vec{B}) \cdot \vec{j} \\ \frac{d}{d t} (\vec{P} \cdot \vec{j}) & = q (\vec{E} + \vec{v} \times \vec{B}) \cdot \vec{j} \\ \frac{d}{d t} (P_{y}) & = q \vec{E} \cdot \vec{j} + q (\vec{v} \times \vec{B}) \cdot \vec{j} \\ m \frac{d v_{y}}{d t} & = q E + q (\vec{v} \times \vec{B}) \cdot \vec{i} \\ m \frac{d v_{y}}{d t} & = q E - q (\vec{v} \cdot (\vec{B} \times \vec{j})) \\ m \frac{d v_{y}}{d t} & = q E + q (\vec{v} \cdot B \vec{i}) \end{aligned}

\begin{array}{r} (2) & m \frac{d v_{y}}{d t} = q E + q B v_{x} \\ (3) \end{array}

We have to solve Eq.( $1$ ) and Eq.( $2$ ). Differentiating Eq.( $2$ ) with respect to $t$ we get.

\begin{aligned} (4) & m \frac{d^{2} v_{y}}{d t} & = \frac{q B}{m} (m \frac{d v_{x}}{d t}) \frac{d^{2} v_{y}}{d t} \\ (5) \\ (6) & = - {(\frac{q B}{m})}^{2} v_{y} \end{aligned}

The solution of Eq.( $4$ ) is

\begin{array}{r} (7) & v_{y} (t) = C_{1} \cos (\frac{q B}{m} t) + C_{2} \sin (\frac{q B}{m} t) \end{array}

v_{y} (0) = 0; \frac{d v_{y}}{d t} |_{t = 0} = a_{y} (0) = q \frac{E}{m}

Applying these conditions we get:

\begin{array}{r} 0 = C_{1} + 0; \Rightarrow C_{1} = 0 \end{array}

Differentiating Eq.( $7$ ) we get.

\begin{array}{r} a_{y} (t) = - C_{1} \frac{q B}{m} \sin (\frac{q B}{m} t) + C_{2} \frac{q B}{m} \cos (\frac{q B}{m} t) \end{array}

\begin{array}{r} \frac{q E}{m} = C_{2} \frac{q B}{m} \end{array}

On simplification

\begin{array}{r} C_{2} = \frac{E}{B} \end{array}

Thus finally:

\begin{array}{r} v_{y} (t) = \frac{E}{B} \sin (\frac{q B}{m} t) \end{array}

From Eq.( $2$ ) we get

\begin{aligned} q B v_{x} & = m \frac{d v_{y}}{d t} - q E \\ = m \frac{E}{B} \frac{q B}{m} \cos (\frac{q B}{m} t) - q E v_{x} (t) \\ v_{x} (t) & = \frac{E}{B} [\cos (\frac{q B}{m} t) - 1] \end{aligned}

Now,

\begin{aligned} x (t) & = \int_{0}^{t} v_{x} (t) d t \\ = \int_{0}^{t} q E [\cos (\frac{q B}{m} t) - 1] d t \\ = \frac{E}{B} {[\frac{m}{q B} \sin (\frac{q B}{m} t) - t]}_{0}^{t} \end{aligned}

\begin{array}{r} x (t) = \frac{m E}{q B^{2}} [\sin (\frac{q B}{m} t) - \frac{q B}{m} t] \end{array}

Also similarly,

\begin{aligned} y (t) & = \int_{0}^{t} v_{y} (t) d t \\ = \int_{0}^{t} \frac{E}{B} \sin (\frac{q B}{m} t) d t \\ = - \frac{E}{B} {[\frac{m}{q B} \cos (\frac{q B}{m} t)]}_{0}^{t} y (t) \\ = \frac{m E}{q B^{2}} [\cos (\frac{q B}{m} t) - 1] \end{aligned}

\begin{array}{r} y (t) = \frac{m E}{q B^{2}} [\cos (\frac{q B}{m} t) - 1] \end{array}

Let us substitute $\frac{q B}{m}$ with $ω$ . Thus our final ralations become

\begin{aligned} (8) & x (t) & = \frac{E}{B ω} [\sin (ω t) - ω t] \end{aligned}

\begin{aligned} (9) & y (t) & = \frac{m E}{q B^{2}} [\cos (ω t) - 1] \end{aligned}

So,

Eqs.( $8$ ) and ( $9$ ) together serve as the parametric equations for the trajectory of the particle in the cross field. The cartesian plot of these equations is:

Conclusion Thus we have successfully solved the problem of finding trajectory of a charged particle in cross field. The PDF versionof this article is available.

Introduction#

Problem#

Introduction

Problem