Mutually perpendicular electric and magnetic field is generally eqreferred 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.
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
\[
\begin{aligned}\vec{F} = q(\vec{E}+\vec{v}\times \vec{B})\end{aligned}
\]
From Newton’s second law:
\[
\begin{aligned}\vec{F} = q(\vec{E}+\vec{v}\times \vec{B})
\frac{d}{dt}\vec{P} = \vec{F}\end{aligned}
\]
Combining these two.
\[
\begin{aligned}\frac{d\vec{P}}{dt} = q(\vec{E}+\vec{v}\times \vec{B})\end{aligned}
\]
Taking the component along x axis.
\[
\begin{aligned}\frac{d\vec{P}}{dt} \cdot \vec{i} &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{i} \\
\frac{d}{dt}\left(\vec{P}\cdot \vec{i}\right) &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{i} \\
\frac{d}{dt}\left(P_x\right) &= q\vec{E} \cdot \vec{i} + q( \vec{v}\times \vec{B}) \cdot \vec{i} \\
m\frac{dv_x}{dt} &= 0 + q(\vec{v}\times \vec{B}) \cdot \vec{i} \\
m\frac{dv_x}{dt} &= -q(\vec{v}\cdot (\vec{B} \times \vec{i})) \\
m\frac{dv_x}{dt} &= -q(\vec{v}\cdot B\vec{j})\end{aligned}
\]
\[
\begin{aligned}m\frac{dv_x}{dt} &= -qBv_y\end{aligned}
\]
Similarly taking the y component we get
\[
\begin{aligned}\frac{d\vec{P}}{dt} \cdot \vec{j} &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{j} \\
\frac{d}{dt}\left(\vec{P}\cdot \vec{j}\right) &= q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{j} \\
\frac{d}{dt}\left(P_y\right) &= q\vec{E} \cdot \vec{j} + q( \vec{v}\times \vec{B}) \cdot \vec{j} \\
m\frac{dv_y}{dt} &= qE + q(\vec{v}\times \vec{B}) \cdot \vec{i} \\
m\frac{dv_y}{dt} &= qE – q(\vec{v}\cdot (\vec{B} \times \vec{j})) \\
m\frac{dv_y}{dt} &= qE + q(\vec{v}\cdot B\vec{i})\end{aligned}
\]
\[
\begin{aligned}m\frac{dv_y}{dt} = qE + qBv_x \\\end{aligned}
\]
We have to solve Eq.((1)) and Eq.((2)). Differentiating Eq.((2)) with respect to \(t\) we get.
\[
\begin{aligned}m\frac{d^2v_y}{dt} &= \frac{qB}{m}\left(m\frac{dv_x}{dt}\right) \frac{d^2v_y}{dt}\\
&= -\left(\frac{qB}{m}\right)^2v_y\end{aligned}
\]
The solution of Eq.((3)) is
\[
\begin{aligned}v_y(t) = C_1 \cos\left(\frac{qB}{m}t\right) + C_2\sin\left(\frac{qB}{m}t\right)\end{aligned}
\]
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:
\[
\begin{aligned}0 = C_1 + 0; \qquad \Rightarrow C_1 = 0\end{aligned}
\]
Differentiating Eq.((4)) we get.
\[
\begin{aligned}a_y(t) = -C_1\frac{qB}{m}\sin\left(\frac{qB}{m}t\right) + C_2\frac{qB}{m}\cos\left(\frac{qB}{m}t\right)\end{aligned}
\]
\[
\begin{aligned}\frac{qE}{m} = C_2\frac{qB}{m}\end{aligned}
\]
On simplification
\[
\begin{aligned}C_2 = \frac{E}{B}\end{aligned}
\]
Thus finally:
\[
\begin{aligned}v_y(t) = \frac{E}{B}\sin\left(\frac{qB}{m}t\right)\end{aligned}
\]
From Eq.((2)) we get
\[
\begin{aligned}qBv_x &= m\frac{dv_y}{dt} – qE{} \\\
&= m\frac{E}{B}\frac{qB}{m}\cos\left(\frac{qB}{m}t\right) – qEv_x(t) \\\
v_x(t) &= \frac{E}{B}\left[\cos\left(\frac{qB}{m}t\right)-1\right]\end{aligned}
\]
Now,
\[
\begin{aligned}x(t) &= \int_0^t v_x(t)dt \\
{} &= \int_0^tqE\left[\cos\left(\frac{qB}{m}t\right)-1\right]dt \\
&= \frac{E}{B}\left[\frac{m}{qB}\sin\left(\frac{qB}{m}t\right)-t\right]_0^t \\\end{aligned}
\]
\[
\begin{aligned}x(t) = \frac{mE}{qB^2} \left[\sin\left(\frac{qB}{m}t\right)-\frac{qB}{m}t\right]\end{aligned}
\]
Also similarly,
\[
\begin{aligned}y(t) &= \int_0^t v_y(t)dt {} \\
&= \int_0^t\frac{E}{B}\sin\left(\frac{qB}{m}t\right) dt {} \\
&= -\frac{E}{B}\left[\frac{m}{qB}\cos\left(\frac{qB}{m}t\right)\right]_0^t y(t) \\
&= \frac{mE}{qB^2}\left[\cos\left(\frac{qB}{m}t\right)-1\right]\end{aligned}
\]
\[
\begin{aligned}y(t) = \frac{mE}{qB^2}\left[\cos\left(\frac{qB}{m}t\right)-1\right]\end{aligned}
\]
Let us substitute \(\frac{qB}{m}\) with \(\omega\). Thus our final ralations become
\[
\begin{aligned}x(t) &= \frac{E}{B\omega} \left[\sin\left(\omega t\right)-\omega t\right]\end{aligned}
\]
\[
\begin{aligned}y(t) &= \frac{mE}{qB^2} \left[\cos\left(\omega t\right) - 1 \right]\end{aligned}
\]
So,
Eqs.((6)) and ((7)) 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.
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