Problem Analysis

The problem describes the motion of a positive point charge q in a region with simultaneous uniform electric and magnetic fields. Both fields are directed along the y-axis:

• Electric field: $\overrightarrow E = E_0\widehat j$
• Magnetic field: $\overrightarrow B = B_0\widehat j$

At time t = 0, the charge has an initial velocity $\overrightarrow v$ in the xy-plane, making an angle $\theta$ with the x-axis. The initial velocity vector is: $\overrightarrow v(0) = v \cos\theta \widehat i + v \sin\theta \widehat j$

The force acting on the charge is the Lorentz force, which is the sum of the electric force and the magnetic force: $\overrightarrow F = \overrightarrow F_e + \overrightarrow F_m = q\overrightarrow E + q(\overrightarrow v \times \overrightarrow B)$

Let's express the velocity at any time t as $\overrightarrow v(t) = v_x \widehat i + v_y \widehat j + v_z \widehat k$. Substituting the fields and velocity into the Lorentz force equation: $\overrightarrow F = q(E_0 \widehat j) + q((v_x \widehat i + v_y \widehat j + v_z \widehat k) \times (B_0 \widehat j))$ $\overrightarrow F = qE_0 \widehat j + qB_0 (v_x (\widehat i \times \widehat j) + v_y (\widehat j \times \widehat j) + v_z (\widehat k \times \widehat j))$ $\overrightarrow F = qE_0 \widehat j + qB_0 (v_x \widehat k + 0 - v_z \widehat i)$ $\overrightarrow F = -qB_0 v_z \widehat i + qE_0 \widehat j + qB_0 v_x \widehat k$

From Newton's second law, $\overrightarrow F = m\overrightarrow a$, the components of acceleration are:

• $a_x = \frac{dv_x}{dt} = -\frac{qB_0}{m} v_z$
• $a_y = \frac{dv_y}{dt} = \frac{qE_0}{m}$
• $a_z = \frac{dv_z}{dt} = \frac{qB_0}{m} v_x$

Motion Decomposition

We can analyze the motion by decomposing it into two parts: motion parallel to the fields (along the y-axis) and motion perpendicular to the fields (in the xz-plane).

1. Motion along the y-axis: The acceleration $a_y = \frac{qE_0}{m}$ is constant. This is a uniformly accelerated linear motion. The velocity along the y-axis at time t is: $v_y(t) = v_y(0) + a_y t = v \sin\theta + \frac{qE_0}{m} t$ The magnetic force has no component along the y-axis, so it does not affect this part of the motion.

2. Motion in the xz-plane: The equations for the perpendicular components are: $\frac{dv_x}{dt} = -\omega v_z \quad \text{and} \quad \frac{dv_z}{dt} = \omega v_x$ where $\omega = \frac{qB_0}{m}$ is the cyclotron frequency. These equations describe uniform circular motion in the xz-plane. The velocity components are: $v_x(t) = v_\perp \cos(\omega t + \phi) \quad \text{and} \quad v_z(t) = v_\perp \sin(\omega t + \phi)$ The initial conditions at t=0 are $v_x(0) = v \cos\theta$ and $v_z(0) = 0$. This implies that the speed in the xz-plane is $v_\perp = v\cos\theta$ and the phase angle is $\phi=0$ (or $-\pi/2$ depending on the exact form, but the motion is circular). The radius of this circular path is $R = \frac{v_\perp}{\omega} = \frac{mv\cos\theta}{qB_0}$.

Overall Motion

The overall motion is a superposition of the two: uniform circular motion in the xz-plane and uniformly accelerated motion along the y-axis. This combined motion results in a helical path whose axis is the y-axis.

The pitch of the helix is the distance traveled along the y-axis during one period of the circular motion, $T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB_0}$. Since the velocity $v_y(t)$ is not constant, the pitch is not constant. The pitch for a cycle starting at time t is: $P(t) = \int_t^{t+T} v_y(\tau) d\tau = \int_t^{t+T} (v \sin\theta + a_y \tau) d\tau$ $P(t) = [v \sin\theta \cdot \tau + \frac{1}{2}a_y \tau^2]_t^{t+T} = (v \sin\theta)T + \frac{1}{2}a_y((t+T)^2 - t^2)$ $P(t) = (v \sin\theta)T + \frac{1}{2}a_y(2tT + T^2) = (v \sin\theta)T + a_y t T + \frac{1}{2}a_y T^2$ Since $a_y = \frac{qE_0}{m} > 0$, the pitch $P(t)$ increases linearly with time t.

Evaluating the Options

A: If $\theta = 0^\circ$, the charge moves in a circular path in the xy-plane. If $\theta = 0^\circ$, $\overrightarrow v(0) = v \widehat i$. There is a velocity component perpendicular to $\overrightarrow B$, so it will move in a circle in the xz-plane. Also, there is an electric force $qE_0 \widehat j$ causing acceleration along the y-axis. The charge will not stay in the xy-plane. Its path is a helix with increasing pitch. So, A is incorrect.

B: If $\theta = 0^\circ$, the charge undergoes helical motion with constant pitch along the y-axis. As derived above, the motion is helical. However, because of the constant acceleration $a_y$ along the y-axis, the pitch increases with time. So, B is incorrect.

C: If $\theta = 10^\circ$, the charge undergoes helical motion with its pitch increasing with time, along the y-axis. If $\theta = 10^\circ$, the initial velocity has both x and y components. The x-component ($v\cos 10^\circ$) causes circular motion in the xz-plane. The y-component ($v\sin 10^\circ$) is the initial velocity for the accelerated motion along the y-axis. As shown in the general analysis, the combination is a helical path with a pitch that increases with time. So, C is correct.

D: If $\theta = 90^\circ$, the charge undergoes linear but accelerated motion along the y-axis. If $\theta = 90^\circ$, the initial velocity is $\overrightarrow v(0) = v \widehat j$. This velocity is parallel to the magnetic field $\overrightarrow B$. Therefore, the magnetic force is zero: $\overrightarrow F_m = q(\overrightarrow v \times \overrightarrow B) = q((v\widehat j) \times (B_0\widehat j)) = 0$ The only force acting on the charge is the electric force: $\overrightarrow F = q\overrightarrow E = qE_0 \widehat j$ This force causes a constant acceleration $\overrightarrow a = \frac{qE_0}{m} \widehat j$. Since the initial velocity and the acceleration are both along the y-axis, the charge will continue to move in a straight line along the y-axis with constant acceleration. This is linear accelerated motion. So, D is correct.