Motion of Charged Particle in Parallel Electric and Magnetic Fields

To analyze the motion of the positive point charge $q > 0$ under the simultaneous action of the uniform collinear electric field $\vec{E} = E_0 \hat{j}$ and magnetic field $\vec{B} = B_0 \hat{j}$, we consider the Lorentz force acting on the particle:

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

Given that at $t = 0$, the velocity vector lies in the $x-y$ plane making an angle $\theta$ with the $x$-axis, we have:

$\vec{v}(0) = v \cos\theta \, \hat{i} + v \sin\theta \, \hat{j}$

Substituting $\vec{E}$, $\vec{B}$, and $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ into the force expression:

$\vec{F} = q E_0 \hat{j} + q (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times (B_0 \hat{j})$ $\vec{F} = q E_0 \hat{j} + q B_0 (v_x \hat{k} - v_z \hat{i})$

Separating the force components along each axis:

$F_x = m \frac{dv_x}{dt} = -q B_0 v_z$ $F_y = m \frac{dv_y}{dt} = q E_0$ $F_z = m \frac{dv_z}{dt} = q B_0 v_x$

1. Motion along the $y$-axis:

The equation along the $y$-axis is uncoupled from $v_x$ and $v_z$:

$a_y = \frac{dv_y}{dt} = \frac{q E_0}{m}$

Integrating with respect to time $t$ gives:

$v_y(t) = v \sin\theta + \left(\frac{q E_0}{m}\right)t$ $y(t) = (v \sin\theta) t + \frac{1}{2} \left(\frac{q E_0}{m}\right) t^2$

Since $E_0 > 0$, the velocity $v_y(t)$ along the $y$-axis continuously increases with time $t$.

2. Motion in the $x-z$ plane:

The magnetic force acts entirely perpendicular to the $y$-axis, causing circular motion in the $x-z$ plane with a constant speed $v_{\perp} = \sqrt{v_x^2 + v_z^2} = v \cos\theta$.

The radius of the circular path in the $x-z$ plane is:

$R = \frac{m v_{\perp}}{q B_0} = \frac{m v \cos\theta}{q B_0}$

The time period of one revolution in the $x-z$ plane is:

$T = \frac{2\pi m}{q B_0}$

The pitch $p$ of the helical motion is defined as the distance traveled along the $y$-axis in one full revolution $T$:

$p(t) = \int_{t}^{t+T} v_y(t') \, dt'$

Since $v_y(t)$ increases with time due to the acceleration $a_y = \frac{q E_0}{m}$, the pitch $p(t)$ increases continuously over time.

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Analysis of the Options:

• For $\theta = 0^\circ$: $v_{\perp} = v \cos 0^\circ = v \neq 0$, and $v_y(0) = 0$. However, because $E_0 \neq 0$, $v_y(t) = \frac{q E_0}{m} t$ increases for $t > 0$. The charge undergoes helical motion along the $y$-axis with continuously increasing pitch, not a simple circular path in the $x-z$ plane nor a helix with constant pitch. (Options A and B are incorrect)

• For $\theta = 10^\circ$: $v_{\perp} = v \cos 10^\circ \neq 0$, so the charge undergoes circular motion in the $x-z$ plane. Simultaneously, $v_y(t) = v \sin 10^\circ + \frac{q E_0}{m} t$ increases over time. Hence, the charge undergoes helical motion with its pitch continuously increasing along the $y$-axis. (Option C is correct)

• For $\theta = 90^\circ$: $v_{\perp} = v \cos 90^\circ = 0$. Thus, $v_x(0) = 0$ and $v_z(0) = 0$. The magnetic force is $\vec{F}_B = q(\vec{v} \times \vec{B}) = 0$. The charge experiences only the electric force along the $y$-axis, leading to linear accelerated motion along the $y$-axis. (Option D is correct)

Conclusion:

The correct options are C and D.