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:

$$ \frac{d}{dt}\vec{P} = \vec{F} $$

Combining these two.

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

Taking the component along x axis.

$$ \begin{align*} \frac{d\vec{P}}{dt} \cdot \vec{i} &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{i} \\\\ \frac{d}{dt}\left(\vec{P}\cdot \vec{i}\right) &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{i} \\\\ \frac{d}{dt}\left(P_x\right) &= q\vec{E} \cdot \vec{i} + q( \vec{v}\times \vec{B}) \cdot \vec{i} \\\\ m\frac{dv_x}{dt} &= 0 + q(\vec{v}\times \vec{B}) \cdot \vec{i} \\\\ m\frac{dv_x}{dt} &= -q(\vec{v}\cdot (\vec{B} \times \vec{i})) \\\\ m\frac{dv_x}{dt} &= -q(\vec{v}\cdot B\vec{j}) \end{align*} $$$$ \begin{align} \label{eq:dvx} m\frac{dv_x}{dt} &= -qBv_y \end{align} $$

Similarly taking the y component we get

$$ \begin{align*} \frac{d\vec{P}}{dt} \cdot \vec{j} &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{j} \\\\ \frac{d}{dt}\left(\vec{P}\cdot \vec{j}\right) &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{j} \\\\ \frac{d}{dt}\left(P_y\right) &= q\vec{E} \cdot \vec{j} + q( \vec{v}\times \vec{B}) \cdot \vec{j} \\\\ m\frac{dv_y}{dt} &= qE + q(\vec{v}\times \vec{B}) \cdot \vec{i} \\\\ m\frac{dv_y}{dt} &= qE – q(\vec{v}\cdot (\vec{B} \times \vec{j})) \\\\ m\frac{dv_y}{dt} &= qE + q(\vec{v}\cdot B\vec{i}) \end{align*} $$$$ \begin{align} \label{eq:dvy} m\frac{dv_y}{dt} = qE + qBv_x \\\\ \end{align} $$

We have to solve Eq.(\ref{eq:dvx}) and Eq.(\ref{eq:dvy}). Differentiating Eq.(\ref{eq:dvy}) with respect to $t$ we get.

$$ \begin{align} \label{eq:d2vy} m\frac{d^2v_y}{dt} &= \frac{qB}{m}\left(m\frac{dv_x}{dt}\right) \frac{d^2v_y}{dt}\\\\ &= -\left(\frac{qB}{m}\right)^2v_y \end{align} $$

The solution of Eq.(\ref{eq:d2vy}) is

$$ \begin{align} \label{eq:vyt} v_y(t) = C_1 \cos\left(\frac{qB}{m}t\right) + C_2\sin\left(\frac{qB}{m}t\right) \end{align} $$

Applying the conditions:

$$ v_y(0) = 0; \qquad \frac{dv_y}{dt}|_{t = 0} = a_y(0) = q\frac{E}{m} $$

Applying these conditions we get:

$$ \begin{align*} 0 = C_1 + 0; \qquad \Rightarrow C_1 = 0 \end{align*} $$

Differentiating Eq.(\ref{eq:vyt}) we get.

$$ \begin{align*} a_y(t) = -C_1\frac{qB}{m}\sin\left(\frac{qB}{m}t\right) + C_2\frac{qB}{m}\cos\left(\frac{qB}{m}t\right) \end{align*} $$$$ \begin{align*} \frac{qE}{m} = C_2\frac{qB}{m} \end{align*} $$

On simplification

$$ \begin{align*} C_2 = \frac{E}{B} \end{align*} $$

Thus finally:

$$ \begin{align*} v_y(t) = \frac{E}{B}\sin\left(\frac{qB}{m}t\right) \end{align*} $$

From Eq.(\ref{eq:dvy}) we get

$$ \begin{align*} qBv_x &= m\frac{dv_y}{dt} – qE{} \\\ &= m\frac{E}{B}\frac{qB}{m}\cos\left(\frac{qB}{m}t\right) – qEv_x(t) \\\ v_x(t) &= \frac{E}{B}\left[\cos\left(\frac{qB}{m}t\right)-1\right] \end{align*} $$

Now,

$$ \begin{align*} x(t) &= \int_0^t v_x(t)dt \\\\ {} &= \int_0^tqE\left[\cos\left(\frac{qB}{m}t\right)-1\right]dt \\\\ &= \frac{E}{B}\left[\frac{m}{qB}\sin\left(\frac{qB}{m}t\right)-t\right]_0^t \\\\ \end{align*} $$$$ \begin{align*} x(t) = \frac{mE}{qB^2} \left[\sin\left(\frac{qB}{m}t\right)-\frac{qB}{m}t\right] \end{align*} $$

Also similarly,

$$ \begin{align*} y(t) &= \int_0^t v_y(t)dt {} \\\\ &= \int_0^t\frac{E}{B}\sin\left(\frac{qB}{m}t\right) dt {} \\\\ &= -\frac{E}{B}\left[\frac{m}{qB}\cos\left(\frac{qB}{m}t\right)\right]_0^t y(t) \\\\ &= \frac{mE}{qB^2}\left[\cos\left(\frac{qB}{m}t\right)-1\right] \end{align*} $$$$ \begin{align*} y(t) = \frac{mE}{qB^2}\left[\cos\left(\frac{qB}{m}t\right)-1\right] \end{align*} $$

Let us substitute $\frac{qB}{m}$ with $\omega$. Thus our final ralations become

$$ \begin{align} \label{eq:xt} x(t) &= \frac{E}{B\omega} \left[\sin\left(\omega t\right)-\omega t\right] \end{align} $$$$ \begin{align} \label{eq:yt} y(t) &= \frac{mE}{qB^2} \left[\cos\left(\omega t\right) - 1 \right] \end{align} $$

So,

Eqs.(\ref{eq:xt}) and (\ref{eq:yt}) 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.